Class 12: Vector or Cross Product – Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

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

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

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

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

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

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

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

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