Class 12: Vector or Cross Product – CBSE Board Problems

$\displaystyle \textbf{Question 1:}\ \ \text{If }\overrightarrow{a},\ \overrightarrow{b}\text{ are any two vectors, then}$
$\displaystyle \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2} = \left|\overrightarrow{a}\right|^{2}\left|\overrightarrow{b}\right|^{2} - \left(\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2} = \begin{vmatrix} \overrightarrow{a}\cdot\overrightarrow{a} & \overrightarrow{a}\cdot\overrightarrow{b} \\ \overrightarrow{b}\cdot\overrightarrow{a} & \overrightarrow{b}\cdot\overrightarrow{b} \end{vmatrix}$
$\displaystyle \text{or,}$
$\displaystyle \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2} + \left(\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2} = \left|\overrightarrow{a}\right|^{2}\left|\overrightarrow{b}\right|^{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that}$
$\displaystyle \left|\overrightarrow{a}\times\overrightarrow{b}\right| = \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right| \sin\theta$
$\displaystyle \therefore \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2} = \left( \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right| \sin\theta \right)^{2}$
$\displaystyle \Rightarrow \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2} = \left|\overrightarrow{a}\right|^{2} \left|\overrightarrow{b}\right|^{2} \sin^{2}\theta$
$\displaystyle \Rightarrow \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2} = \left|\overrightarrow{a}\right|^{2} \left|\overrightarrow{b}\right|^{2} \left(1-\cos^{2}\theta\right)$
$\displaystyle \Rightarrow \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2} = \left|\overrightarrow{a}\right|^{2} \left|\overrightarrow{b}\right|^{2} - \left|\overrightarrow{a}\right|^{2} \left|\overrightarrow{b}\right|^{2} \cos^{2}\theta$
$\displaystyle \Rightarrow \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2} = \left|\overrightarrow{a}\right|^{2} \left|\overrightarrow{b}\right|^{2} - \left( \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right| \cos\theta \right)^{2}$
$\displaystyle \Rightarrow \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2} = \left|\overrightarrow{a}\right|^{2} \left|\overrightarrow{b}\right|^{2} - \left(\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2}$
$\displaystyle \Rightarrow \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2} = \begin{vmatrix} \overrightarrow{a}\cdot\overrightarrow{a} & \overrightarrow{a}\cdot\overrightarrow{b} \\ \overrightarrow{a}\cdot\overrightarrow{b} & \overrightarrow{b}\cdot\overrightarrow{b} \end{vmatrix} = \begin{vmatrix} \overrightarrow{a}\cdot\overrightarrow{a} & \overrightarrow{a}\cdot\overrightarrow{b} \\ \overrightarrow{b}\cdot\overrightarrow{a} & \overrightarrow{b}\cdot\overrightarrow{b} \end{vmatrix}$
$\displaystyle \text{Hence,} \ \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2} = \left|\overrightarrow{a}\right|^{2} \left|\overrightarrow{b}\right|^{2} - \left(\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2}$
$\displaystyle \Rightarrow \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2} + \left(\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2} = \left|\overrightarrow{a}\right|^{2} \left|\overrightarrow{b}\right|^{2}$
$\displaystyle \text{Hence Proved.}$

$\displaystyle \textbf{Question 2: }\ \text{If }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}, \text{ find the value of }\left(\overrightarrow{r}\times\widehat{i}\right)\cdot \left(\overrightarrow{r}\times\widehat{j}\right)+xy.$
$\displaystyle \text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Proceeding as in Example 4, we obtain}$
$\displaystyle \overrightarrow{r}\times\widehat{i} =-y\widehat{k}+z\widehat{j} =0\widehat{i}+z\widehat{j}-y\widehat{k} \text{ and } \overrightarrow{r}\times\widehat{j} =x\widehat{k}-z\widehat{i} =-z\widehat{i}+0\widehat{j}+x\widehat{k}$
$\displaystyle \therefore \left(\overrightarrow{r}\times\widehat{i}\right)\cdot \left(\overrightarrow{r}\times\widehat{j}\right) =\left(0\widehat{i}+z\widehat{j}-y\widehat{k}\right)\cdot \left(-z\widehat{i}+0\widehat{j}+x\widehat{k}\right) =0\times(-z)+z\times0+(-y)\times x=-yx$
$\displaystyle \Rightarrow \left(\overrightarrow{r}\times\widehat{i}\right)\cdot \left(\overrightarrow{r}\times\widehat{j}\right)+xy =-xy+xy=0$

$\displaystyle \textbf{Question 3: } \text{Show that the area of a parallelogram having diagonals} \\ 3\widehat{i}+\widehat{j}-2\widehat{k}\text{ and }\widehat{i}-3\widehat{j}+4\widehat{k} \text{ is }5\sqrt{3}\text{ square units.} \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=3\widehat{i}+\widehat{j}-2\widehat{k} \text{ and }\overrightarrow{b}=\widehat{i}-3\widehat{j}+4\widehat{k}.\ \text{Then,}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b} =\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k}\\ 3 & 1 & -2\\ 1 & -3 & 4 \end{vmatrix} =(4-6)\widehat{i}-(12+2)\widehat{j}+(-9-1)\widehat{k} =-2\widehat{i}-14\widehat{j}-10\widehat{k}$
$\displaystyle \Rightarrow \left|\overrightarrow{a}\times\overrightarrow{b}\right| =\sqrt{(-2)^{2}+(-14)^{2}+(-10)^{2}} =\sqrt{300}$
$\displaystyle \therefore\ \text{Area of the parallelogram} =\frac{1}{2}\left|\overrightarrow{a}\times\overrightarrow{b}\right| =\frac{1}{2}\sqrt{300}=5\sqrt{3}\ \text{sq. units.}$

$\displaystyle \textbf{Question 4:} \text{Find the area of the triangle whose vertices are } A(3,-1,2),\ B(1,-1,-3) \\ \text{and }C(4,-3,1). \ \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a},\overrightarrow{b}\text{ and } \overrightarrow{c}\text{ be the position vectors of points }A,B\text{ and }C\text{ respectively. Then,}$
$\displaystyle \overrightarrow{a}=3\widehat{i}-\widehat{j}+2\widehat{k},\ \overrightarrow{b}=\widehat{i}-\widehat{j}-3\widehat{k}\text{ and } \overrightarrow{c}=4\widehat{i}-3\widehat{j}+\widehat{k}$
$\displaystyle \text{We know that:}$
$\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a} =\left(\widehat{i}-\widehat{j}-3\widehat{k}\right) -\left(3\widehat{i}-\widehat{j}+2\widehat{k}\right) =-2\widehat{i}+0\widehat{j}-5\widehat{k}$
$\displaystyle \text{and,}$
$\displaystyle \overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{a} =\left(4\widehat{i}-3\widehat{j}+\widehat{k}\right) -\left(3\widehat{i}-\widehat{j}+2\widehat{k}\right) =\widehat{i}-2\widehat{j}-\widehat{k}$
$\displaystyle \therefore \overrightarrow{AB}\times\overrightarrow{AC} =\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k}\\ -2 & 0 & -5\\ 1 & -2 & -1 \end{vmatrix} =(0-10)\widehat{i}-(2+5)\widehat{j}+(4-0)\widehat{k} =-10\widehat{i}-7\widehat{j}+4\widehat{k}$
$\displaystyle \Rightarrow \left|\overrightarrow{AB}\times\overrightarrow{AC}\right| =\sqrt{(-10)^{2}+(-7)^{2}+4^{2}} =\sqrt{165}$
$\displaystyle \text{Hence, area of }\triangle ABC =\frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right| =\frac{1}{2}\sqrt{165}.$

$\displaystyle \textbf{Question 5:}\ \ \text{If }\overrightarrow{a}\times\overrightarrow{b} =\overrightarrow{c}\times\overrightarrow{d}\text{ and } \overrightarrow{a}\times\overrightarrow{c} =\overrightarrow{b}\times\overrightarrow{d},\text{ show that } \overrightarrow{a}-\overrightarrow{d}\text{ is parallel to }\overrightarrow{b}-\overrightarrow{c}, \text{where }\overrightarrow{a}\neq\overrightarrow{d} \text{ and }\overrightarrow{b}\neq\overrightarrow{c}.\ \text{[CBSE 2001, 2009, 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Recall that two non-zero vectors are parallel iff their} \text{ cross-product is zero vector.}$
$\displaystyle \text{Therefore, to prove that }\overrightarrow{a}-\overrightarrow{d} \text{ is parallel to }\overrightarrow{b}-\overrightarrow{c},\text{ it is sufficient to show that}$
$\displaystyle \left(\overrightarrow{a}-\overrightarrow{d}\right)\times \left(\overrightarrow{b}-\overrightarrow{c}\right)=\overrightarrow{0}.$
$\displaystyle \text{Now,}\ \left(\overrightarrow{a}-\overrightarrow{d}\right)\times \left(\overrightarrow{b}-\overrightarrow{c}\right) =\overrightarrow{a}\times\left(\overrightarrow{b}-\overrightarrow{c}\right) -\overrightarrow{d}\times\left(\overrightarrow{b}-\overrightarrow{c}\right)$
$\displaystyle =\overrightarrow{a}\times\overrightarrow{b} -\overrightarrow{a}\times\overrightarrow{c} -\overrightarrow{d}\times\overrightarrow{b} +\overrightarrow{d}\times\overrightarrow{c}\ \ \ \ \ [\text{Using distributive law}]$
$\displaystyle =\overrightarrow{c}\times\overrightarrow{d} -\overrightarrow{b}\times\overrightarrow{d} -\overrightarrow{d}\times\overrightarrow{b} +\overrightarrow{d}\times\overrightarrow{c}\ \ \ \ \ [\because\ \overrightarrow{a}\times\overrightarrow{b} =\overrightarrow{c}\times\overrightarrow{d},\ \overrightarrow{a}\times\overrightarrow{c} =\overrightarrow{b}\times\overrightarrow{d}]$
$\displaystyle =\overrightarrow{c}\times\overrightarrow{d} -\overrightarrow{b}\times\overrightarrow{d} +\overrightarrow{b}\times\overrightarrow{d} -\overrightarrow{c}\times\overrightarrow{d}\ \ \ \ \ [\because\ -\left(\overrightarrow{d}\times\overrightarrow{b}\right) =\overrightarrow{b}\times\overrightarrow{d}\ \&\ \overrightarrow{d}\times\overrightarrow{c} =-\left(\overrightarrow{c}\times\overrightarrow{d}\right)]$
$\displaystyle =\overrightarrow{0}$
$\displaystyle \text{Hence, }\left(\overrightarrow{a}-\overrightarrow{d}\right) \text{ is parallel to }\left(\overrightarrow{b}-\overrightarrow{c}\right)$

$\displaystyle \textbf{Question 6:}\ \ \text{If }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \text{ are three vectors such that }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} =\overrightarrow{0}, \\ \text{then prove that} \overrightarrow{a}\times\overrightarrow{b} =\overrightarrow{b}\times\overrightarrow{c} =\overrightarrow{c}\times\overrightarrow{a}\ \ \ \ \ [\text{CBSE 2001, 2004}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}\ \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} =\overrightarrow{0}$
$\displaystyle \Rightarrow\ \overrightarrow{a}\times\left(\overrightarrow{a} +\overrightarrow{b}+\overrightarrow{c}\right) =\overrightarrow{a}\times\overrightarrow{0}\ \ \ \ \ [\text{Taking cross-product on left with }\overrightarrow{a}]$
$\displaystyle \Rightarrow\ \overrightarrow{a}\times\overrightarrow{a} +\overrightarrow{a}\times\overrightarrow{b} +\overrightarrow{a}\times\overrightarrow{c} =\overrightarrow{0}\ \ \ \ \ [\text{Using distributive law}]$
$\displaystyle \Rightarrow\ \overrightarrow{a}\times\overrightarrow{b} -\overrightarrow{c}\times\overrightarrow{a} =\overrightarrow{0}\ \ \ \ \ [\because\ \overrightarrow{a}\times\overrightarrow{a}=\overrightarrow{0} \text{ and }\overrightarrow{a}\times\overrightarrow{c}=-\overrightarrow{c}\times\overrightarrow{a}]$
$\displaystyle \Rightarrow\ \overrightarrow{a}\times\overrightarrow{b} =\overrightarrow{c}\times\overrightarrow{a}\ \ \ \ \ ...(i)$
$\displaystyle \text{Again,}\ \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} =\overrightarrow{0}$
$\displaystyle \Rightarrow\ \overrightarrow{b}\times\left(\overrightarrow{a} +\overrightarrow{b}+\overrightarrow{c}\right) =\overrightarrow{b}\times\overrightarrow{0}\ \ \ \ \ [\text{Taking cross-product on left with }\overrightarrow{b}]$
$\displaystyle \Rightarrow\ \overrightarrow{b}\times\overrightarrow{a} +\overrightarrow{b}\times\overrightarrow{b} +\overrightarrow{b}\times\overrightarrow{c} =\overrightarrow{0}$
$\displaystyle \Rightarrow\ -\overrightarrow{a}\times\overrightarrow{b} +\overrightarrow{0} +\overrightarrow{b}\times\overrightarrow{c} =\overrightarrow{0}\ \ \ \ \ [\because\ \overrightarrow{b}\times\overrightarrow{b}=\overrightarrow{0} \text{ and }\overrightarrow{b}\times\overrightarrow{a}=-\overrightarrow{a}\times\overrightarrow{b}]$
$\displaystyle \Rightarrow\ \overrightarrow{a}\times\overrightarrow{b} =\overrightarrow{b}\times\overrightarrow{c}\ \ \ \ \ ...(ii)$
$\displaystyle \text{From (i) and (ii), we obtain}\ \overrightarrow{a}\times\overrightarrow{b} =\overrightarrow{b}\times\overrightarrow{c} =\overrightarrow{c}\times\overrightarrow{a}.$

$\displaystyle \textbf{Question 7:}\ \ \text{Prove that the normal to the plane containing three points} \text{ whose position} \\ \text{vectors are }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \text{ lies in the direction }\overrightarrow{b}\times\overrightarrow{c} +\overrightarrow{c}\times\overrightarrow{a} +\overrightarrow{a}\times\overrightarrow{b}.\ \ \ \ [\text{CBSE 2001C}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A,B\text{ and }C\text{ be the points having position vectors } \overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ respectively. Then,}$
$\displaystyle \overrightarrow{AB}\times\overrightarrow{AC}\text{ is a vector normal to the plane} \text{ containing the points }A,B\text{ and }C.$
$\displaystyle \text{Now,}\ \overrightarrow{AB}\times\overrightarrow{AC} =\left(\overrightarrow{b}-\overrightarrow{a}\right)\times \left(\overrightarrow{c}-\overrightarrow{a}\right)$
$\displaystyle \Rightarrow\ \overrightarrow{AB}\times\overrightarrow{AC} =\overrightarrow{b}\times\left(\overrightarrow{c}-\overrightarrow{a}\right) -\overrightarrow{a}\times\left(\overrightarrow{c}-\overrightarrow{a}\right)\ \ \ \ \ [\text{By distributivity}]$
$\displaystyle \Rightarrow\ \overrightarrow{AB}\times\overrightarrow{AC} =\overrightarrow{b}\times\overrightarrow{c} -\overrightarrow{b}\times\overrightarrow{a} -\overrightarrow{a}\times\overrightarrow{c} +\overrightarrow{a}\times\overrightarrow{a}$
$\displaystyle \Rightarrow\ \overrightarrow{AB}\times\overrightarrow{AC} =\overrightarrow{b}\times\overrightarrow{c} +\overrightarrow{a}\times\overrightarrow{b} +\overrightarrow{c}\times\overrightarrow{a} +\overrightarrow{0}\ \ \ \ \ [\because\ \overrightarrow{a}\times\overrightarrow{a}=\overrightarrow{0}]$
$\displaystyle \Rightarrow\ \overrightarrow{AB}\times\overrightarrow{AC} =\overrightarrow{b}\times\overrightarrow{c} +\overrightarrow{c}\times\overrightarrow{a} +\overrightarrow{a}\times\overrightarrow{b}$
$\displaystyle \text{Hence, }\overrightarrow{b}\times\overrightarrow{c} +\overrightarrow{c}\times\overrightarrow{a} +\overrightarrow{a}\times\overrightarrow{b}\text{ is a vector normal to the plane containing points} \\ \text{having position vectors }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}.$

$\displaystyle \textbf{Question 8:}\ \ \text{For any two vectors }\overrightarrow{a}\text{ and } \overrightarrow{b},\text{ show that:}$
$\displaystyle \left(1+\left|\overrightarrow{a}\right|^{2}\right) \left(1+\left|\overrightarrow{b}\right|^{2}\right) =\left(1-\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2} +\left|\overrightarrow{a}+\overrightarrow{b} +\left(\overrightarrow{a}\times\overrightarrow{b}\right)\right|^{2} \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \left(1-\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2} +\left|\overrightarrow{a}+\overrightarrow{b} +\left(\overrightarrow{a}\times\overrightarrow{b}\right)\right|^{2}$
$\displaystyle =\left\{1-2\left(\overrightarrow{a}\cdot\overrightarrow{b}\right) +\left(\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2}\right\} +\left\{\left(\overrightarrow{a}+\overrightarrow{b} +\overrightarrow{a}\times\overrightarrow{b}\right)\cdot \left(\overrightarrow{a}+\overrightarrow{b} +\overrightarrow{a}\times\overrightarrow{b}\right)\right\}$
$\displaystyle =\left\{1-2\left(\overrightarrow{a}\cdot\overrightarrow{b}\right) +\left(\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2}\right\} +\left\{\left(\overrightarrow{a}+\overrightarrow{b}\right)\cdot \left(\overrightarrow{a}+\overrightarrow{b}\right) +\left(\overrightarrow{a}+\overrightarrow{b}\right)\cdot \left(\overrightarrow{a}\times\overrightarrow{b}\right) +\left(\overrightarrow{a}\times\overrightarrow{b}\right)\cdot \left(\overrightarrow{a}+\overrightarrow{b}\right) +\left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2}\right\}$
$\displaystyle =\left\{1-2\left(\overrightarrow{a}\cdot\overrightarrow{b}\right) +\left(\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2}\right\} +\left\{\left|\overrightarrow{a}+\overrightarrow{b}\right|^{2} +\overrightarrow{a}\cdot\left(\overrightarrow{a}\times\overrightarrow{b}\right) +\overrightarrow{b}\cdot\left(\overrightarrow{a}\times\overrightarrow{b}\right) +\left(\overrightarrow{a}\times\overrightarrow{b}\right)\cdot\overrightarrow{a} +\left(\overrightarrow{a}\times\overrightarrow{b}\right)\cdot\overrightarrow{b} +\left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2}\right\}$
$\displaystyle =\left\{1-2\left(\overrightarrow{a}\cdot\overrightarrow{b}\right) +\left(\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2}\right\} +\left\{\left|\overrightarrow{a}+\overrightarrow{b}\right|^{2} +\left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2}\right\}$
$\displaystyle =1-2\left(\overrightarrow{a}\cdot\overrightarrow{b}\right) +\left(\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2} +\left|\overrightarrow{a}\right|^{2} +\left|\overrightarrow{b}\right|^{2} +2\left(\overrightarrow{a}\cdot\overrightarrow{b}\right) +\left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2}$
$\displaystyle =1+\left|\overrightarrow{a}\right|^{2} +\left|\overrightarrow{b}\right|^{2} +\left(\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2} +\left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2}$
$\displaystyle =1+\left|\overrightarrow{a}\right|^{2} +\left|\overrightarrow{b}\right|^{2} +\left|\overrightarrow{a}\right|^{2}\left|\overrightarrow{b}\right|^{2}$
$\displaystyle =\left(1+\left|\overrightarrow{a}\right|^{2}\right) \left(1+\left|\overrightarrow{b}\right|^{2}\right)$
$\displaystyle \text{Hence, }\left(1+\left|\overrightarrow{a}\right|^{2}\right) \left(1+\left|\overrightarrow{b}\right|^{2}\right) =\left(1-\overrightarrow{a}\cdot\overrightarrow{b}\right)^{2} +\left|\overrightarrow{a}+\overrightarrow{b} +\overrightarrow{a}\times\overrightarrow{b}\right|^{2}$

$\displaystyle \textbf{Question 9:}\ \ \text{If }\overrightarrow{a} =\widehat{i}+\widehat{j}+\widehat{k},\ \overrightarrow{c} =\widehat{j}-\widehat{k}\text{ are given vectors, then find a vector } \\ \overrightarrow{b}\text{ satisfying the equations } \overrightarrow{a}\times\overrightarrow{b} =\overrightarrow{c}\text{ and }\overrightarrow{a}\cdot\overrightarrow{b}=3. \text{[CBSE 2008, 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{b} =x\widehat{i}+y\widehat{j}+z\widehat{k}.\ \text{Then,}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b} =\overrightarrow{c}$
$\displaystyle \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k}\\ 1 & 1 & 1\\ x & y & z \end{vmatrix} =\widehat{j}-\widehat{k}$
$\displaystyle \Rightarrow (z-y)\widehat{i}+(x-z)\widehat{j} +(y-x)\widehat{k}=0\widehat{i}+\widehat{j}-\widehat{k}$
$\displaystyle \Rightarrow z-y=0,\ x-z=1,\ y-x=-1$
$\displaystyle \Rightarrow y=z,\ x-z=1,\ x-y=1\ \ ...(i)$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=3$
$\displaystyle \Rightarrow (\widehat{i}+\widehat{j}+\widehat{k}) \cdot(x\widehat{i}+y\widehat{j}+z\widehat{k})=3$
$\displaystyle \Rightarrow x+y+z=3$
$\displaystyle \Rightarrow x+x-1+x-1=3$
$\displaystyle \Rightarrow 3x=5\Rightarrow x=\frac{5}{3}$
$\displaystyle \therefore y=x-1=\frac{5}{3}-1=\frac{2}{3} \text{ and }z=y=\frac{2}{3}$
$\displaystyle \text{Hence, }\overrightarrow{b} =\frac{5}{3}\widehat{i}+\frac{2}{3}\widehat{j} +\frac{2}{3}\widehat{k}$

$\displaystyle \textbf{Question 10. }\text{The sine of the angle between the vectors}\overrightarrow{a}=3\widehat{i}+\widehat{j}+2\widehat{k} \text{ and }$
$\displaystyle \overrightarrow{b}=\widehat{i}+\widehat{j}+2\widehat{k}\text{ is} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }\sqrt{\frac{5}{21}} \qquad \text{(b) }\frac{5}{\sqrt{21}} \qquad \text{(c) }\sqrt{\frac{3}{21}} \qquad \text{(d) }\frac{4}{\sqrt{21}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Given vectors are, } \overrightarrow{a}=3\widehat{i}+\widehat{j}+2\widehat{k}\text{ and}$
$\displaystyle \overrightarrow{b}=\widehat{i}+\widehat{j}+2\widehat{k}$
$\displaystyle \text{Let }\theta\text{ be the angle between the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b},\text{ then}$
$\displaystyle \sin\theta=\frac{|\overrightarrow{a}\times\overrightarrow{b}|}{|\overrightarrow{a}||\overrightarrow{b}|}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&1&2\\1&1&2\end{vmatrix}$
$\displaystyle =\widehat{i}(2-2)-\widehat{j}(6-2)+\widehat{k}(3-1)$
$\displaystyle =-4\widehat{j}+2\widehat{k}$
$\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{(-4)^2+(2)^2}=\sqrt{20}$
$\displaystyle \therefore \sin\theta=\frac{\sqrt{20}}{\sqrt{14}\sqrt6}=\frac{2\sqrt5}{(\sqrt2\times\sqrt7)(\sqrt2\times\sqrt3)}$
$\displaystyle =\frac{\sqrt5}{\sqrt{21}}=\sqrt{\frac{5}{21}}$
$\\$

$\displaystyle \textbf{Question 11. }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are two non-zero vectors such that the projection of }$
$\displaystyle \overrightarrow{a}\text{ on }\overrightarrow{b}\text{ is }0.\text{ The angle between }\overrightarrow{a}\text{ and }\overrightarrow{b} \text{is} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }\frac{\pi}{2} \qquad \text{(b) }\pi$
$\displaystyle \text{(c) }\frac{\pi}{4} \qquad \text{(d) }0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Given, projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}\text{ is }0.$
$\displaystyle \frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}=0\Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \text{Now, let }\theta\text{ be the angle between the two vectors then,}$
$\displaystyle \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=0$
$\displaystyle \Rightarrow \cos\theta=\cos\frac{\pi}{2}$
$\displaystyle \Rightarrow \theta=\frac{\pi}{2}$
$\\$

$\displaystyle \textbf{Question 12. }\text{If the angle between the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }\frac{\pi}{4}\text{ and }$
$\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|=1,\text{ then }\overrightarrow{a}\cdot\overrightarrow{b}\text{ is equal to} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }-\frac{1}{\sqrt{2}} \qquad \text{(b) }1 \qquad \text{(c) }\frac{1}{\sqrt{2}} \qquad \text{(d) }\sqrt{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given, the angle between the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }\frac{\pi}{4}$
$\displaystyle \text{and }|\overrightarrow{a}\times\overrightarrow{b}|=1$
$\displaystyle \text{We know that, }\sin\theta=\frac{|\overrightarrow{a}\times\overrightarrow{b}|}{|\overrightarrow{a}||\overrightarrow{b}|}$
$\displaystyle \Rightarrow \sin\frac{\pi}{4}=\frac{1}{|\overrightarrow{a}||\overrightarrow{b}|}$
$\displaystyle \Rightarrow \frac{1}{\sqrt2}=\frac{1}{|\overrightarrow{a}||\overrightarrow{b}|}$
$\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|=\sqrt2\qquad ...(i)$
$\displaystyle \text{and }\cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos\frac{\pi}{4}$
$\displaystyle =\sqrt2\cdot\frac{1}{\sqrt2}\qquad [\text{using Eq. (i)}]$
$\displaystyle =1$
$\\$

$\displaystyle \textbf{Question 13. }\text{If }\theta\text{ is the angle between two vectors }\overrightarrow{a}\text{ and }\overrightarrow{b},\text{ then }$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}\geq0\text{ only when} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }0<\theta<\frac{\pi}{2} \qquad \text{(b) }0\leq\theta\leq\frac{\pi}{2} \qquad \text{(c) }0<\theta<\pi \qquad \text{(d) }0\leq\theta\leq\pi$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) We have, }\overrightarrow{a}\cdot\overrightarrow{b}\geq0$
$\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\cos\theta\geq0$
$\displaystyle \Rightarrow \cos\theta\geq0$
$\displaystyle \Rightarrow 0\leq\theta\leq\frac{\pi}{2}$
$\\$

$\displaystyle \textbf{Question 14. }\text{If the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are such that }|\overrightarrow{a}|=3,\ |\overrightarrow{b}|=\frac{2}{3}\text{ and}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b}\text{ is a unit vector, then find the angle between }\overrightarrow{a} \text{ and }\overrightarrow{b}.$
$\displaystyle \hspace{2.2cm}\text{[CBSE 2023; CBSE 2014; CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}|=3\text{ and }|\overrightarrow{b}|=\frac{2}{3}$
$\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}.$
$\displaystyle \text{Also, given }|\overrightarrow{a}\times\overrightarrow{b}|=1$
$\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\sin\theta=1$
$\displaystyle \Rightarrow 3\times\frac{2}{3}\sin\theta=1$
$\displaystyle \Rightarrow 2\sin\theta=1$
$\displaystyle \Rightarrow \sin\theta=\frac{1}{2}$
$\displaystyle \Rightarrow \theta=\frac{\pi}{6}$
$\\$

$\displaystyle \textbf{Question 15. }\text{Find the area of a parallelogram whose adjacent sides}$
$\displaystyle \text{are determined by the vectors }\overrightarrow{a}=\widehat{i}-\widehat{j}+3\widehat{k}\text{ and } \overrightarrow{b}=2\widehat{i}-7\widehat{j}+\widehat{k}. \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, the adjacent sides of a parallelogram are}$
$\displaystyle \overrightarrow{a}=\widehat{i}-\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}-7\widehat{j}+\widehat{k}$
$\displaystyle \therefore \text{Area of parallelogram}=|\overrightarrow{a}\times\overrightarrow{b}|$
$\displaystyle \text{Now, }\overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-1&3\\2&-7&1\end{vmatrix}$
$\displaystyle =\widehat{i}(-1+21)-\widehat{j}(1-6)+\widehat{k}(-7+2)$
$\displaystyle =20\widehat{i}+5\widehat{j}-5\widehat{k}$
$\displaystyle \therefore |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{(20)^2+(5)^2+(-5)^2}$
$\displaystyle =\sqrt{400+25+25}=\sqrt{450}=15\sqrt2$
$\displaystyle \text{Hence, the area of parallelogram is }15\sqrt2\text{ sq units.}$
$\\$

$\displaystyle \textbf{Question 16. }\text{If }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k},\text{ then find a unit}$
$\displaystyle \text{vector perpendicular to both }\overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{a}-\overrightarrow{b}. \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{and }\overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}$
$\displaystyle \text{Let the required unit vector be}$
$\displaystyle \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k},$
$\displaystyle \text{then}$
$\displaystyle \sqrt{x^2+y^2+z^2}=1$
$\displaystyle \Rightarrow x^2+y^2+z^2=1\qquad ...(i)$
$\displaystyle \text{Now, }\overrightarrow{a}+\overrightarrow{b}=(\widehat{i}+\widehat{j}+\widehat{k})+(\widehat{i}+2\widehat{j}+3\widehat{k})$
$\displaystyle =2\widehat{i}+3\widehat{j}+4\widehat{k}$
$\displaystyle \text{and }\overrightarrow{a}-\overrightarrow{b}=(\widehat{i}+\widehat{j}+\widehat{k})-(\widehat{i}+2\widehat{j}+3\widehat{k})$
$\displaystyle =-\widehat{j}-2\widehat{k}$
$\displaystyle \text{Since, }\overrightarrow{r}\text{ is perpendicular to }\overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{a}-\overrightarrow{b},$
$\displaystyle \therefore \overrightarrow{r}\cdot(\overrightarrow{a}+\overrightarrow{b})=0\text{ and }\overrightarrow{r}\cdot(\overrightarrow{a}-\overrightarrow{b})=0$
$\displaystyle \text{i.e. }(x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(2\widehat{i}+3\widehat{j}+4\widehat{k})=0$
$\displaystyle \Rightarrow 2x+3y+4z=0\qquad ...(ii)$
$\displaystyle \text{and }(x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(-\widehat{j}-2\widehat{k})=0$
$\displaystyle \Rightarrow -y-2z=0$
$\displaystyle \Rightarrow y=-2z$
$\displaystyle \text{On putting the value of }y\text{ in Eq. (ii), we get}$
$\displaystyle 2x+3(-2z)+4z=0$
$\displaystyle \Rightarrow x=z$
$\displaystyle \text{On substituting the value of }x\text{ and }y\text{ in Eq. (i), we get}$
$\displaystyle z^2+4z^2+z^2=1$
$\displaystyle \Rightarrow 6z^2=1\Rightarrow z=\pm\frac{1}{\sqrt6}$
$\displaystyle \text{then, }x=\pm\frac{1}{\sqrt6}\text{ and }y=\mp\frac{2}{\sqrt6}$
$\displaystyle \text{Hence, the required vectors are}$
$\displaystyle \frac{1}{\sqrt6}\widehat{i}-\frac{2}{\sqrt6}\widehat{j}+\frac{1}{\sqrt6}\widehat{k}\text{ and }-\frac{1}{\sqrt6}\widehat{i}+\frac{2}{\sqrt6}\widehat{j}-\frac{1}{\sqrt6}\widehat{k}.$
$\\$

$\displaystyle \textbf{Question 17. }\text{If the projection of the vector }\widehat{i}+\widehat{j}+\widehat{k}\text{ on the vector}$
$\displaystyle p\widehat{i}+\widehat{j}-2\widehat{k}\text{ is }\frac{1}{3},\text{ then find the value of }p. \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let, the given vectors be }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}\text{ and}$
$\displaystyle \overrightarrow{b}=p\widehat{i}+\widehat{j}-2\widehat{k}$
$\displaystyle \text{Also, given projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{1}{3}.$
$\displaystyle \therefore \text{Projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}$
$\displaystyle \Rightarrow \frac{1}{3}=\frac{(\widehat{i}+\widehat{j}+\widehat{k})\cdot(p\widehat{i}+\widehat{j}-2\widehat{k})}{\sqrt{p^2+1+4}}$
$\displaystyle \Rightarrow \frac{1}{3}=\frac{p+1-2}{\sqrt{p^2+5}}$
$\displaystyle \Rightarrow \sqrt{p^2+5}=3p-3$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle p^2+5=9p^2+9-18p$
$\displaystyle \Rightarrow 8p^2-18p+4=0\Rightarrow 4p^2-9p+2=0$
$\displaystyle \Rightarrow (4p-1)(p-2)=0$
$\displaystyle \Rightarrow p=\frac{1}{4}\text{ or }p=2$
$\\$

$\displaystyle \textbf{Question 18. }\text{If }\overrightarrow{a}=4\widehat{i}-\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}-2\widehat{j}+\widehat{k},\text{ then find a unit}$
$\displaystyle \text{vector along the vector }\overrightarrow{a}\times\overrightarrow{b}. \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=4\widehat{i}-\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\4&-1&1\\2&-2&1\end{vmatrix}$
$\displaystyle =\widehat{i}(-1+2)-\widehat{j}(4-2)+\widehat{k}(-8+2)$
$\displaystyle =\widehat{i}-2\widehat{j}-6\widehat{k}$
$\displaystyle \therefore \text{Required unit vector along the vector}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\frac{\overrightarrow{a}\times\overrightarrow{b}}{|\overrightarrow{a}\times\overrightarrow{b}|}$
$\displaystyle =\frac{\widehat{i}-2\widehat{j}-6\widehat{k}}{\sqrt{1^2+(-2)^2+(-6)^2}}$
$\displaystyle =\frac{\widehat{i}-2\widehat{j}-6\widehat{k}}{\sqrt{1+4+36}}$
$\displaystyle =\frac{\widehat{i}-2\widehat{j}-6\widehat{k}}{\sqrt{41}}=\frac{1}{\sqrt{41}}\widehat{i}-\frac{2}{\sqrt{41}}\widehat{j}-\frac{6}{\sqrt{41}}\widehat{k}$
$\\$

$\displaystyle \textbf{Question 19. }\text{If }\overrightarrow{r}=3\widehat{i}-2\widehat{j}+6\widehat{k},\text{ then find the value of }$
$\displaystyle (\overrightarrow{r}\times\widehat{j})\cdot(\overrightarrow{r}\times\widehat{k})-12. \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{r}=3\widehat{i}-2\widehat{j}+6\widehat{k}$
$\displaystyle \text{Now, we have to find }(\overrightarrow{r}\times\widehat{j})\cdot(\overrightarrow{r}\times\widehat{k})-12$
$\displaystyle \overrightarrow{r}\times\widehat{j}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&-2&6\\0&1&0\end{vmatrix}$
$\displaystyle =\widehat{i}(0-6)-\widehat{j}(0-0)+\widehat{k}(3-0)$
$\displaystyle =-6\widehat{i}+3\widehat{k}$
$\displaystyle \text{and }\overrightarrow{r}\times\widehat{k}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&-2&6\\0&0&1\end{vmatrix}$
$\displaystyle =\widehat{i}(-2-0)-\widehat{j}(3-0)+\widehat{k}(0-0)$
$\displaystyle =-2\widehat{i}-3\widehat{j}$
$\displaystyle \therefore (\overrightarrow{r}\times\widehat{j})\cdot(\overrightarrow{r}\times\widehat{k})-12$
$\displaystyle =(-6\widehat{i}+3\widehat{k})\cdot(-2\widehat{i}-3\widehat{j})-12$
$\displaystyle =12-12=0$
$\\$

$\displaystyle \textbf{Question 20. }\text{If }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are three non-zero unequal vectors such that}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{a}\cdot\overrightarrow{c},\text{ then find the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}-\overrightarrow{c}. \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are three non-zero unequal vectors}$
$\displaystyle \text{such that }\overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{a}\cdot\overrightarrow{c}$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}-\overrightarrow{a}\cdot\overrightarrow{c}=0$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot(\overrightarrow{b}-\overrightarrow{c})=0\qquad [\text{subtracting }\overrightarrow{a}\cdot\overrightarrow{c}\text{ from both sides}]$
$\displaystyle \text{Now, let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}-\overrightarrow{c}.$
$\displaystyle \therefore \cos\theta=\frac{\overrightarrow{a}\cdot(\overrightarrow{b}-\overrightarrow{c})}{|\overrightarrow{a}||\overrightarrow{b}-\overrightarrow{c}|}$
$\displaystyle \Rightarrow \cos\theta=\frac{0}{|\overrightarrow{a}||\overrightarrow{b}-\overrightarrow{c}|}\qquad [\text{using Eq. (i)}]$
$\displaystyle \Rightarrow \cos\theta=0$
$\displaystyle \Rightarrow \cos\theta=\cos\frac{\pi}{2}\Rightarrow \theta=\frac{\pi}{2}$
$\\$

$\displaystyle \textbf{Question 21. }\text{Three vectors }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ satisfy the condition }$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0.\text{ Evaluate the quantity } \mu=\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}.$
$\displaystyle \text{ If }|\overrightarrow{a}|=3,\ |\overrightarrow{b}|=4\text{ and }|\overrightarrow{c}|=2. \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}|=3,\ |\overrightarrow{b}|=4\text{ and }|\overrightarrow{c}|=2$
$\displaystyle \text{Now, }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})^2=(\overrightarrow{0})^2$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \Rightarrow (3)^2+(4)^2+(2)^2+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \Rightarrow 9+16+4+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \Rightarrow 29+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \Rightarrow 2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=-29$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}=-\frac{29}{2}$
$\\$

$\displaystyle \textbf{Question 22. }\text{Write the projection of vector }(\overrightarrow{b}+\overrightarrow{c})\text{ on the vector}$
$\displaystyle \overrightarrow{a},\text{ where }\overrightarrow{a}=2\widehat{i}-2\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}+2\widehat{j}-2\widehat{k}\text{ and} \overrightarrow{c}=2\widehat{i}-\widehat{j}+4\widehat{k}.$
$\displaystyle \hspace{2.2cm}\text{[CBSE 2022 Term II; CBSE 2013C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{To find projection of }(\overrightarrow{b}+\overrightarrow{c})\text{ on }\overrightarrow{a}.$
$\displaystyle \text{Given, }\overrightarrow{a}=2\widehat{i}-2\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}+2\widehat{j}-2\widehat{k}$
$\displaystyle \text{and }\overrightarrow{c}=2\widehat{i}-\widehat{j}+4\widehat{k}$
$\displaystyle \text{Consider, }(\overrightarrow{b}+\overrightarrow{c})=(\widehat{i}+2\widehat{j}-2\widehat{k})+(2\widehat{i}-\widehat{j}+4\widehat{k})$
$\displaystyle =3\widehat{i}+\widehat{j}+2\widehat{k}$
$\displaystyle \text{Now, the projection of }\overrightarrow{b}+\overrightarrow{c}\text{ on }\overrightarrow{a}\text{ is given by}$
$\displaystyle \frac{(\overrightarrow{b}+\overrightarrow{c})\cdot\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{(3\widehat{i}+\widehat{j}+2\widehat{k})(2\widehat{i}-2\widehat{j}+\widehat{k})}{\sqrt{2^2+(-2)^2+1^2}}$
$\displaystyle =\frac{6-2+2}{\sqrt{4+4+1}}=\frac{6}{3}=2$
$\\$

$\displaystyle \textbf{Question 23. }\text{If }|\overrightarrow{a}|=3,\ |\overrightarrow{b}|=5,\ |\overrightarrow{c}|=4\text{ and }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0},\text{ then find}$
$\displaystyle \text{the value of }(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}). \hspace{2.2cm}\text{[CBSE 2022 (Term II)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})^2=0$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \Rightarrow 9+25+16+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \Rightarrow \frac{50}{2}=-(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})$
$\displaystyle \therefore (\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=-25$
$\\$

$\displaystyle \textbf{Question 24. }\text{Find the projection of the vector }\widehat{i}-\widehat{j}\text{ on the vector } \widehat{i}+\widehat{j}. \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=\widehat{i}-\widehat{j}\text{ and }\overrightarrow{b}=\widehat{i}+\widehat{j}$
$\displaystyle \text{We know that projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}\text{ is }\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}$
$\displaystyle \therefore \frac{(\widehat{i}-\widehat{j})\cdot(\widehat{i}+\widehat{j})}{|\widehat{i}+\widehat{j}|}=\frac{1-1}{\sqrt2}=0$
$\\$

$\displaystyle \textbf{Question 25. }\text{If }\overrightarrow{a}\text{ is a non-zero vector, then find the value of}$
$\displaystyle (\overrightarrow{a}\cdot\widehat{i})\widehat{i}+(\overrightarrow{a}\cdot\widehat{j})\widehat{j}+(\overrightarrow{a}\cdot\widehat{k})\widehat{k}. \hspace{2.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}$
$\displaystyle \therefore (\overrightarrow{a}\cdot\widehat{i})=(x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot\widehat{i}=x,\quad (\overrightarrow{a}\cdot\widehat{j})=y$
$\displaystyle \text{and }(\overrightarrow{a}\cdot\widehat{k})=z$
$\displaystyle \text{Now, }(\overrightarrow{a}\cdot\widehat{i})\widehat{i}+(\overrightarrow{a}\cdot\widehat{j})\widehat{j}+(\overrightarrow{a}\cdot\widehat{k})\widehat{k}=x\widehat{i}+y\widehat{j}+z\widehat{k}=\overrightarrow{a}$
$\\$

$\displaystyle \textbf{Question 26. }\text{Find the unit vector perpendicular to each of the}$
$\displaystyle \text{vectors }\overrightarrow{a}=4\widehat{i}+3\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}-\widehat{j}+2\widehat{k}. \hspace{2.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, vectors are }\overrightarrow{a}=4\widehat{i}+3\widehat{j}+\widehat{k}$
$\displaystyle \text{and }\overrightarrow{b}=2\widehat{i}-\widehat{j}+2\widehat{k}$
$\displaystyle \text{Now, perpendicular vector to the given vector is}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\4&3&1\\2&-1&2\end{vmatrix}$
$\displaystyle =\widehat{i}(6+1)-\widehat{j}(8-2)+\widehat{k}(-4-6)$
$\displaystyle =7\widehat{i}-6\widehat{j}-10\widehat{k}$
$\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{7^2+(-6)^2+(-10)^2}$
$\displaystyle =\sqrt{49+36+100}=\sqrt{185}$
$\displaystyle \therefore \text{Required unit vector }=\frac{\overrightarrow{a}\times\overrightarrow{b}}{|\overrightarrow{a}\times\overrightarrow{b}|}$
$\displaystyle =\frac{7\widehat{i}-6\widehat{j}-10\widehat{k}}{\sqrt{185}}$
$\\$

$\displaystyle \textbf{Question 27. }\text{Find }|\overrightarrow{a}|\text{ and }|\overrightarrow{b}|,\text{ if }|\overrightarrow{a}|=2|\overrightarrow{b}|\text{ and }$
$\displaystyle (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=12. \hspace{2.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }(\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=12\text{ and }|\overrightarrow{a}|=2|\overrightarrow{b}|$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2-|\overrightarrow{b}|^2=12$
$\displaystyle \Rightarrow (2|\overrightarrow{b}|)^2-|\overrightarrow{b}|^2=12$
$\displaystyle \Rightarrow 4|\overrightarrow{b}|^2-|\overrightarrow{b}|^2=12$
$\displaystyle \Rightarrow 3|\overrightarrow{b}|^2=12$
$\displaystyle \Rightarrow |\overrightarrow{b}|^2=4\Rightarrow |\overrightarrow{b}|=2$
$\displaystyle \therefore |\overrightarrow{a}|=2|\overrightarrow{b}|=2(2)=4$
$\\$

$\displaystyle \textbf{Question 28. }\text{If }\overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+4\widehat{j}-5\widehat{k}\text{ represent two}$
$\displaystyle \text{adjacent sides of a parallelogram, find unit vectors parallel to the diagonals }$
$\displaystyle \text{of the parallelogram.} \hspace{2.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }\overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+4\widehat{j}-5\widehat{k}$
$\displaystyle \text{So, the diagonals of the parallelogram whose adjacent}$
$\displaystyle \text{sides are }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are given by}$
$\displaystyle \overrightarrow{p}=\overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{q}=\overrightarrow{a}-\overrightarrow{b}$
$\displaystyle \text{Now, }\overrightarrow{p}=(\widehat{i}+2\widehat{j}+3\widehat{k})+(2\widehat{i}+4\widehat{j}-5\widehat{k})$
$\displaystyle =3\widehat{i}+6\widehat{j}-2\widehat{k}$
$\displaystyle \text{and }\overrightarrow{q}=(\widehat{i}+2\widehat{j}+3\widehat{k})-(2\widehat{i}+4\widehat{j}-5\widehat{k})$
$\displaystyle =-\widehat{i}-2\widehat{j}+8\widehat{k}$
$\displaystyle \therefore \widehat{p}=\frac{\overrightarrow{p}}{|\overrightarrow{p}|}=\frac{3\widehat{i}+6\widehat{j}-2\widehat{k}}{\sqrt{9+36+4}}$
$\displaystyle =\frac{3}{7}\widehat{i}+\frac{6}{7}\widehat{j}-\frac{2}{7}\widehat{k}$
$\displaystyle \text{and }\widehat{q}=\frac{\overrightarrow{q}}{|\overrightarrow{q}|}=\frac{-\widehat{i}-2\widehat{j}+8\widehat{k}}{\sqrt{1+4+64}}$
$\displaystyle =-\frac{1}{\sqrt{69}}\widehat{i}-\frac{2}{\sqrt{69}}\widehat{j}+\frac{8}{\sqrt{69}}\widehat{k}$
$\\$

$\displaystyle \textbf{Question 29. }\text{Using vectors, find the area of the }\triangle ABC\text{ with vertices}$
$\displaystyle A(1,2,3),\ B(2,-1,4)\text{ and }C(4,5,-1). \hspace{2.2cm}\text{[CBSE 2020, 2013; CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ be the position vectors of points }A,B$
$\displaystyle \text{and }C,\text{ respectively. Then, }\overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k},$
$\displaystyle \overrightarrow{b}=2\widehat{i}-\widehat{j}+4\widehat{k}\text{ and }\overrightarrow{c}=4\widehat{i}+5\widehat{j}-\widehat{k}.$
$\displaystyle \text{Clearly, the area of }\triangle ABC=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|$
$\displaystyle \text{Now, }\overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A$
$\displaystyle =\overrightarrow{b}-\overrightarrow{a}=2\widehat{i}-\widehat{j}+4\widehat{k}-(\widehat{i}+2\widehat{j}+3\widehat{k})$
$\displaystyle =\widehat{i}-3\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{AC}=\text{Position vector of }C-\text{Position vector of }A$
$\displaystyle =\overrightarrow{c}-\overrightarrow{a}=4\widehat{i}+5\widehat{j}-\widehat{k}-(\widehat{i}+2\widehat{j}+3\widehat{k})$
$\displaystyle =3\widehat{i}+3\widehat{j}-4\widehat{k}$
$\displaystyle \therefore \overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-3&1\\3&3&-4\end{vmatrix}$
$\displaystyle =(12-3)\widehat{i}-(-4-3)\widehat{j}+(3+9)\widehat{k}$
$\displaystyle =9\widehat{i}+7\widehat{j}+12\widehat{k}$
$\displaystyle \text{and }|\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{(9)^2+(7)^2+(12)^2}$
$\displaystyle =\sqrt{81+49+144}=\sqrt{274}$
$\displaystyle \text{So, area of }\triangle ABC=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|=\frac{1}{2}\sqrt{274}$
$\\$

$\displaystyle \textbf{Question 30. }\text{If the sum of two unit vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is a unit vector,}$
$\displaystyle \text{then show that the magnitude of their difference is }\sqrt{3}. \hspace{0.2cm}\text{[CBSE 2019, 2012C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}.\text{ Then, according to given condition }\overrightarrow{c}\text{ is}$
$\displaystyle \text{a unit vector i.e., }|\overrightarrow{c}|=1$
$\displaystyle \text{To show }|\overrightarrow{a}-\overrightarrow{b}|=\sqrt3$
$\displaystyle \text{Consider, }\overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}\Rightarrow |\overrightarrow{c}|=|\overrightarrow{a}+\overrightarrow{b}|$
$\displaystyle \Rightarrow 1=|\overrightarrow{a}+\overrightarrow{b}|\Rightarrow |\overrightarrow{a}+\overrightarrow{b}|^2=1$
$\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})=1$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2+2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^2=1$
$\displaystyle \Rightarrow 1+2\overrightarrow{a}\cdot\overrightarrow{b}+1=1\Rightarrow 2\overrightarrow{a}\cdot\overrightarrow{b}=-1\qquad ...(i)$
$\displaystyle \text{Now consider, }|\overrightarrow{a}-\overrightarrow{b}|^2=(\overrightarrow{a}-\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})$
$\displaystyle =|\overrightarrow{a}|^2-2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^2$
$\displaystyle =1-(-1)+1\qquad [\text{using Eq. (i)}]$
$\displaystyle \Rightarrow |\overrightarrow{a}-\overrightarrow{b}|^2=3$
$\displaystyle \therefore |\overrightarrow{a}-\overrightarrow{b}|=\sqrt3$
$\displaystyle \text{[taking positive square root, as magnitude cannot be negative]}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 31. }\text{If }|\overrightarrow{a}|=2,\ |\overrightarrow{b}|=7\text{ and }\overrightarrow{a}\times\overrightarrow{b}=3\widehat{i}+2\widehat{j}+6\widehat{k},\text{ then find}$
$\displaystyle \text{the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}. \hspace{2.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}.$
$\displaystyle \text{We have, }\overrightarrow{a}\times\overrightarrow{b}=3\widehat{i}+2\widehat{j}+6\widehat{k}$
$\displaystyle \text{Now, }|\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{3^2+2^2+6^2}=\sqrt{49}=7$
$\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\sin\theta=7\qquad [\because |\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta]$
$\displaystyle \Rightarrow \sin\theta=\frac{7}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{7}{2\times7}=\frac{1}{2}$
$\displaystyle \Rightarrow \sin\theta=\sin\left(\frac{\pi}{6}\right)\Rightarrow \theta=\frac{\pi}{6}$
$\displaystyle \text{Hence, the required angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }\frac{\pi}{6}.$
$\\$

$\displaystyle \textbf{Question 32. }\text{Show that the points }A(-2\widehat{i}+3\widehat{j}+5\widehat{k}),\ B(\widehat{i}+2\widehat{j}+3\widehat{k})$
$\displaystyle \text{and }C(7\widehat{i}-\widehat{k})\text{ are collinear.} \hspace{2.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, points are }A(-2\widehat{i}+3\widehat{j}+5\widehat{k}),\ B(\widehat{i}+2\widehat{j}+3\widehat{k})$
$\displaystyle \text{and }C(7\widehat{i}-\widehat{k}).$
$\displaystyle \text{Here, }\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}=(\widehat{i}+2\widehat{j}+3\widehat{k})-(-2\widehat{i}+3\widehat{j}+5\widehat{k})$
$\displaystyle =3\widehat{i}-\widehat{j}-2\widehat{k}$
$\displaystyle \text{and }\overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b}=(7\widehat{i}-\widehat{k})-(\widehat{i}+2\widehat{j}+3\widehat{k})$
$\displaystyle =6\widehat{i}-2\widehat{j}-4\widehat{k}=2(3\widehat{i}-\widehat{j}-2\widehat{k})$
$\displaystyle \text{Since, }\overrightarrow{AB}=\lambda\overrightarrow{BC},\text{ where }\lambda=2$
$\displaystyle \text{So, the given points are collinear.}$
$\\$

$\displaystyle \textbf{Question 33. }\text{Find }|\overrightarrow{a}\times\overrightarrow{b}|,\text{ if }\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{b}=3\widehat{i}+5\widehat{j}-2\widehat{k}.$
$\displaystyle \hspace{2.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{b}=3\widehat{i}+5\widehat{j}-2\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&1&3\\3&5&-2\end{vmatrix}$
$\displaystyle =\widehat{i}(-2-15)-\widehat{j}(-4-9)+\widehat{k}(10-3)$
$\displaystyle =-17\widehat{i}+13\widehat{j}+7\widehat{k}$
$\displaystyle \therefore |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{(-17)^2+(13)^2+(7)^2}$
$\displaystyle =\sqrt{289+169+49}=\sqrt{507}$
$\\$

$\displaystyle \textbf{Question 34. }\text{If }\widehat{i}+\widehat{j}+\widehat{k},\ 2\widehat{i}+5\widehat{j},\ 3\widehat{i}+2\widehat{j}-3\widehat{k}\text{ and }\widehat{i}-6\widehat{j}-\widehat{k} \text{ respectively, }$
$\displaystyle \text{are the position vectors of points }A,\ B,\ C \text{and }D,\text{ then find the angle between}$
$\displaystyle \text{the straight lines } AB\text{ and }CD.\text{ Find whether }\overrightarrow{AB}\text{ and }\overrightarrow{CD} \text{ are collinear or not.} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }\overrightarrow{OA}=(\widehat{i}+\widehat{j}+\widehat{k}),\ \overrightarrow{OB}=(2\widehat{i}+5\widehat{j}),$
$\displaystyle \overrightarrow{OC}=(3\widehat{i}+2\widehat{j}-3\widehat{k})\text{ and }\overrightarrow{OD}=(\widehat{i}-6\widehat{j}-\widehat{k})$
$\displaystyle \text{Angle between }\overrightarrow{AB}\text{ and }\overrightarrow{CD}\text{ is given by}$
$\displaystyle \cos\theta=\frac{\overrightarrow{AB}\cdot\overrightarrow{CD}}{|\overrightarrow{AB}||\overrightarrow{CD}|}\qquad ...(i)$
$\displaystyle \text{Here, }\overrightarrow{AB}=(2-1)\widehat{i}+(5-1)\widehat{j}+(0-1)\widehat{k}$
$\displaystyle =\widehat{i}+4\widehat{j}-\widehat{k}$
$\displaystyle \overrightarrow{CD}=(1-3)\widehat{i}+(-6-2)\widehat{j}+(-1-(-3))\widehat{k}$
$\displaystyle =-2\widehat{i}-8\widehat{j}+2\widehat{k}$
$\displaystyle |\overrightarrow{AB}|=\sqrt{1^2+4^2+(-1)^2}=\sqrt{18}=3\sqrt2$
$\displaystyle \text{and }|\overrightarrow{CD}|=\sqrt{(-2)^2+(-8)^2+2^2}=\sqrt{72}=6\sqrt2$
$\displaystyle \text{Now, }\cos\theta=\frac{(\widehat{i}+4\widehat{j}-\widehat{k})\cdot(-2\widehat{i}-8\widehat{j}+2\widehat{k})}{3\sqrt2\times6\sqrt2}$
$\displaystyle =\frac{1(-2)+4(-8)+(-1)(2)}{3\times6\times2}$
$\displaystyle =\frac{-36}{36}=-1$
$\displaystyle \Rightarrow \cos\theta=-1\Rightarrow \theta=180^\circ=\pi$
$\displaystyle \text{So, angle between }\overrightarrow{AB}\text{ and }\overrightarrow{CD}\text{ is }\pi.$
$\displaystyle \text{Also, since angle between }\overrightarrow{AB}\text{ and }\overrightarrow{CD}\text{ is }180^\circ,\text{ they are in opposite directions.}$
$\displaystyle \text{Since, }\overrightarrow{AB}\text{ and }\overrightarrow{CD}\text{ are parallel to the same line }m,$
$\displaystyle \text{they are collinear.}$
$\\$

$\displaystyle \textbf{Question 35. }\text{Find the magnitude of each of the two vectors }\overrightarrow{a}\text{ and } \overrightarrow{b},$
$\displaystyle \text{ having the same magnitude such that the angle } \text{between them is }60^\circ \text{ and their}$
$\displaystyle \text{scalar product is }\frac{9}{2}. \hspace{2.2cm}\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, two vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ such that }|\overrightarrow{a}|=|\overrightarrow{b}|,$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=\frac{9}{2}\text{ and angle between them is }60^\circ.$
$\displaystyle \text{We know that }\overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos\theta,$
$\displaystyle \text{where }\theta\text{ is angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}.$
$\displaystyle \therefore \frac{9}{2}=|\overrightarrow{a}||\overrightarrow{a}|\cos60^\circ$
$\displaystyle \Rightarrow \frac{1}{2}|\overrightarrow{a}|^2=\frac{9}{2}\qquad \left[\because \cos60^\circ=\frac{1}{2}\right]$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2=9\Rightarrow |\overrightarrow{a}|=3$
$\displaystyle [\because \text{magnitude cannot be negative}]$
$\displaystyle \text{Thus, }|\overrightarrow{a}|=|\overrightarrow{b}|=3$
$\\$

$\displaystyle \textbf{Question 36. }\text{If }\theta\text{ is the angle between two vectors }\widehat{i}-2\widehat{j}+3\widehat{k}\text{ and}$
$\displaystyle 3\widehat{i}-2\widehat{j}+\widehat{k},\text{ then find }\sin\theta. \hspace{2.2cm}\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{b}=3\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \text{Then, }\sin\theta=\frac{|\overrightarrow{a}\times\overrightarrow{b}|}{|\overrightarrow{a}||\overrightarrow{b}|}\qquad ...(i)$
$\displaystyle [\because |\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta]$
$\displaystyle \text{Here, }|\overrightarrow{a}|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$
$\displaystyle |\overrightarrow{b}|=\sqrt{3^2+(-2)^2+1^2}=\sqrt{9+4+1}=\sqrt{14}$
$\displaystyle \text{and }\overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-2&3\\3&-2&1\end{vmatrix}$
$\displaystyle =\widehat{i}(-2+6)-\widehat{j}(1-9)+\widehat{k}(-2+6)$
$\displaystyle =4\widehat{i}+8\widehat{j}+4\widehat{k}=4(\widehat{i}+2\widehat{j}+\widehat{k})$
$\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=4\sqrt{1^2+2^2+1^2}$
$\displaystyle =4\sqrt{1+4+1}=4\sqrt6$
$\displaystyle \text{Now, from Eq. (i), we get}$
$\displaystyle \sin\theta=\frac{4\sqrt6}{\sqrt{14}\cdot\sqrt{14}}=\frac{4\sqrt6}{14}=\frac{2\sqrt6}{7}$
$\\$

$\displaystyle \textbf{Question 37. }\text{If }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\text{ and }|\overrightarrow{a}|=5,\ |\overrightarrow{b}|=6\text{ and }|\overrightarrow{c}|=9,\text{ then}$
$\displaystyle \text{find the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}. \hspace{2.2cm}\text{[CBSE 2018C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$
$\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b})^2=(-\overrightarrow{c})^2\qquad [\text{on squaring both sides}]$
$\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})=(-\overrightarrow{c})\cdot(-\overrightarrow{c})$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{a}+\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{b}=\overrightarrow{c}\cdot\overrightarrow{c}$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2+2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^2=|\overrightarrow{c}|^2\qquad ...(i)$
$\displaystyle [\because \overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos\theta]$
$\displaystyle \text{Putting the values of }|\overrightarrow{a}|=5,\ |\overrightarrow{b}|=6\text{ and }|\overrightarrow{c}|=9\text{ in Eq. (i), we get}$
$\displaystyle (5)^2+2\times5\times6\cos\theta+(6)^2=(9)^2$
$\displaystyle \Rightarrow 25+60\cos\theta+36=81$
$\displaystyle \Rightarrow 60\cos\theta=81-61=20$
$\displaystyle \Rightarrow \cos\theta=\frac{20}{60}=\frac{1}{3}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{1}{3}\right)$
$\\$

$\displaystyle \textbf{Question 38. }\text{Let }\overrightarrow{a}=4\widehat{i}+5\widehat{j}-\widehat{k},\ \overrightarrow{b}=\widehat{i}-4\widehat{j}+5\widehat{k} \text{and }\overrightarrow{c}=3\widehat{i}+\widehat{j}-\widehat{k}.$
$\displaystyle \text{ Find a vector }\overrightarrow{d}\text{ which is } \text{perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ and }\overrightarrow{d}\cdot\overrightarrow{a}=21. \hspace{0.2cm}\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }\overrightarrow{a}=4\widehat{i}+5\widehat{j}-\widehat{k},\ \overrightarrow{b}=\widehat{i}-4\widehat{j}+5\widehat{k}$
$\displaystyle \text{and }\overrightarrow{c}=3\widehat{i}+\widehat{j}-\widehat{k}$
$\displaystyle \text{Since, }\overrightarrow{d}\text{ is perpendicular to both }\overrightarrow{c}\text{ and }\overrightarrow{b}.$
$\displaystyle \therefore \overrightarrow{d}=\lambda(\overrightarrow{c}\times\overrightarrow{b})=\lambda\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&1&-1\\1&-4&5\end{vmatrix}$
$\displaystyle =\lambda[\widehat{i}(5-4)-\widehat{j}(15+1)+\widehat{k}(-12-1)]$
$\displaystyle =\lambda(\widehat{i}-16\widehat{j}-13\widehat{k})\qquad ...(i)$
$\displaystyle \text{Also, it is given that }\overrightarrow{d}\cdot\overrightarrow{a}=21$
$\displaystyle \therefore \lambda(\widehat{i}-16\widehat{j}-13\widehat{k})\cdot(4\widehat{i}+5\widehat{j}-\widehat{k})=21$
$\displaystyle \Rightarrow \lambda(4-80+13)=21$
$\displaystyle \Rightarrow \lambda(-63)=21$
$\displaystyle \Rightarrow \lambda=-\frac{1}{3}$
$\displaystyle \text{Now from Eq. (i), we get}$
$\displaystyle \overrightarrow{d}=-\frac{1}{3}(\widehat{i}-16\widehat{j}-13\widehat{k})$
$\\$

$\displaystyle \textbf{Question 39. }\text{If }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are three mutually perpendicular vectors of the}$
$\displaystyle \text{same magnitude, then prove that }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\text{ is equally inclined with the vectors }$
$\displaystyle \text{}\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}. \hspace{2.2cm}\text{[CBSE 2017, 2013C, 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{According to given condition, }|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=\lambda\text{ (say)}$
$\displaystyle \text{and }\overrightarrow{a}\cdot\overrightarrow{b}=0,\ \overrightarrow{b}\cdot\overrightarrow{c}=0\text{ and }\overrightarrow{c}\cdot\overrightarrow{a}=0$
$\displaystyle \text{Now, }|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2$
$\displaystyle +2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})$
$\displaystyle \Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=\lambda^2+\lambda^2+\lambda^2+2(0+0+0)=3\lambda^2$
$\displaystyle \Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt3\lambda$
$\displaystyle \text{[length cannot be negative]}$
$\displaystyle \text{Suppose, }(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\text{ is inclined at angles }\theta_1,\theta_2\text{ and }\theta_3$
$\displaystyle \text{respectively with vectors }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c},\text{ then}$
$\displaystyle (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot\overrightarrow{a}=|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{a}|\cos\theta_1$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2+\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{a}\cdot\overrightarrow{c}=\sqrt3\lambda\times\lambda\cos\theta_1$
$\displaystyle \Rightarrow \lambda^2+0+0=\sqrt3\lambda^2\cos\theta_1$
$\displaystyle \therefore \cos\theta_1=\frac{1}{\sqrt3}$
$\displaystyle \text{Similarly, }(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot\overrightarrow{b}=|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{b}|\cos\theta_2$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^2+\overrightarrow{c}\cdot\overrightarrow{b}=\sqrt3\lambda^2\cos\theta_2$
$\displaystyle \Rightarrow 0+\lambda^2+0=\sqrt3\lambda^2\cos\theta_2$
$\displaystyle \therefore \cos\theta_2=\frac{1}{\sqrt3}$
$\displaystyle \text{Similarly, }(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot\overrightarrow{c}=|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{c}|\cos\theta_3$
$\displaystyle \Rightarrow \cos\theta_3=\frac{1}{\sqrt3}$
$\displaystyle \text{Thus, }\cos\theta_1=\cos\theta_2=\cos\theta_3=\frac{1}{\sqrt3}$
$\displaystyle \text{Hence, it is proved that }(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\text{ is equally inclined}$
$\displaystyle \text{with the vectors }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}.\text{ Hence proved.}$
$\\$

$\displaystyle \textbf{Question 40. }\text{Show that the points }A,\ B,\ C\text{ with position vectors}$
$\displaystyle 2\widehat{i}-\widehat{j}+\widehat{k},\ \widehat{i}-3\widehat{k}\text{ and }3\widehat{i}-4\widehat{j}-4\widehat{k} \text{ respectively, are the vertices of a right-angled triangle.}$
$\displaystyle \text{Hence, find the area of the triangle.} \hspace{2.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \overrightarrow{AB}=(\text{position vector of }B)-(\text{position vector of }A)$
$\displaystyle =(\widehat{i}-3\widehat{j}-5\widehat{k})-(2\widehat{i}-\widehat{j}+\widehat{k})=-\widehat{i}-2\widehat{j}-6\widehat{k}$
$\displaystyle \overrightarrow{BC}=(3\widehat{i}-4\widehat{j}-4\widehat{k})-(\widehat{i}-3\widehat{j}-5\widehat{k})=2\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \text{and }\overrightarrow{CA}=(2\widehat{i}-\widehat{j}+\widehat{k})-(3\widehat{i}-4\widehat{j}-4\widehat{k})$
$\displaystyle =-\widehat{i}+3\widehat{j}+5\widehat{k}$
$\displaystyle \text{Here, }\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=0$
$\displaystyle \Rightarrow A,B\text{ and }C\text{ are the vertices of a triangle.}$
$\displaystyle \text{Now, }\overrightarrow{BC}\cdot\overrightarrow{CA}=(2\widehat{i}-\widehat{j}+\widehat{k})\cdot(-\widehat{i}+3\widehat{j}+5\widehat{k})$
$\displaystyle =-2-3+5=0$
$\displaystyle \Rightarrow \overrightarrow{BC}\perp\overrightarrow{CA}\Rightarrow \angle C=90^\circ$
$\displaystyle \text{Now, area of }\triangle ABC=\frac{1}{2}|\overrightarrow{CA}\times\overrightarrow{BC}|$
$\displaystyle =\frac{1}{2}\left|\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-1&3&5\\2&-1&1\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}|(8\widehat{i}+11\widehat{j}-5\widehat{k})|$
$\displaystyle =\frac{1}{2}\sqrt{64+121+25}=\frac{1}{2}\sqrt{210}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 41. }\text{Find }\lambda\text{ and }\mu,\text{ if }(\widehat{i}+3\widehat{j}+9\widehat{k})\times(3\widehat{i}-\lambda\widehat{j}+\mu\widehat{k})=\overrightarrow{0}. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }(\widehat{i}+3\widehat{j}+9\widehat{k})\times(3\widehat{i}-\lambda\widehat{j}+\mu\widehat{k})=0$
$\displaystyle \Rightarrow \begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&3&9\\3&-\lambda&\mu\end{vmatrix}=0$
$\displaystyle \Rightarrow \widehat{i}(3\mu+9\lambda)-\widehat{j}(\mu-27)+\widehat{k}(-\lambda-9)=0$
$\displaystyle \text{On comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k},\text{ we get}$
$\displaystyle 3\mu+9\lambda=0,\ -\mu+27=0\text{ and }-\lambda-9=0$
$\displaystyle \Rightarrow \mu=27\text{ and }\lambda=-9$
$\displaystyle \text{Also, the values of }\mu\text{ and }\lambda\text{ satisfy the equation}$
$\displaystyle 3\mu+9\lambda=0$
$\displaystyle \text{Hence, }\mu=27\text{ and }\lambda=-9$
$\\$

$\displaystyle \textbf{Question 42. }\text{Write the number of vectors of unit length } \text{perpendicular to both}$
$\displaystyle \text{the vectors }\overrightarrow{a}=2\widehat{i}+\widehat{j}+2\widehat{k}\text{ and} \overrightarrow{b}=\widehat{j}+\widehat{k}. \hspace{2.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that unit vectors perpendicular to }\overrightarrow{a}\text{ and }\overrightarrow{b}$
$\displaystyle \text{are }\pm\frac{\overrightarrow{a}\times\overrightarrow{b}}{|\overrightarrow{a}\times\overrightarrow{b}|}$
$\displaystyle \text{So, there are two unit vectors perpendicular to the given vectors.}$
$\\$

$\displaystyle \textbf{Question 43. }\text{If }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are unit vectors such that }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0},\text{ then}$
$\displaystyle \text{write the value of }\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}. \hspace{2.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1\text{ and }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\displaystyle \text{Consider,}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})^2=(0)^2$
$\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=0$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{b}+\overrightarrow{c}\cdot\overrightarrow{c}+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \Rightarrow 1+1+1+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}=-\frac{3}{2}$
$\\$

$\displaystyle \textbf{Question 44. }\text{If }|\overrightarrow{a}\times\overrightarrow{b}|^{2}+|\overrightarrow{a}\cdot\overrightarrow{b}|^{2}=400\text{ and }|\overrightarrow{a}|=5,\text{ then write the}$
$\displaystyle \text{value of }|\overrightarrow{b}|. \hspace{2.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}\times\overrightarrow{b}|^2+|\overrightarrow{a}\cdot\overrightarrow{b}|^2=400\text{ and }|\overrightarrow{a}|=5$
$\displaystyle \text{Consider,}$
$\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|^2+|\overrightarrow{a}\cdot\overrightarrow{b}|^2=400$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2|\overrightarrow{b}|^2\sin^2\theta+|\overrightarrow{a}|^2|\overrightarrow{b}|^2\cos^2\theta=400$
$\displaystyle \qquad [\because |\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta\text{ and }|\overrightarrow{a}\cdot\overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\cos\theta]$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2|\overrightarrow{b}|^2(\sin^2\theta+\cos^2\theta)=400$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2|\overrightarrow{b}|^2=400$
$\displaystyle \Rightarrow 25|\overrightarrow{b}|^2=400\qquad [\because |\overrightarrow{a}|=5]$
$\displaystyle \Rightarrow |\overrightarrow{b}|^2=16\Rightarrow |\overrightarrow{b}|=4$
$\displaystyle \qquad [\because \text{length cannot be negative}]$
$\\$

$\displaystyle \textbf{Question 45. }\text{The two adjacent sides of a parallelogram are vectors}$
$\displaystyle 2\widehat{i}-4\widehat{j}-5\widehat{k}\text{ and }2\widehat{i}+2\widehat{j}+3\widehat{k}.\text{ Find the two unit vectors parallel to its diagonals. }$
$\displaystyle \text{Using the diagonal } \text{vectors, find the area of the parallelogram.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }ABCD\text{ be the given parallelogram with}$
$\displaystyle \overrightarrow{AB}=2\widehat{i}-4\widehat{j}-5\widehat{k}\text{ and }\overrightarrow{AD}=2\widehat{i}+2\widehat{j}+3\widehat{k}$
$\displaystyle \text{Clearly, diagonal }\overrightarrow{AC}\text{ is given by }\overrightarrow{AB}+\overrightarrow{AD}$
$\displaystyle \therefore \overrightarrow{AC}=4\widehat{i}-2\widehat{j}-2\widehat{k}$
$\displaystyle \text{and the diagonal }\overrightarrow{BD}\text{ is given by }\overrightarrow{BC}+\overrightarrow{BA}$
$\displaystyle =\overrightarrow{AD}-\overrightarrow{AB}=6\widehat{j}+8\widehat{k}$
$\displaystyle \text{[using parallelogram law of addition]}$
$\displaystyle \text{Since, the unit vector along }\overrightarrow{AC}\text{ is given by}$
$\displaystyle \frac{\overrightarrow{AC}}{|\overrightarrow{AC}|}=\frac{4\widehat{i}-2\widehat{j}-2\widehat{k}}{\sqrt{16+4+4}}$
$\displaystyle =\frac{4\widehat{i}-2\widehat{j}-2\widehat{k}}{\sqrt{24}}=\frac{1}{\sqrt6}(2\widehat{i}-\widehat{j}-\widehat{k})$
$\displaystyle \text{and the unit vector along }\overrightarrow{BD}\text{ is given by}$
$\displaystyle \frac{\overrightarrow{BD}}{|\overrightarrow{BD}|}=\frac{6\widehat{j}+8\widehat{k}}{\sqrt{36+64}}$
$\displaystyle =\frac{6\widehat{j}+8\widehat{k}}{10}=\frac{1}{5}(3\widehat{j}+4\widehat{k})$
$\displaystyle \text{Since, area of parallelogram }ABCD$
$\displaystyle =\frac{1}{2}|\overrightarrow{AC}\times\overrightarrow{BD}|$
$\displaystyle \text{Here, }\overrightarrow{AC}\times\overrightarrow{BD}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\4&-2&-2\\0&6&8\end{vmatrix}$
$\displaystyle =\widehat{i}(-16+12)-\widehat{j}(32-0)+\widehat{k}(24-0)$
$\displaystyle =-4\widehat{i}-32\widehat{j}+24\widehat{k}$
$\displaystyle \text{and }|\overrightarrow{AC}\times\overrightarrow{BD}|=\sqrt{(-4)^2+(-32)^2+24^2}$
$\displaystyle =\sqrt{4^2(1+8^2+6^2)}=4\sqrt{101}$
$\displaystyle \therefore \text{Area of parallelogram }ABCD=\frac{1}{2}\times4\sqrt{101}=2\sqrt{101}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 46. }\text{If }\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{c}\times\overrightarrow{d}\text{ and }\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b}\times\overrightarrow{d},\text{ then show that}$
$\displaystyle \overrightarrow{a}-\overrightarrow{d}\text{ is parallel to }\overrightarrow{b}-\overrightarrow{c},\text{ where }\overrightarrow{a}\neq\overrightarrow{d}\text{ and }\overrightarrow{b}\neq\overrightarrow{c}. \hspace{0.2cm}\text{[CBSE 2016; CBSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given,}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{c}\times\overrightarrow{d}\qquad ...(i)$
$\displaystyle \text{and}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b}\times\overrightarrow{d}\qquad ...(ii)$
$\displaystyle \text{On subtracting Eq. (ii) from Eq. (i), we get}$
$\displaystyle (\overrightarrow{a}\times\overrightarrow{b})-(\overrightarrow{a}\times\overrightarrow{c})=(\overrightarrow{c}\times\overrightarrow{d})-(\overrightarrow{b}\times\overrightarrow{d})$
$\displaystyle \Rightarrow (\overrightarrow{a}\times\overrightarrow{b})-(\overrightarrow{a}\times\overrightarrow{c})+(\overrightarrow{b}\times\overrightarrow{d})-(\overrightarrow{c}\times\overrightarrow{d})=0$
$\displaystyle \Rightarrow \overrightarrow{a}\times(\overrightarrow{b}-\overrightarrow{c})+(\overrightarrow{b}-\overrightarrow{c})\times\overrightarrow{d}=0$
$\displaystyle \Rightarrow \overrightarrow{a}\times(\overrightarrow{b}-\overrightarrow{c})-\overrightarrow{d}\times(\overrightarrow{b}-\overrightarrow{c})=0$
$\displaystyle \Rightarrow (\overrightarrow{a}-\overrightarrow{d})\times(\overrightarrow{b}-\overrightarrow{c})=0$
$\displaystyle \text{[}\because \overrightarrow{a}\neq\overrightarrow{d}\text{ and }\overrightarrow{b}\neq\overrightarrow{c},\text{ given]}$
$\displaystyle \text{Thus, we have that cross-product of vectors }\overrightarrow{a}-\overrightarrow{d}\text{ and }\overrightarrow{b}-\overrightarrow{c}\text{ is zero vector,}$
$\displaystyle \text{so }\overrightarrow{a}-\overrightarrow{d}\text{ is parallel to }\overrightarrow{b}-\overrightarrow{c}.$
$\\$

$\displaystyle \textbf{Question 47. }\text{If }\overrightarrow{a}=7\widehat{i}+\widehat{j}-4\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+6\widehat{j}+3\widehat{k},\text{ then find the}$
$\displaystyle \text{projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}. \hspace{2.2cm}\text{[CBSE 2015, 2013C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, vectors are }\overrightarrow{a}=7\widehat{i}+\widehat{j}-4\widehat{k}$
$\displaystyle \text{and }\overrightarrow{b}=2\widehat{i}+6\widehat{j}+3\widehat{k}$
$\displaystyle \text{Now, the projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}$
$\displaystyle =\frac{(7\widehat{i}+\widehat{j}-4\widehat{k})\cdot(2\widehat{i}+6\widehat{j}+3\widehat{k})}{\sqrt{2^2+6^2+3^2}}$
$\displaystyle =\frac{14+6-12}{\sqrt{49}}=\frac{8}{7}$
$\\$

$\displaystyle \textbf{Question 48. }\text{If }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are mutually perpendicular unit vectors,}$
$\displaystyle \text{then find the value of }|2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|. \hspace{2.2cm}\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are mutually perpendicular unit vectors.}$
$\displaystyle \therefore \overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot\overrightarrow{c}=\overrightarrow{c}\cdot\overrightarrow{a}=0 \qquad \text{(i)}$
$\displaystyle \text{and }|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1 \qquad \text{(ii)}$
$\displaystyle \text{Now, }|2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=(2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot(2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})$
$\displaystyle =4(\overrightarrow{a}\cdot\overrightarrow{a})+2(2\overrightarrow{a}\cdot\overrightarrow{b})+2(2\overrightarrow{a}\cdot\overrightarrow{c})+(\overrightarrow{b}\cdot\overrightarrow{b})+2(\overrightarrow{b}\cdot\overrightarrow{c})+(\overrightarrow{c}\cdot\overrightarrow{c})$
$\displaystyle \qquad [\because \text{dot product is distributive over addition}]$
$\displaystyle =4(1)+2(0)+2(0)+1+2(0)+1 \qquad [\text{from Eqs. (i) and (ii)}]$
$\displaystyle =6$
$\displaystyle \therefore |2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt6$
$\displaystyle \qquad [\because \text{length cannot be negative}]$
$\\$

$\displaystyle \textbf{Question 49. }\text{Write a unit vector perpendicular to both the vectors}$
$\displaystyle \overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}+\widehat{j}. \hspace{2.2cm}\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, vectors are }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}+\widehat{j}$
$\displaystyle \text{As, we know that the vectors }\overrightarrow{a}\times\overrightarrow{b}\text{ is perpendicular to both the vectors,}$
$\displaystyle \text{so let us first evaluate }\overrightarrow{a}\times\overrightarrow{b}.$
$\displaystyle \text{Then, }\overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&1&1\\1&1&0\end{vmatrix}$
$\displaystyle =\widehat{i}(0-1)-\widehat{j}(0-1)+\widehat{k}(1-1)$
$\displaystyle =-\widehat{i}+\widehat{j}$
$\displaystyle \text{Then, the unit vector perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is given by}$
$\displaystyle \frac{\overrightarrow{a}\times\overrightarrow{b}}{|\overrightarrow{a}\times\overrightarrow{b}|}=\frac{-\widehat{i}+\widehat{j}}{\sqrt{(-1)^2+(1)^2}}=-\frac{1}{\sqrt2}\widehat{i}+\frac{1}{\sqrt2}\widehat{j}$
$\\$

$\displaystyle \textbf{Question 50. }\text{Find the area of a parallelogram whose adjacent sides}$
$\displaystyle \text{are represented by the vectors }2\widehat{i}-3\widehat{k}\text{ and }4\widehat{j}+2\widehat{k}. \hspace{1.2cm}\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let adjacent sides of a parallelogram be }\overrightarrow{a}=2\widehat{i}-3\widehat{k}$
$\displaystyle \text{and }\overrightarrow{b}=4\widehat{j}+2\widehat{k}.$
$\displaystyle \therefore \text{Area of parallelogram}=|\overrightarrow{a}\times\overrightarrow{b}|$
$\displaystyle =(2\widehat{i}-3\widehat{k})\times(4\widehat{j}+2\widehat{k})$
$\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&0&-3\\0&4&2\end{vmatrix}$
$\displaystyle =\widehat{i}(0+12)-\widehat{j}(4+0)+\widehat{k}(8-0)$
$\displaystyle =12\widehat{i}-4\widehat{j}+8\widehat{k}$
$\displaystyle =\sqrt{(12)^2+(-4)^2+(8)^2}$
$\displaystyle =\sqrt{144+16+64}=\sqrt{224}=4\sqrt{14}\text{ sq units.}$
$\\$

$\displaystyle \textbf{Question 51. }\text{If }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k},\text{ then find }(\overrightarrow{r}\times\widehat{i})\cdot(\overrightarrow{r}\times\widehat{j})+xy. \hspace{0.2cm}\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{r}\times\widehat{i}=(x\widehat{i}+y\widehat{j}+z\widehat{k})\times\widehat{i}$
$\displaystyle =x(\widehat{i}\times\widehat{i})+y(\widehat{j}\times\widehat{i})+z(\widehat{k}\times\widehat{i})$
$\displaystyle =x\cdot0+y(-\widehat{k})+z\widehat{j}$
$\displaystyle =-y\widehat{k}+z\widehat{j}\qquad [\because \widehat{i}\times\widehat{i}=0,\ \widehat{j}\times\widehat{i}=-\widehat{k},\ \widehat{k}\times\widehat{i}=\widehat{j}]$
$\displaystyle \text{and }(\overrightarrow{r}\times\widehat{j})=(x\widehat{i}+y\widehat{j}+z\widehat{k})\times\widehat{j}$
$\displaystyle =x(\widehat{i}\times\widehat{j})+y(\widehat{j}\times\widehat{j})+z(\widehat{k}\times\widehat{j})$
$\displaystyle =x\widehat{k}+y\cdot0+z(-\widehat{i})=x\widehat{k}-z\widehat{i}$
$\displaystyle [\because \widehat{i}\times\widehat{j}=\widehat{k},\ \widehat{j}\times\widehat{j}=0,\ \widehat{k}\times\widehat{j}=-\widehat{i}]$
$\displaystyle \therefore (\overrightarrow{r}\times\widehat{i})\cdot(\overrightarrow{r}\times\widehat{j})=(-y\widehat{k}+z\widehat{j})\cdot(x\widehat{k}-z\widehat{i})$
$\displaystyle =0+0-yx-0$
$\displaystyle =-xy$
$\displaystyle \therefore (\overrightarrow{r}\times\widehat{i})\cdot(\overrightarrow{r}\times\widehat{j})+xy=-xy+xy=0$
$\\$

$\displaystyle \textbf{Question 52. }\text{If }\overrightarrow{a}=\widehat{i}+2\widehat{j}+\widehat{k},\ \overrightarrow{b}=2\widehat{i}+\widehat{j}\text{ and }\overrightarrow{c}=3\widehat{i}-4\widehat{j}-5\widehat{k},$
$\displaystyle \text{then find a unit vector perpendicular to both of the } \text{vectors }(\overrightarrow{a}-\overrightarrow{b})$
$\displaystyle \text{ and }(\overrightarrow{c}-\overrightarrow{b}). \hspace{2.2cm}\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, vectors are }\overrightarrow{a}=\widehat{i}+2\widehat{j}+\widehat{k},\ \overrightarrow{b}=2\widehat{i}+\widehat{j}$
$\displaystyle \text{and }\overrightarrow{c}=3\widehat{i}-4\widehat{j}-5\widehat{k}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{a}-\overrightarrow{b}=(\widehat{i}+2\widehat{j}+\widehat{k})-(2\widehat{i}+\widehat{j})$
$\displaystyle =-\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{and}$
$\displaystyle \overrightarrow{c}-\overrightarrow{b}=(3\widehat{i}-4\widehat{j}-5\widehat{k})-(2\widehat{i}+\widehat{j})$
$\displaystyle =\widehat{i}-5\widehat{j}-5\widehat{k}$
$\displaystyle \text{Now, a vector perpendicular to }(\overrightarrow{a}-\overrightarrow{b})\text{ and }(\overrightarrow{c}-\overrightarrow{b})\text{ is given by}$
$\displaystyle (\overrightarrow{a}-\overrightarrow{b})\times(\overrightarrow{c}-\overrightarrow{b})=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-1&1&1\\1&-5&-5\end{vmatrix}$
$\displaystyle =\widehat{i}(-5+5)-\widehat{j}(5-1)+\widehat{k}(5-1)$
$\displaystyle =0\widehat{i}-4\widehat{j}+4\widehat{k}$
$\displaystyle \text{and unit vector along }(\overrightarrow{a}-\overrightarrow{b})\times(\overrightarrow{c}-\overrightarrow{b})\text{ is given by}$
$\displaystyle \frac{-4\widehat{j}+4\widehat{k}}{| -4\widehat{j}+4\widehat{k}|}=\frac{-4\widehat{j}+4\widehat{k}}{\sqrt{(-4)^2+4^2}}$
$\displaystyle =\frac{-4\widehat{j}+4\widehat{k}}{4\sqrt2}=-\frac{1}{\sqrt2}\widehat{j}+\frac{1}{\sqrt2}\widehat{k}$
$\\$

$\displaystyle \textbf{Question 53. }\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are perpendicular vectors, }|\overrightarrow{a}+\overrightarrow{b}|=13\text{ and}$
$\displaystyle |\overrightarrow{a}|=5,\text{ then find the value of }|\overrightarrow{b}|. \hspace{2.2cm}\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}+\overrightarrow{b}|=13\text{ and }|\overrightarrow{a}|=5$
$\displaystyle \text{Now, }(\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{a}\cdot\overrightarrow{a}+\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{b}$
$\displaystyle \Rightarrow |\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}|^2+2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^2$
$\displaystyle \Rightarrow |\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}|^2+0+|\overrightarrow{b}|^2\qquad [\because \overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot\overrightarrow{a}=0\text{ as }\overrightarrow{a}\perp\overrightarrow{b}]$
$\displaystyle \Rightarrow (13)^2=(5)^2+|\overrightarrow{b}|^2$
$\displaystyle \Rightarrow 169=25+|\overrightarrow{b}|^2$
$\displaystyle \Rightarrow 144=|\overrightarrow{b}|^2\Rightarrow |\overrightarrow{b}|=12$
$\displaystyle [\because \text{length cannot be negative}]$
$\\$

$\displaystyle \textbf{Question 54. }\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are two unit vectors such that }\overrightarrow{a}+\overrightarrow{b}\text{ is also}$
$\displaystyle \text{a unit vector, then find the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}. \hspace{2.2cm}\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}|=1,\ |\overrightarrow{b}|=1\text{ and }|\overrightarrow{a}+\overrightarrow{b}|=1$
$\displaystyle \text{Now, }|\overrightarrow{a}+\overrightarrow{b}|^2=(\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})$
$\displaystyle =\overrightarrow{a}\cdot\overrightarrow{a}+\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{b}$
$\displaystyle \Rightarrow |\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}|^2+2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^2$
$\displaystyle \Rightarrow 1=1+2\overrightarrow{a}\cdot\overrightarrow{b}+1\qquad [\text{given}]$
$\displaystyle \Rightarrow 2\overrightarrow{a}\cdot\overrightarrow{b}=-1$
$\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\cos\theta=-\frac{1}{2}\qquad [\because \overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos\theta]$
$\displaystyle \Rightarrow \cos\theta=-\frac{1}{2}\qquad [\because |\overrightarrow{a}|=|\overrightarrow{b}|=1]$
$\displaystyle \Rightarrow \cos\theta=\cos\frac{2\pi}{3}\Rightarrow \theta=\frac{2\pi}{3}$
$\displaystyle \text{Hence, the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }\frac{2\pi}{3}.$
$\\$

$\displaystyle \textbf{Question 55. }\text{Find the projection of the vector }\widehat{i}+3\widehat{j}+7\widehat{k}\text{ on the vector } \\ 2\widehat{i}-3\widehat{j}+6\widehat{k}.\hspace{2.2cm}\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=\widehat{i}+3\widehat{j}+7\widehat{k},\qquad \overrightarrow{b}=2\widehat{i}-3\widehat{j}+6\widehat{k}$
$\displaystyle \text{Projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(1)(2)+(3)(-3)+(7)(6)=35$
$\displaystyle |\overrightarrow{b}|=\sqrt{2^{2}+(-3)^{2}+6^{2}}=\sqrt{49}=7$
$\displaystyle \therefore \text{Projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{35}{7}=5$
$\\$

$\displaystyle \textbf{Question 56. }\text{Write the projection of vector }\widehat{i}+\widehat{j}+\widehat{k}\text{ along the vector }\widehat{j}.\hspace{2.2cm}\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\qquad \overrightarrow{b}=\widehat{j}$
$\displaystyle \text{Projection of }\overrightarrow{a}\text{ along }\overrightarrow{b}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}$
$\displaystyle =\frac{(\widehat{i}+\widehat{j}+\widehat{k})\cdot\widehat{j}}{1}$
$\displaystyle =1$
$\displaystyle \therefore \text{The required projection is }1.$
$\\$

$\displaystyle \textbf{Question 57. }\text{Write the value of the following}$
$\displaystyle \widehat{i}\times(\widehat{j}+\widehat{k})+\widehat{j}\times(\widehat{k}+\widehat{i})+\widehat{k}\times(\widehat{i}+\widehat{j}) \hspace{2.2cm}\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }\widehat{i}\times(\widehat{j}+\widehat{k})+\widehat{j}\times(\widehat{k}+\widehat{i})+\widehat{k}\times(\widehat{i}+\widehat{j})$
$\displaystyle =\widehat{i}\times\widehat{j}+\widehat{i}\times\widehat{k}+\widehat{j}\times\widehat{k}+\widehat{j}\times\widehat{i}+\widehat{k}\times\widehat{i}+\widehat{k}\times\widehat{j}$
$\displaystyle \text{[cross product is distributive over addition]}$
$\displaystyle =\widehat{k}-\widehat{j}+\widehat{i}-\widehat{k}+\widehat{j}-\widehat{i}=0$
$\displaystyle [\because \widehat{i}\times\widehat{j}=\widehat{k},\ \widehat{i}\times\widehat{k}=-\widehat{j},\ \widehat{j}\times\widehat{k}=\widehat{i},\ \widehat{j}\times\widehat{i}=-\widehat{k},\ \widehat{k}\times\widehat{i}=\widehat{j},\ \widehat{k}\times\widehat{j}=-\widehat{i}]$
$\\$

$\displaystyle \textbf{Question 58. }\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are unit vectors, then find the angle between}$
$\displaystyle \overrightarrow{a}\text{ and }\overrightarrow{b},\text{ given that }(\sqrt{3}\overrightarrow{a}-\overrightarrow{b})\text{ is a unit} \text{vector.} \hspace{2.2cm}\text{[CBSE 2014C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are two unit vectors, then }|\overrightarrow{a}|=|\overrightarrow{b}|=1$
$\displaystyle \text{Also, }(\sqrt3\overrightarrow{a}-\overrightarrow{b})\text{ is a unit vector.}$
$\displaystyle \therefore |\sqrt3\overrightarrow{a}-\overrightarrow{b}|=1\Rightarrow |\sqrt3\overrightarrow{a}-\overrightarrow{b}|^2=1^2$
$\displaystyle \Rightarrow (\sqrt3\overrightarrow{a}-\overrightarrow{b})\cdot(\sqrt3\overrightarrow{a}-\overrightarrow{b})=1$
$\displaystyle \Rightarrow 3(\overrightarrow{a}\cdot\overrightarrow{a})-\sqrt3(\overrightarrow{a}\cdot\overrightarrow{b})-\sqrt3(\overrightarrow{b}\cdot\overrightarrow{a})+\overrightarrow{b}\cdot\overrightarrow{b}=1$
$\displaystyle \Rightarrow 3|\overrightarrow{a}|^2-\sqrt3|\overrightarrow{a}||\overrightarrow{b}|\cos\theta-\sqrt3|\overrightarrow{b}||\overrightarrow{a}|\cos\theta+|\overrightarrow{b}|^2=1$
$\displaystyle \text{where }\theta\text{ is the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}.$
$\displaystyle \Rightarrow 3\times1-\sqrt3\times1\times1\cos\theta-\sqrt3\times1\times1\cos\theta+1=1$
$\displaystyle \Rightarrow 3=2\sqrt3\cos\theta\Rightarrow \cos\theta=\frac{3}{2\sqrt3}$
$\displaystyle \Rightarrow \cos\theta=\frac{\sqrt3}{2}\Rightarrow \theta=\frac{\pi}{6}$
$\displaystyle \text{Hence, the required angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }\frac{\pi}{6}.$
$\\$

$\displaystyle \textbf{Question 59. }\text{If }|\overrightarrow{a}|=8,\ |\overrightarrow{b}|=3\text{ and }|\overrightarrow{a}\times\overrightarrow{b}|=12,\text{ then find the angle}$
$\displaystyle \text{between }\overrightarrow{a}\text{ and }\overrightarrow{b}. \hspace{2.2cm}\text{[CBSE 2014C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}.$
$\displaystyle \text{Given, }|\overrightarrow{a}|=8,\ |\overrightarrow{b}|=3\text{ and }|\overrightarrow{a}\times\overrightarrow{b}|=12$
$\displaystyle \text{We know that }|\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta$
$\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\sin\theta=12$
$\displaystyle \Rightarrow \sin\theta=\frac{12}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{12}{8\times3}$
$\displaystyle \Rightarrow \sin\theta=\frac{1}{2}\Rightarrow \theta=\frac{\pi}{6}$
$\displaystyle \text{Hence, the required angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }\frac{\pi}{6}.$
$\\$

$\displaystyle \textbf{Question 60. }\text{Write the projection of the vector }\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k}\text{ on the vector } \\ \overrightarrow{b}=\widehat{i}+2\widehat{j}+2\widehat{k}.\hspace{2.2cm}\text{[CBSE 2014 C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(2)(1)+(-1)(2)+(1)(2)=2$
$\displaystyle |\overrightarrow{b}|=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{9}=3$
$\displaystyle \therefore \text{Projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{2}{3}$
$\\$

$\displaystyle \textbf{Question 61. }\text{If }\overrightarrow{a}=2\widehat{i}-3\widehat{j}+\widehat{k},\ \overrightarrow{b}=-\widehat{i}+\widehat{k},\ \overrightarrow{c}=2\widehat{j}-\widehat{k}\text{ are three vectors, then find}$
$\displaystyle \text{the area of the parallelogram having } \text{diagonals }(\overrightarrow{a}+\overrightarrow{b})\text{ and }(\overrightarrow{b}+\overrightarrow{c}). \hspace{0.2cm}\text{[CBSE 2014C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=2\widehat{i}-3\widehat{j}+\widehat{k},\ \overrightarrow{b}=-\widehat{i}+\widehat{k}\text{ and }\overrightarrow{c}=2\widehat{j}-\widehat{k}$
$\displaystyle \text{Let }\overrightarrow{d_1}=\overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{d_2}=\overrightarrow{b}+\overrightarrow{c}.$
$\displaystyle \text{Then, }\overrightarrow{d_1}=(2\widehat{i}-3\widehat{j}+\widehat{k})+(-\widehat{i}+\widehat{k})=\widehat{i}-3\widehat{j}+2\widehat{k}$
$\displaystyle \text{and }\overrightarrow{d_2}=(-\widehat{i}+\widehat{k})+(2\widehat{j}-\widehat{k})=-\widehat{i}+2\widehat{j}$
$\displaystyle \text{Clearly, area of given parallelogram with diagonals }\overrightarrow{d_1}$
$\displaystyle \text{and }\overrightarrow{d_2}\text{ is given by }\frac{1}{2}|\overrightarrow{d_1}\times\overrightarrow{d_2}|$
$\displaystyle \text{Here, }\overrightarrow{d_1}\times\overrightarrow{d_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-3&2\\-1&2&0\end{vmatrix}$
$\displaystyle =\widehat{i}(-4)-\widehat{j}(0+2)+\widehat{k}(2-3)$
$\displaystyle =-4\widehat{i}-2\widehat{j}-\widehat{k}$
$\displaystyle \text{So, area of parallelogram}=\frac{1}{2}|-4\widehat{i}-2\widehat{j}-\widehat{k}|$
$\displaystyle =\frac{1}{2}\sqrt{(-4)^2+(-2)^2+(-1)^2}$
$\displaystyle =\frac{1}{2}\sqrt{16+4+1}$
$\displaystyle =\frac{1}{2}\sqrt{21}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 62. }\text{Vectors }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are such that }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\text{ and}$
$\displaystyle |\overrightarrow{a}|=3,|\overrightarrow{b}|=5\text{ and }|\overrightarrow{c}|=7. \text{Find the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}. \\ \hspace{0.2cm}\text{[CBSE 2014, 2008; CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}|^{2}=|\overrightarrow{c}|^{2}$
$\displaystyle |\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{c}|^{2}$
$\displaystyle 3^{2}+5^{2}+2(3)(5)\cos\theta=7^{2}$
$\displaystyle 9+25+30\cos\theta=49$
$\displaystyle 30\cos\theta=15$
$\displaystyle \cos\theta=\frac{1}{2}$
$\displaystyle \theta=\frac{\pi}{3}=60^{\circ}$
$\displaystyle \therefore \text{The angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }60^{\circ}.$
$\\$

$\displaystyle \textbf{Question 63. }\text{Find the vector }\overrightarrow{p}\text{ which is perpendicular to both}$
$\displaystyle \overrightarrow{a}=4\widehat{i}+5\widehat{j}-\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}-4\widehat{j}+5\widehat{k}\text{ and }\overrightarrow{p}\cdot\overrightarrow{q}=21, \text{ where }\overrightarrow{q}=3\widehat{i}+\widehat{j}-\widehat{k}.$
$\displaystyle \hspace{2.2cm}\text{[CBSE 2014C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{\alpha}=4\widehat{i}+5\widehat{j}-\widehat{k},\ \overrightarrow{\beta}=\widehat{i}-4\widehat{j}+5\widehat{k}$
$\displaystyle \text{and }\overrightarrow{q}=3\widehat{i}+\widehat{j}-\widehat{k}$
$\displaystyle \text{Also, vector }\overrightarrow{p}\text{ is perpendicular to }\overrightarrow{\alpha}\text{ and }\overrightarrow{\beta}.$
$\displaystyle \text{Then, }\overrightarrow{p}=\lambda(\overrightarrow{\alpha}\times\overrightarrow{\beta})\qquad ...(i)$
$\displaystyle \text{Now, }\overrightarrow{\alpha}\times\overrightarrow{\beta}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\4&5&-1\\1&-4&5\end{vmatrix}$
$\displaystyle =\widehat{i}(25-4)-\widehat{j}(20+1)+\widehat{k}(-16-5)$
$\displaystyle =21\widehat{i}-21\widehat{j}-21\widehat{k}$
$\displaystyle \Rightarrow \overrightarrow{\alpha}\times\overrightarrow{\beta}=21\widehat{i}-21\widehat{j}-21\widehat{k}$
$\displaystyle \text{So, }\overrightarrow{p}=21\lambda\widehat{i}-21\lambda\widehat{j}-21\lambda\widehat{k}\qquad [\text{from Eq. (i)}]\qquad ...(ii)$
$\displaystyle \text{Also, given that }\overrightarrow{p}\cdot\overrightarrow{q}=21$
$\displaystyle \therefore (21\lambda\widehat{i}-21\lambda\widehat{j}-21\lambda\widehat{k})\cdot(3\widehat{i}+\widehat{j}-\widehat{k})=21$
$\displaystyle \Rightarrow 63\lambda-21\lambda+21\lambda=21$
$\displaystyle \Rightarrow 63\lambda=21$
$\displaystyle \Rightarrow \lambda=\frac{1}{3}$
$\displaystyle \text{On putting }\lambda=\frac{1}{3}\text{ in Eq. (ii), we get}$
$\displaystyle \overrightarrow{p}=21\times\frac{1}{3}\widehat{i}-21\times\frac{1}{3}\widehat{j}-21\times\frac{1}{3}\widehat{k}$
$\displaystyle \therefore \overrightarrow{p}=7\widehat{i}-7\widehat{j}-7\widehat{k}$
$\displaystyle \text{which is the required vector.}$
$\\$

$\displaystyle \textbf{Question 64. }\text{Find the unit vector perpendicular to the plane }ABC, \text{where the position}$
$\displaystyle \text{vectors of }A,B\text{ and }C\text{ are } 2\widehat{i}-\widehat{j}+\widehat{k},\ \widehat{i}+\widehat{j}+2\widehat{k}\text{ and }2\widehat{i}+3\widehat{k},\text{ respectively.}$
$\displaystyle \hspace{2.2cm}\text{[CBSE 2014C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }O\text{ be the origin of reference.}$
$\displaystyle \text{Then, given}$
$\displaystyle \overrightarrow{OA}=2\widehat{i}-\widehat{j}+\widehat{k},$
$\displaystyle \overrightarrow{OB}=\widehat{i}+\widehat{j}+2\widehat{k}$
$\displaystyle \text{and}$
$\displaystyle \overrightarrow{OC}=2\widehat{i}+3\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$\displaystyle =(\widehat{i}+\widehat{j}+2\widehat{k})-(2\widehat{i}-\widehat{j}+\widehat{k})$
$\displaystyle =-\widehat{i}+2\widehat{j}+\widehat{k}$
$\displaystyle \text{and}$
$\displaystyle \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}$
$\displaystyle =(2\widehat{i}+3\widehat{k})-(2\widehat{i}-\widehat{j}+\widehat{k})$
$\displaystyle =\widehat{j}+2\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-1&2&1\\0&1&2\end{vmatrix}$
$\displaystyle =\widehat{i}(4-1)-\widehat{j}(-2-0)+\widehat{k}(-1-0)$
$\displaystyle =3\widehat{i}+2\widehat{j}-\widehat{k}$
$\displaystyle \text{and }|\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{3^2+2^2+(-1)^2}$
$\displaystyle =\sqrt{9+4+1}=\sqrt{14}$
$\displaystyle \therefore \text{Unit vector perpendicular to the plane }ABC$
$\displaystyle =\frac{\overrightarrow{AB}\times\overrightarrow{AC}}{|\overrightarrow{AB}\times\overrightarrow{AC}|}$
$\displaystyle =\frac{3\widehat{i}+2\widehat{j}-\widehat{k}}{\sqrt{14}}$
$\displaystyle =\frac{3}{\sqrt{14}}\widehat{i}+\frac{2}{\sqrt{14}}\widehat{j}-\frac{1}{\sqrt{14}}\widehat{k}$
$\\$

$\displaystyle \textbf{Question 65. }\text{Write the value of }\lambda,\text{ so that the vectors } \overrightarrow{a}=2\widehat{i}+\lambda\widehat{j}+\widehat{k}$
$\displaystyle \text{ and }\overrightarrow{b}=\widehat{i}-2\widehat{j}+3\widehat{k}\text{ are perpendicular } \text{to each other.} \hspace{2.2cm}\text{[CBSE 2013C, 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, vectors are }\overrightarrow{a}=2\widehat{i}+\lambda\widehat{j}+\widehat{k}$
$\displaystyle \text{and }\overrightarrow{b}=\widehat{i}-2\widehat{j}+3\widehat{k}$
$\displaystyle \text{Since, vectors are perpendicular,}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow (2\widehat{i}+\lambda\widehat{j}+\widehat{k})\cdot(\widehat{i}-2\widehat{j}+3\widehat{k})=0$
$\displaystyle \Rightarrow 2-2\lambda+3=0$
$\displaystyle \Rightarrow \lambda=\frac{5}{2}$
$\\$

$\displaystyle \textbf{Question 66. }\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are two vectors such that }|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|,\text{ then}$
$\displaystyle \text{prove that vector }2\overrightarrow{a}+\overrightarrow{b}\text{ is perpendicular to vector }\overrightarrow{b}. \hspace{2.2cm}\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{To prove, }(2\overrightarrow{a}+\overrightarrow{b})\perp\overrightarrow{b}$
$\displaystyle \text{Given, }|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}|^2$
$\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{a}\cdot\overrightarrow{a}$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2+2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^2=|\overrightarrow{a}|^2$
$\displaystyle \Rightarrow 2\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow (2\overrightarrow{a}+\overrightarrow{b})\cdot\overrightarrow{b}=0$
$\displaystyle \therefore (2\overrightarrow{a}+\overrightarrow{b})\perp\overrightarrow{b}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 67. }\text{Find }|x|,\text{ if for a unit vector }\overrightarrow{a},\ (\overrightarrow{x}-\overrightarrow{a})\cdot(\overrightarrow{x}+\overrightarrow{a})=15.$
$\displaystyle \hspace{2.2cm}\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}\text{ is a unit vector. Then, }|\overrightarrow{a}|=1.$
$\displaystyle \text{Now, we have }(\overrightarrow{x}-\overrightarrow{a})\cdot(\overrightarrow{x}+\overrightarrow{a})=15$
$\displaystyle \Rightarrow \overrightarrow{x}\cdot\overrightarrow{x}-\overrightarrow{a}\cdot\overrightarrow{x}+\overrightarrow{x}\cdot\overrightarrow{a}-\overrightarrow{a}\cdot\overrightarrow{a}=15$
$\displaystyle \Rightarrow \overrightarrow{x}\cdot\overrightarrow{x}-\overrightarrow{a}\cdot\overrightarrow{x}+\overrightarrow{a}\cdot\overrightarrow{x}-\overrightarrow{a}\cdot\overrightarrow{a}=15$
$\displaystyle \qquad [\because \text{scalar product is commutative i.e. }\overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot\overrightarrow{a}]$
$\displaystyle \Rightarrow |\overrightarrow{x}|^2-|\overrightarrow{a}|^2=15$
$\displaystyle \Rightarrow |\overrightarrow{x}|^2-1=15\qquad [\text{given, }|\overrightarrow{a}|=1]$
$\displaystyle \Rightarrow |\overrightarrow{x}|^2=16$
$\displaystyle \Rightarrow |\overrightarrow{x}|=4$
$\displaystyle \qquad [\because \text{length cannot be negative}]$
$\\$

$\displaystyle \textbf{Question 68. }\text{Dot product of a vector with vectors }\widehat{i}-\widehat{j}+\widehat{k}, \ 2\widehat{i}+\widehat{j}-3\widehat{k}$
$\displaystyle \text{ and }\widehat{i}+\widehat{j}+\widehat{k}\text{ are respectively }4,0\text{ and }2. \text{ Find the vector.} \hspace{1.2cm}\text{[CBSE 2013C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required vector is }\overrightarrow{a}=a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k}$
$\displaystyle \text{Also, let }\overrightarrow{b}=\widehat{i}-\widehat{j}+\widehat{k},$
$\displaystyle \overrightarrow{c}=2\widehat{i}+\widehat{j}-3\widehat{k}\text{ and }\overrightarrow{d}=\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{Given, }\overrightarrow{a}\cdot\overrightarrow{b}=4, \ \overrightarrow{a}\cdot\overrightarrow{c}=0\text{ and }\overrightarrow{a}\cdot\overrightarrow{d}=2$
$\displaystyle \text{Now, }\overrightarrow{a}\cdot\overrightarrow{b}=4\Rightarrow a_1-a_2+a_3=4\qquad ...(i)$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{c}=0\Rightarrow 2a_1+a_2-3a_3=0\qquad ...(ii)$
$\displaystyle \text{and }\overrightarrow{a}\cdot\overrightarrow{d}=2\Rightarrow a_1+a_2+a_3=2\qquad ...(iii)$
$\displaystyle \text{On subtracting Eq. (iii) from Eq. (i), we get}$
$\displaystyle -2a_2=2\Rightarrow a_2=-1$
$\displaystyle \text{On substituting }a_2=-1\text{ in Eq. (ii) and (iii), we get}$
$\displaystyle 2a_1-3a_3=1\qquad ...(iv)$
$\displaystyle a_1+a_3=3\qquad ...(v)$
$\displaystyle \text{On multiplying Eq. (v) by 3 and then adding with Eq. (iv), we get}$
$\displaystyle 5a_1=1+9=10\Rightarrow a_1=2$
$\displaystyle \text{On substituting }a_1=2\text{ in Eq. (v), we get}$
$\displaystyle a_3=1$
$\displaystyle \therefore \text{Required vector is }\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k}$
$\\$

$\displaystyle \textbf{Question 69. }\text{Find the values of }\lambda\text{ for which the angle between the}$
$\displaystyle \text{vectors }\overrightarrow{a}=2\lambda^{2}\widehat{i}+4\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=7\widehat{i}-2\widehat{j}+\lambda\widehat{k}\text{ is} \text{ obtuse.} \hspace{0.2cm}\text{[CBSE 2013C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\theta\text{ be the obtuse angle between the vectors}$
$\displaystyle \overrightarrow{a}=2\lambda^2\widehat{i}+4\lambda\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=7\widehat{i}-2\widehat{j}+\lambda\widehat{k}$
$\displaystyle \text{Then, }\cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$
$\displaystyle \Rightarrow \cos\theta=\frac{14\lambda^2-8\lambda+\lambda}{\sqrt{4\lambda^4+16\lambda^2+1}\sqrt{49+4+\lambda^2}}$
$\displaystyle \text{Since }\theta\text{ is an obtuse angle,}$
$\displaystyle \cos\theta<0$
$\displaystyle \Rightarrow \frac{14\lambda^2-7\lambda}{\sqrt{4\lambda^4+16\lambda^2+1}\sqrt{53+\lambda^2}}<0$
$\displaystyle \Rightarrow 14\lambda^2-7\lambda<0$
$\displaystyle \Rightarrow 7\lambda(2\lambda-1)<0$
$\displaystyle \Rightarrow \lambda(2\lambda-1)<0$
$\displaystyle \Rightarrow \text{Either }\lambda<0,\ 2\lambda-1>0\text{ or }\lambda>0,\ 2\lambda-1<0$
$\displaystyle \Rightarrow \text{Either }\lambda<0,\ \lambda>\frac12\text{ or }\lambda>0,\ \lambda<\frac12$
$\displaystyle \text{Clearly, first option is impossible.}$
$\displaystyle \therefore \lambda>0,\ \lambda<\frac12$
$\displaystyle \Rightarrow \lambda\in\left(0,\frac12\right)$
$\\$

$\displaystyle \textbf{Question 70. }\text{If }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are three vectors such that each one is}$
$\displaystyle \text{perpendicular to the vector obtained by sum of the other two and }$
$\displaystyle |\overrightarrow{a}|=3,\ |\overrightarrow{b}|=4\text{ and }|\overrightarrow{c}|=5, \text{ then prove that }|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=5\sqrt{2}. \\ \hspace{0.2cm}\text{[CBSE 2013C, 2010C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }\overrightarrow{a}\perp(\overrightarrow{b}+\overrightarrow{c}),\ \overrightarrow{b}\perp(\overrightarrow{c}+\overrightarrow{a}),\ \overrightarrow{c}\perp(\overrightarrow{a}+\overrightarrow{b})$
$\displaystyle \text{and }|\overrightarrow{a}|=3,\ |\overrightarrow{b}|=4,\ |\overrightarrow{c}|=5$
$\displaystyle \text{To prove }|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=5\sqrt2$
$\displaystyle \text{Consider, }|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})$
$\displaystyle =\overrightarrow{a}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{b}+\overrightarrow{c}\cdot\overrightarrow{c}+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})$
$\displaystyle =|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})$
$\displaystyle [\because \overrightarrow{a}\perp(\overrightarrow{b}+\overrightarrow{c}),\ \therefore \overrightarrow{a}\cdot(\overrightarrow{b}+\overrightarrow{c})=0]$
$\displaystyle \text{Similarly, }\overrightarrow{b}\cdot(\overrightarrow{a}+\overrightarrow{c})=0\text{ and }\overrightarrow{c}\cdot(\overrightarrow{a}+\overrightarrow{b})=0$
$\displaystyle \therefore |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=3^2+4^2+5^2=9+16+25=50$
$\displaystyle \therefore |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt{50}=5\sqrt2$
$\\$

$\displaystyle \textbf{Question 71. }\text{If }\overrightarrow{a}=3\widehat{i}-\widehat{j}\text{ and }\overrightarrow{b}=2\widehat{i}+\widehat{j}-3\widehat{k},\text{ then express}$
$\displaystyle \overrightarrow{b}\text{ in the form }\overrightarrow{b}=\overrightarrow{b}_{1}+\overrightarrow{b}_{2},\text{ where }\overrightarrow{b}_{1}\parallel\overrightarrow{a}\text{ and }\overrightarrow{b}_{2}\perp\overrightarrow{a}. \hspace{2.2cm}\text{[CBSE 2013C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=3\widehat{i}-\widehat{j}\text{ and }\overrightarrow{b}=2\widehat{i}+\widehat{j}-3\widehat{k}$
$\displaystyle \text{Let }\overrightarrow{b_1}=x_1\widehat{i}+y_1\widehat{j}+z_1\widehat{k}\text{ and }\overrightarrow{b_2}=x_2\widehat{i}+y_2\widehat{j}+z_2\widehat{k}$
$\displaystyle \text{be two vectors such that }\overrightarrow{b_1}+\overrightarrow{b_2}=\overrightarrow{b},\ \overrightarrow{b_1}\parallel\overrightarrow{a}\text{ and }\overrightarrow{b_2}\perp\overrightarrow{a}$
$\displaystyle \text{Consider, }\overrightarrow{b_1}+\overrightarrow{b_2}=\overrightarrow{b}$
$\displaystyle \Rightarrow (x_1+x_2)\widehat{i}+(y_1+y_2)\widehat{j}+(z_1+z_2)\widehat{k}=2\widehat{i}+\widehat{j}-3\widehat{k}$
$\displaystyle \text{On comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{ both sides, we get}$
$\displaystyle x_1+x_2=2\qquad ...(i)$
$\displaystyle y_1+y_2=1\qquad ...(ii)$
$\displaystyle z_1+z_2=-3\qquad ...(iii)$
$\displaystyle \text{Now, consider, }\overrightarrow{b_1}\parallel\overrightarrow{a}$
$\displaystyle \Rightarrow \frac{x_1}{3}=\frac{y_1}{-1}=\frac{z_1}{0}=\lambda\ (\text{say})$
$\displaystyle \Rightarrow x_1=3\lambda,\ y_1=-\lambda\text{ and }z_1=0\qquad ...(iv)$
$\displaystyle \text{On substituting the values of }x_1,y_1\text{ and }z_1\text{ from Eq. (iv) to Eq. (i), (ii) and (iii), respectively, we get}$
$\displaystyle x_2=2-3\lambda,\ y_2=1+\lambda\text{ and }z_2=-3\qquad ...(v)$
$\displaystyle \text{Since, }\overrightarrow{b_2}\perp\overrightarrow{a},\text{ therefore }\overrightarrow{b_2}\cdot\overrightarrow{a}=0$
$\displaystyle \Rightarrow 3(2-3\lambda)-1(1+\lambda)=0\qquad [\text{from Eq. (v)}]$
$\displaystyle \Rightarrow 6-9\lambda-1-\lambda=0$
$\displaystyle \Rightarrow 5-10\lambda=0$
$\displaystyle \Rightarrow \lambda=\frac12$
$\displaystyle \text{On substituting }\lambda=\frac12\text{ in Eq. (iv) and (v), we get}$
$\displaystyle x_1=\frac32,\ y_1=-\frac12,\ z_1=0$
$\displaystyle \text{and }x_2=\frac12,\ y_2=\frac32,\ z_2=-3$
$\displaystyle \therefore \overrightarrow{b_1}+\overrightarrow{b_2}=\left(\frac32\widehat{i}-\frac12\widehat{j}\right)+\left(\frac12\widehat{i}+\frac32\widehat{j}-3\widehat{k}\right)$
$\displaystyle =2\widehat{i}+\widehat{j}-3\widehat{k}=\overrightarrow{b}$
$\\$

$\displaystyle \textbf{Question 72. }\text{If }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=\widehat{j}-\widehat{k},\text{ then find a vector }\overrightarrow{c},\text{ such}$
$\displaystyle \text{that }\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b}\text{ and }\overrightarrow{a}\cdot\overrightarrow{c}=3. \hspace{2.2cm}\text{[CBSE 2013, 08]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=\widehat{j}-\widehat{k}$
$\displaystyle \text{Let }\overrightarrow{c}=x\widehat{i}+y\widehat{j}+z\widehat{k}\qquad ...(i)$
$\displaystyle \text{Now, }\overrightarrow{a}\times\overrightarrow{c}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&1&1\\x&y&z\end{vmatrix}$
$\displaystyle =\widehat{i}(z-y)-\widehat{j}(z-x)+\widehat{k}(y-x)$
$\displaystyle \text{Now, }\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b}$
$\displaystyle \Rightarrow \widehat{i}(z-y)+\widehat{j}(x-z)+\widehat{k}(y-x)=0\widehat{i}+1\widehat{j}+(-1)\widehat{k}$
$\displaystyle \text{On comparing the coefficients from both sides, we get}$
$\displaystyle z-y=0,\ x-z=1,\ y-x=-1$
$\displaystyle \Rightarrow y=z\text{ and }x=y+1\qquad ...(ii)$
$\displaystyle \text{Also given, }\overrightarrow{a}\cdot\overrightarrow{c}=3$
$\displaystyle \Rightarrow (\widehat{i}+\widehat{j}+\widehat{k})\cdot(x\widehat{i}+y\widehat{j}+z\widehat{k})=3$
$\displaystyle \Rightarrow x+y+z=3\qquad ...(iii)$
$\displaystyle \text{On substituting Eq. (ii) in Eq. (iii), we get}$
$\displaystyle x+2y=3$
$\displaystyle \Rightarrow y=\frac23=z$
$\displaystyle \therefore x=1+y=1+\frac23=\frac53$
$\displaystyle \therefore \overrightarrow{c}=\frac53\widehat{i}+\frac23\widehat{j}+\frac23\widehat{k}$
$\\$

$\displaystyle \textbf{Question 73. }\text{If }\overrightarrow{a}=\widehat{i}-\widehat{j}+7\widehat{k}\text{ and }\overrightarrow{b}=5\widehat{i}-\widehat{j}+\lambda\widehat{k},\text{ then find the}$
$\displaystyle \text{value of }\lambda,\text{ so that }\overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{a}-\overrightarrow{b}\text{ are perpendicular} \text{ vectors.} \hspace{0.2cm}\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=\widehat{i}-\widehat{j}+7\widehat{k}\text{ and }\overrightarrow{b}=5\widehat{i}-\widehat{j}+\lambda\widehat{k}$
$\displaystyle \text{Then, }\overrightarrow{a}+\overrightarrow{b}=(\widehat{i}-\widehat{j}+7\widehat{k})+(5\widehat{i}-\widehat{j}+\lambda\widehat{k})$
$\displaystyle =6\widehat{i}-2\widehat{j}+(7+\lambda)\widehat{k}$
$\displaystyle \text{and}$
$\displaystyle \overrightarrow{a}-\overrightarrow{b}=(\widehat{i}-\widehat{j}+7\widehat{k})-(5\widehat{i}-\widehat{j}+\lambda\widehat{k})$
$\displaystyle =-4\widehat{i}+(7-\lambda)\widehat{k}$
$\displaystyle \text{Since, }(\overrightarrow{a}+\overrightarrow{b})\text{ and }(\overrightarrow{a}-\overrightarrow{b})\text{ are perpendicular vectors,}$
$\displaystyle (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=0$
$\displaystyle \Rightarrow [6\widehat{i}-2\widehat{j}+(7+\lambda)\widehat{k}]\cdot[-4\widehat{i}+(7-\lambda)\widehat{k}]=0$
$\displaystyle \Rightarrow -24+(7+\lambda)(7-\lambda)=0$
$\displaystyle \Rightarrow 49-\lambda^2=24$
$\displaystyle \Rightarrow \lambda^2=25$
$\displaystyle \therefore \lambda=\pm5$
$\\$

$\displaystyle \textbf{Question 74. }\text{If }\overrightarrow{p}=5\widehat{i}+\lambda\widehat{j}-3\widehat{k}\text{ and }\overrightarrow{q}=\widehat{i}+3\widehat{j}-5\widehat{k},\text{ then find the value of }\lambda,$
$\displaystyle \text{so that }\overrightarrow{p}+\overrightarrow{q}\text{ and }\overrightarrow{p}-\overrightarrow{q}\text{ are perpendicular vectors.}\hspace{2.2cm}\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{p}+\overrightarrow{q}=(5+1)\widehat{i}+(\lambda+3)\widehat{j}+(-3-5)\widehat{k}$
$\displaystyle =6\widehat{i}+(\lambda+3)\widehat{j}-8\widehat{k}$
$\displaystyle \overrightarrow{p}-\overrightarrow{q}=(5-1)\widehat{i}+(\lambda-3)\widehat{j}+(-3+5)\widehat{k}$
$\displaystyle =4\widehat{i}+(\lambda-3)\widehat{j}+2\widehat{k}$
$\displaystyle \text{Since the vectors are perpendicular,}$
$\displaystyle (\overrightarrow{p}+\overrightarrow{q})\cdot(\overrightarrow{p}-\overrightarrow{q})=0$
$\displaystyle (6)(4)+(\lambda+3)(\lambda-3)+(-8)(2)=0$
$\displaystyle 24+\lambda^{2}-9-16=0$
$\displaystyle \lambda^{2}-1=0$
$\displaystyle \lambda=\pm1$
$\displaystyle \therefore \text{The required values of }\lambda\text{ are }1\text{ and }-1.$
$\\$

$\displaystyle \textbf{Question 75. }\text{Find }\lambda,\text{ when projection of }\overrightarrow{a}=\lambda\widehat{i}+\widehat{j}+4\widehat{k}\text{ on}$
$\displaystyle \overrightarrow{b}=2\widehat{i}+6\widehat{j}+3\widehat{k}\text{ is }4\text{ units.} \hspace{2.2cm}\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=\lambda\widehat{i}+\widehat{j}+4\widehat{k},\ \overrightarrow{b}=2\widehat{i}+6\widehat{j}+3\widehat{k}$
$\displaystyle \text{and projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=4.$
$\displaystyle \Rightarrow \frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}=4 \qquad \left[\because \text{projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}\right]$
$\displaystyle \Rightarrow \frac{(\lambda\widehat{i}+\widehat{j}+4\widehat{k})\cdot(2\widehat{i}+6\widehat{j}+3\widehat{k})}{\sqrt{2^2+6^2+3^2}}=4$
$\displaystyle \Rightarrow \frac{2\lambda+6+12}{\sqrt{49}}=4$
$\displaystyle \Rightarrow \frac{2\lambda+18}{7}=4$
$\displaystyle \Rightarrow 2\lambda+18=28$
$\displaystyle \Rightarrow 2\lambda=10$
$\displaystyle \Rightarrow \lambda=5$
$\\$

$\displaystyle \textbf{Question 76. }\text{If }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are three vectors, such that }|\overrightarrow{a}|=5,$
$\displaystyle |\overrightarrow{b}|=12,\ |\overrightarrow{c}|=13\text{ and }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0,\text{ then find the}$
$\displaystyle \text{value of }\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}. \hspace{2.2cm}\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }|\overrightarrow{a}|=5,\ |\overrightarrow{b}|=12\text{ and }|\overrightarrow{c}|=13$
$\displaystyle \text{Now, }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})^2=(\overrightarrow{0})^2$
$\displaystyle \Rightarrow |\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \Rightarrow (5)^2+(12)^2+(13)^2+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \Rightarrow 2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=-(25+144+169)$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}=-\frac{338}{2}=-169$
$\\$

$\displaystyle \textbf{Question 77. }\text{Let }\overrightarrow{a}=\widehat{i}+4\widehat{j}+2\widehat{k},\ \overrightarrow{b}=3\widehat{i}-2\widehat{j}+7\widehat{k}\text{ and } \overrightarrow{c}=2\widehat{i}-\widehat{j}+4\widehat{k}.$
$\displaystyle \text{Find a vector }\overrightarrow{p},\text{ which is } \text{perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ and } \\ \\ \overrightarrow{p}\cdot\overrightarrow{c}=18. \hspace{0.2cm}\text{[CBSE 2012, 10]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given vectors are }\overrightarrow{a}=\widehat{i}+4\widehat{j}+2\widehat{k},$
$\displaystyle \overrightarrow{b}=3\widehat{i}-2\widehat{j}+7\widehat{k}$
$\displaystyle \text{and }\overrightarrow{c}=2\widehat{i}-\widehat{j}+4\widehat{k}$
$\displaystyle \text{Let }\overrightarrow{p}=x\widehat{i}+y\widehat{j}+z\widehat{k}$
$\displaystyle \text{We have, }\overrightarrow{p}\text{ is perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b}$
$\displaystyle \Rightarrow \overrightarrow{p}\cdot\overrightarrow{a}=0$
$\displaystyle \Rightarrow (x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(\widehat{i}+4\widehat{j}+2\widehat{k})=0$
$\displaystyle \Rightarrow x+4y+2z=0\qquad ...(i)$
$\displaystyle \text{and}$
$\displaystyle \Rightarrow \overrightarrow{p}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow (x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(3\widehat{i}-2\widehat{j}+7\widehat{k})=0$
$\displaystyle \Rightarrow 3x-2y+7z=0\qquad ...(ii)$
$\displaystyle \text{Also, given}$
$\displaystyle (x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(2\widehat{i}-\widehat{j}+4\widehat{k})=18$
$\displaystyle \Rightarrow 2x-y+4z=18\qquad ...(iii)$
$\displaystyle \text{On multiplying Eq. (i) by 3 and subtracting it from Eq. (ii), we get}$
$\displaystyle -14y+z=0\qquad ...(iv)$
$\displaystyle \text{Now, multiplying Eq. (i) by 2 and subtracting it from Eq. (iii), we get}$
$\displaystyle -9y=18$
$\displaystyle \Rightarrow y=-2$
$\displaystyle \text{On putting }y=-2\text{ in Eq. (iv), we get}$
$\displaystyle -14(-2)+z=0$
$\displaystyle \Rightarrow z=-28$
$\displaystyle \text{On putting }y=-2\text{ and }z=-28\text{ in Eq. (i), we get}$
$\displaystyle x+4(-2)+2(-28)=0$
$\displaystyle \Rightarrow x-8-56=0$
$\displaystyle \Rightarrow x=64$
$\displaystyle \text{Hence, the required vector is}$
$\displaystyle \overrightarrow{p}=x\widehat{i}+y\widehat{j}+z\widehat{k}$
$\displaystyle \therefore \overrightarrow{p}=64\widehat{i}-2\widehat{j}-28\widehat{k}$
$\\$

$\displaystyle \textbf{Question 78. }\text{If }\overrightarrow{a}\cdot\overrightarrow{a}=0\text{ and }\overrightarrow{a}\cdot\overrightarrow{b}=0,\text{ then what can be concluded}$
$\displaystyle \text{about the vector }\overrightarrow{b}\ ? \hspace{2.2cm}\text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}\cdot\overrightarrow{a}=0\Rightarrow |\overrightarrow{a}|^2=0$
$\displaystyle \Rightarrow |\overrightarrow{a}|=0\qquad ...(i)$
$\displaystyle \text{and}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\cos\theta=0\qquad ...(ii)$
$\displaystyle \text{From Eqs. (i) and (ii), it may be concluded that }\overrightarrow{b}\text{ is either zero or non-zero perpendicular vector.}$
$\\$

$\displaystyle \textbf{Question 79. }\text{Write the projection of vector }\widehat{i}-\widehat{j}\text{ on the vector }\widehat{i}+\widehat{j}. \hspace{0.2cm}\text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=\widehat{i}-\widehat{j},\qquad \overrightarrow{b}=\widehat{i}+\widehat{j}$
$\displaystyle \text{Projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(1)(1)+(-1)(1)=0$
$\displaystyle \therefore \text{Projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=0$
$\\$

$\displaystyle \textbf{Question 80. }\text{Find a unit vector perpendicular to each of the vectors }$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{a}-\overrightarrow{b},\text{ where } \overrightarrow{a}=3\widehat{i}+2\widehat{j}+2\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}+2\widehat{j}-2\widehat{k}.\hspace{0.2cm}\text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}=(3+1)\widehat{i}+(2+2)\widehat{j}+(2-2)\widehat{k}$
$\displaystyle =4\widehat{i}+4\widehat{j}$
$\displaystyle \overrightarrow{a}-\overrightarrow{b}=(3-1)\widehat{i}+(2-2)\widehat{j}+(2+2)\widehat{k}$
$\displaystyle =2\widehat{i}+4\widehat{k}$
$\displaystyle \text{A vector perpendicular to both is }(\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{a}-\overrightarrow{b})$
$\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\4&4&0\\2&0&4\end{vmatrix}$
$\displaystyle =16\widehat{i}-16\widehat{j}-8\widehat{k}$
$\displaystyle =8(2\widehat{i}-2\widehat{j}-\widehat{k})$
$\displaystyle \text{Magnitude of }(2\widehat{i}-2\widehat{j}-\widehat{k})=\sqrt{4+4+1}=3$
$\displaystyle \therefore \text{Required unit vector }=\frac{1}{3}(2\widehat{i}-2\widehat{j}-\widehat{k})$
$\displaystyle =\frac{2}{3}\widehat{i}-\frac{2}{3}\widehat{j}-\frac{1}{3}\widehat{k}$
$\\$