Class 12: Vectors – ISC Board Problems

$\displaystyle \textbf{Question 1: } \text{If } \overrightarrow{a} = 3\widehat{i} - 2\widehat{j} + \widehat{k} \text{ and } \overrightarrow{b} = 2\widehat{i} - 4\widehat{j} - 3\widehat{k}, \text{ then the value of } \\ \left| \overrightarrow{a} - 2\overrightarrow{b} \right| \text{ will be } \hspace{1.0cm} \text{ ISC 2024}$
$\displaystyle (a)\ \sqrt{85} \hspace{1cm} (b)\ \sqrt{86} \hspace{1cm} (c)\ \sqrt{87} \hspace{1cm} (d)\ \sqrt{88}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b)}\ \text{We have, }\overrightarrow{a}=3\hat{i}-2\hat{j}+\hat{k}$
$\displaystyle \text{and }\overrightarrow{b}=2\hat{i}-4\hat{j}-3\hat{k}$
$\displaystyle \Rightarrow 2\overrightarrow{b}=2(2\hat{i}-4\hat{j}-3\hat{k})=4\hat{i}-8\hat{j}-6\hat{k}\ \ \text{...(ii)}$
$\displaystyle \text{On subtracting (ii) from (i), we get}$
$\displaystyle \overrightarrow{a}-2\overrightarrow{b}=(3\hat{i}-2\hat{j}+\hat{k})-(4\hat{i}-8\hat{j}-6\hat{k})$
$\displaystyle =(3-4)\hat{i}+(-2+8)\hat{j}+(1+6)\hat{k}=-\hat{i}+6\hat{j}+7\hat{k}$
$\displaystyle \text{Now, }|\overrightarrow{a}-2\overrightarrow{b}|=\sqrt{(-1)^{2}+6^{2}+7^{2}}$
$\displaystyle =\sqrt{1+36+49}=\sqrt{86}$
$\\$

$\displaystyle \textbf{Question 2: } \text{If } \overrightarrow{a} = 2\widehat{i} + \widehat{j} + 2\widehat{k} \text{ and } \overrightarrow{b} = 5\widehat{i} - 3\widehat{j} + \widehat{k}, \text{ find the projection of } \\ \overrightarrow{b} \text{ on } \overrightarrow{a}. \hspace{1.0cm} \text{ ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }\overrightarrow{a}=2\hat{i}+\hat{j}+2\hat{k}$
$\displaystyle \text{and }\overrightarrow{b}=5\hat{i}-3\hat{j}+\hat{k}$
$\displaystyle \text{Projection of }\overrightarrow{b}\text{ on }\overrightarrow{a}\text{ is}$
$\displaystyle \frac{\overrightarrow{b}\cdot\overrightarrow{a}}{|\overrightarrow{a}|}$
$\displaystyle =\frac{(5\hat{i}-3\hat{j}+\hat{k})\cdot(2\hat{i}+\hat{j}+2\hat{k})}{\sqrt{2^{2}+1^{2}+2^{2}}}$
$\displaystyle =\frac{10-3+2}{\sqrt{4+1+4}}=\frac{9}{\sqrt{9}}=3$
$\displaystyle \therefore \text{Projection of }\overrightarrow{b}\text{ on }\overrightarrow{a}\text{ is }3$
$\\$

$\displaystyle \textbf{Question 3: } \text{Find a vector of magnitude } 20 \text{ units parallel to the vector } \\ 2\widehat{i} + 5\widehat{j} + 4\widehat{k}. \hspace{1.0cm} \text{ ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }\overrightarrow{a}=2\hat{i}+5\hat{j}+4\hat{k}$
$\displaystyle |\overrightarrow{a}|=\sqrt{2^{2}+5^{2}+4^{2}}=\sqrt{4+25+16}=\sqrt{45}=3\sqrt{5}$
$\displaystyle \text{Unit vector in direction of }\overrightarrow{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}$
$\displaystyle =\frac{1}{3\sqrt{5}}(2\hat{i}+5\hat{j}+4\hat{k})$
$\displaystyle =\frac{2}{3\sqrt{5}}\hat{i}+\frac{5}{3\sqrt{5}}\hat{j}+\frac{4}{3\sqrt{5}}\hat{k}$
$\displaystyle \text{Thus, vector of magnitude }1\text{ is } \frac{2}{3\sqrt{5}}\hat{i}+\frac{5}{3\sqrt{5}}\hat{j}+\frac{4}{3\sqrt{5}}\hat{k}$
$\displaystyle \text{Vector of magnitude }20=20\left[\frac{2}{3\sqrt{5}}\hat{i}+\frac{5}{3\sqrt{5}}\hat{j}+\frac{4}{3\sqrt{5}}\hat{k}\right]$
$\displaystyle =\frac{40}{3\sqrt{5}}\hat{i}+\frac{100}{3\sqrt{5}}\hat{j}+\frac{80}{3\sqrt{5}}\hat{k}$
$\displaystyle =\frac{8\sqrt{5}}{3}\hat{i}+\frac{20\sqrt{5}}{3}\hat{j}+\frac{16\sqrt{5}}{3}\hat{k}$
$\\$

$\displaystyle \textbf{Question 4: } \text{If } \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{a}\times\overrightarrow{c} \text{ where } \overrightarrow{a},\overrightarrow{b} \text{ and } \overrightarrow{c} \text{ are non-zero}$
$\displaystyle \text{vectors, then prove that either } \overrightarrow{b}=\overrightarrow{c} \text{ or } \overrightarrow{a} \text{ and } (\overrightarrow{b}-\overrightarrow{c}) \text{ are parallel.}$
$\displaystyle \quad \text{ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{a}\times\overrightarrow{c},\text{ where }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are non-zero vectors}$
$\displaystyle \Rightarrow \overrightarrow{a}\times\overrightarrow{b}-\overrightarrow{a}\times\overrightarrow{c}=0$
$\displaystyle \Rightarrow \overrightarrow{a}\times(\overrightarrow{b}-\overrightarrow{c})=0$
$\displaystyle \Rightarrow \overrightarrow{a}\parallel(\overrightarrow{b}-\overrightarrow{c})\text{ or }\overrightarrow{b}-\overrightarrow{c}=0$
$\displaystyle \Rightarrow \overrightarrow{a}\parallel(\overrightarrow{b}-\overrightarrow{c})\text{ or }\overrightarrow{b}=\overrightarrow{c}$
$\\$

$\displaystyle \textbf{Question 5: } \text{If } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are two non-zero} \text{vectors such that } \\ |\overrightarrow{a}\times\overrightarrow{b}|=\overrightarrow{a}\cdot\overrightarrow{b}, \text{find the angle between } \overrightarrow{a} \text{ and } \overrightarrow{b}. \quad \text{ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are non-zero vectors such that}$
$\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|=\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\sin\theta=|\overrightarrow{a}||\overrightarrow{b}|\cos\theta$
$\displaystyle \Rightarrow \sin\theta=\cos\theta$
$\displaystyle \Rightarrow \tan\theta=1$
$\displaystyle \Rightarrow \theta=\tan^{-1}(1)=\frac{\pi}{4}$
$\displaystyle \therefore \text{Angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }\frac{\pi}{4}$
$\\$

$\displaystyle \textbf{Question 6: } \text{If } A(1,2,-3) \text{ and } B(-1,-2,1) \text{ are the end points of a vector } \overrightarrow{AB}, \\ \text{ then find the unit vector in the direction of } \overrightarrow{AB}. \hspace{1.0cm} \text{ ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } A=\widehat{i}+2\widehat{j}-3\widehat{k}\text{ and }B=-\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$\displaystyle =(-\widehat{i}-2\widehat{j}+\widehat{k})-(\widehat{i}+2\widehat{j}-3\widehat{k})$
$\displaystyle =-2\widehat{i}-4\widehat{j}+4\widehat{k}$
$\displaystyle \text{Unit vector in }\overrightarrow{AB}=\frac{-2\widehat{i}-4\widehat{j}+4\widehat{k}}{\sqrt{(-2)^{2}+(-4)^{2}+4^{2}}}$
$\displaystyle =\frac{-2\widehat{i}-4\widehat{j}+4\widehat{k}}{\sqrt{4+16+16}}$
$\displaystyle =\frac{-2\widehat{i}-4\widehat{j}+4\widehat{k}}{6}$
$\displaystyle =-\frac{1}{3}\widehat{i}-\frac{2}{3}\widehat{j}+\frac{2}{3}\widehat{k}$
$\\$

$\displaystyle \textbf{Question 7: } \text{If } |\overrightarrow{a}|=3, |\overrightarrow{b}|=\frac{\sqrt{2}}{3} \text{ and } \overrightarrow{a}\times\overrightarrow{b} \text{ is a unit vector, then the angle }$
$\displaystyle \text{between } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ is } \quad \text{ISC 2023}$
$\displaystyle \text{(a) } \frac{\pi}{6} \quad \text{(b) } \frac{\pi}{4} \quad \text{(c) } \frac{\pi}{3} \quad \text{(d) } \frac{\pi}{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given, }|\overrightarrow{a}|=3\text{ and }|\overrightarrow{b}|=\frac{\sqrt{2}}{3}$
$\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|=1$
$\displaystyle \text{We know that }|\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta$
$\displaystyle \Rightarrow \sin\theta=\frac{|\overrightarrow{a}\times\overrightarrow{b}|}{|\overrightarrow{a}||\overrightarrow{b}|}$
$\displaystyle =\frac{1}{3\times\frac{\sqrt{2}}{3}}=\frac{1}{\sqrt{2}}$
$\displaystyle \Rightarrow \sin\theta=\sin\frac{\pi}{4}$
$\displaystyle \Rightarrow \theta=\frac{\pi}{4}$
$\\$

$\displaystyle \textbf{Question 8: } \text{Find the area of the parallelogram whose diagonals are}$
$\displaystyle \overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k} \text{ and } \overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}. \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of parallelogram whose diagonals are }\overrightarrow{d_{1}}\text{ and }\overrightarrow{d_{2}}$
$\displaystyle \text{is } \frac{1}{2}|\overrightarrow{d_{1}}\times\overrightarrow{d_{2}}|$
$\displaystyle \text{Here, }\overrightarrow{d_{1}}=\widehat{i}-3\widehat{j}+\widehat{k}\text{ and }\overrightarrow{d_{2}}=\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{d_{1}}\times\overrightarrow{d_{2}}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-3&1\\1&1&1\end{vmatrix}$
$\displaystyle =((-3)(1)-1(1))\widehat{i}-(1(1)-1(1))\widehat{j}+(1(1)-(-3)(1))\widehat{k}$
$\displaystyle =-4\widehat{i}+4\widehat{k}$
$\displaystyle \Rightarrow |\overrightarrow{d_{1}}\times\overrightarrow{d_{2}}|=\sqrt{(-4)^{2}+4^{2}}=\sqrt{16+16}=4\sqrt{2}$
$\displaystyle \therefore \text{Area of parallelogram}=\frac{1}{2}\times4\sqrt{2}=2\sqrt{2}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 9: } \text{If the two vectors } 3\overrightarrow{i}+\alpha\overrightarrow{j}+\overrightarrow{k} \text{ and } 2\overrightarrow{i}-\overrightarrow{j}+8\overrightarrow{k}$
$\displaystyle \text{are perpendicular to each other, then find the value of } \alpha. \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, vectors }3\widehat{i}+\alpha\widehat{j}+\widehat{k}\text{ and }2\widehat{i}-\widehat{j}+8\widehat{k}\text{ are perpendicular}$
$\displaystyle \therefore 3\cdot2+\alpha(-1)+1\cdot8=0$
$\displaystyle \Rightarrow 6-\alpha+8=0$
$\displaystyle \Rightarrow \alpha=14$
$\\$

$\displaystyle \textbf{Question 10: } \text{If } \overrightarrow{a} \text{ is unit vector and } (2\overrightarrow{x}-3\overrightarrow{a})\cdot(2\overrightarrow{x}+3\overrightarrow{a})=91,$
$\displaystyle \text{then find the value of } |\overrightarrow{x}|. \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}|=1$
$\displaystyle \Rightarrow (2\overrightarrow{x}-3\overrightarrow{a})\cdot(2\overrightarrow{x}+3\overrightarrow{a})=91$
$\displaystyle \Rightarrow 4|\overrightarrow{x}|^{2}-9|\overrightarrow{a}|^{2}=91$
$\displaystyle \Rightarrow 4|\overrightarrow{x}|^{2}-9=91$
$\displaystyle \Rightarrow 4|\overrightarrow{x}|^{2}=100$
$\displaystyle \Rightarrow |\overrightarrow{x}|^{2}=25$
$\displaystyle \Rightarrow |\overrightarrow{x}|=5$
$\\$

$\displaystyle \textbf{Question 11: } \text{Write a vector of magnitude of } 18 \text{ units in the direction of the} \\ \text{vector } \widehat{i} - 2\widehat{j} - 2\widehat{k}. \hspace{1.0cm} \text{ ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } \overrightarrow{a}=\widehat{i}-2\widehat{j}-2\widehat{k}$
$\displaystyle \text{The unit vector in the direction of } \overrightarrow{a}\text{ is}$
$\displaystyle \widehat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{\widehat{i}-2\widehat{j}-2\widehat{k}}{\sqrt{1+4+4}}=\frac{1}{3}\widehat{i}-\frac{2}{3}\widehat{j}-\frac{2}{3}\widehat{k}$
$\displaystyle \therefore \text{Vector having magnitude }18\text{ in direction of }\overrightarrow{a}\text{ is }18\widehat{a}$
$\displaystyle =18\left(\frac{1}{3}\widehat{i}-\frac{2}{3}\widehat{j}-\frac{2}{3}\widehat{k}\right)$
$\displaystyle =6\widehat{i}-12\widehat{j}-12\widehat{k}$
$\\$

$\displaystyle \textbf{Question 12: } \text{Using vectors, find the area of the triangle whose vertices are}$
$\displaystyle A(3,-1,2), B(1,-1,-3) \text{ and } C(4,-3,1). \quad \text{ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }\overrightarrow{A}=3\widehat{i}-\widehat{j}+2\widehat{k},\ \overrightarrow{B}=\widehat{i}-\widehat{j}-3\widehat{k},\ \overrightarrow{C}=4\widehat{i}-3\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}=-2\widehat{i}-5\widehat{k}$
$\displaystyle \overrightarrow{AC}=\overrightarrow{C}-\overrightarrow{A}=\widehat{i}-2\widehat{j}-\widehat{k}$
$\displaystyle \text{Area of triangle}=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|$
$\displaystyle \overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-2&0&-5\\1&-2&-1\end{vmatrix}$
$\displaystyle =-10\widehat{i}-7\widehat{j}+4\widehat{k}$
$\displaystyle \Rightarrow |\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{100+49+16}=\sqrt{165}$
$\displaystyle \therefore \text{Required area}=\frac{\sqrt{165}}{2}$
$\\$

$\displaystyle \textbf{Question 13: } \text{If } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are non-collinear vectors, then find the value of } x \text{ such} \\ \text{that the vectors } \alpha = (x-2)\overrightarrow{a} + \overrightarrow{b} \text{ and } \beta = (3+2x)\overrightarrow{a} - 2\overrightarrow{b} \text{ are collinear.} \hspace{1.0cm} \text{ ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, vectors }$
$\displaystyle \alpha=(x-2)\overrightarrow{a}+\overrightarrow{b}\text{ and }\beta=(3+2x)\overrightarrow{a}-2\overrightarrow{b}\text{ are collinear}$
$\displaystyle \therefore \alpha=k\beta,\text{ where }k\text{ is a scalar}$
$\displaystyle \Rightarrow (x-2)\overrightarrow{a}+\overrightarrow{b}=k\left[(3+2x)\overrightarrow{a}-2\overrightarrow{b}\right]$
$\displaystyle \text{Comparing coefficients of }\overrightarrow{a}\text{ and }\overrightarrow{b},$
$\displaystyle x-2=k(3+2x)\text{ and }1=-2k$
$\displaystyle \Rightarrow k=-\frac{1}{2}$
$\displaystyle \therefore x-2=-\frac{1}{2}(3+2x)$
$\displaystyle \Rightarrow 2x-4=-3-2x$
$\displaystyle \Rightarrow 4x=1\Rightarrow x=\frac{1}{4}$
$\\$

$\displaystyle \textbf{Question 14: } \text{If } \overrightarrow{a}=\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k}, \overrightarrow{b}=2\overrightarrow{i}+3\overrightarrow{j}-5\overrightarrow{k}, \text{ then prove that } \overrightarrow{a}$
$\displaystyle \text{and } \overrightarrow{a}\times\overrightarrow{b} \text{ are perpendicular.} \quad \text{ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k},\ \overrightarrow{b}=2\widehat{i}+3\widehat{j}-5\widehat{k}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-2&3\\2&3&-5\end{vmatrix}$
$\displaystyle =\widehat{i}(10-9)-\widehat{j}(-5-6)+\widehat{k}(3+4)$
$\displaystyle =\widehat{i}+11\widehat{j}+7\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{a}\cdot(\overrightarrow{a}\times\overrightarrow{b})=(\widehat{i}-2\widehat{j}+3\widehat{k})\cdot(\widehat{i}+11\widehat{j}+7\widehat{k})$
$\displaystyle =1-22+21=0$
$\displaystyle \therefore \overrightarrow{a}\perp(\overrightarrow{a}\times\overrightarrow{b})$
$\\$

$\displaystyle \textbf{Question 15: } \text{If } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are perpendicular vectors, } |\overrightarrow{a}+\overrightarrow{b}|=13$
$\displaystyle \text{and } |\overrightarrow{a}|=5, \text{ then find the value of } |\overrightarrow{b}|. \quad \text{ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}+\overrightarrow{b}|=13,\ |\overrightarrow{a}|=5$
$\displaystyle \text{Since, }\overrightarrow{a}\perp\overrightarrow{b}\Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \text{Now, }|\overrightarrow{a}+\overrightarrow{b}|^{2}=169$
$\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})=169$
$\displaystyle \Rightarrow |\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}=169$
$\displaystyle \Rightarrow 25+|\overrightarrow{b}|^{2}=169$
$\displaystyle \Rightarrow |\overrightarrow{b}|^{2}=144$
$\displaystyle \Rightarrow |\overrightarrow{b}|=12$
$\\$

$\displaystyle \textbf{Question 16: } \text{If } A,B \text{ and } C \text{ are three non-collinear points with position vectors}$
$\displaystyle \overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \text{ respectively, then show that} \text{the length of the perpendicular from } C \text{ on } AB \text{ is}$
$\displaystyle \frac{|(\overrightarrow{a}\times\overrightarrow{b})+(\overrightarrow{b}\times\overrightarrow{c})+(\overrightarrow{c}\times\overrightarrow{a})|}{|\overrightarrow{b}-\overrightarrow{a}|}. \quad \text{ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{OA}=\overrightarrow{a},\ \overrightarrow{OB}=\overrightarrow{b},\ \overrightarrow{OC}=\overrightarrow{c}$
$\displaystyle \text{Let position vector of }D\text{ be }\overrightarrow{d}$ $\displaystyle \text{From the figure, }D \\ \text{ is foot of perpendicular from }C\text{ to }AB$
$\displaystyle \Rightarrow \overrightarrow{CD}\perp\overrightarrow{AB}$
$\displaystyle \Rightarrow (\overrightarrow{d}-\overrightarrow{c})\cdot(\overrightarrow{b}-\overrightarrow{a})=0$
$\displaystyle \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a},\quad \overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{a},\quad \overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b}$
$\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|$
$\displaystyle =\frac{1}{2}|(\overrightarrow{b}-\overrightarrow{a})\times(\overrightarrow{c}-\overrightarrow{a})|$
$\displaystyle =\frac{1}{2}|\overrightarrow{b}\times\overrightarrow{c}-\overrightarrow{a}\times\overrightarrow{c}-\overrightarrow{b}\times\overrightarrow{a}+\overrightarrow{a}\times\overrightarrow{a}|$
$\displaystyle =\frac{1}{2}|\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}+\overrightarrow{a}\times\overrightarrow{b}|\quad ...(i)$
$\displaystyle \text{Also, area of }\triangle ABC=\frac{1}{2}|\overrightarrow{AB}|\cdot d=\frac{1}{2}|\overrightarrow{b}-\overrightarrow{a}|\cdot d\quad ...(ii)$
$\displaystyle \text{From (i) and (ii), }$
$\displaystyle \frac{1}{2}|\overrightarrow{b}-\overrightarrow{a}|\,d=\frac{1}{2}|\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}|$
$\displaystyle \Rightarrow d=\frac{|\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}|}{|\overrightarrow{b}-\overrightarrow{a}|}$
$\displaystyle \therefore \text{Length of perpendicular from }C\text{ on }AB=\frac{|\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}|}{|\overrightarrow{b}-\overrightarrow{a}|}$
$\\$

$\displaystyle \textbf{Question 17: } \text{If } \overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \text{ are three mutually perpendicular vectors of equal magnitude,}$
$\displaystyle \text{ prove that } (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}) \text{ is equally} \text{inclined with vectors } \overrightarrow{a},\overrightarrow{b} \text{ and } \overrightarrow{c}. \quad \text{ISC 2017}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=a$
$\displaystyle \text{and }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are mutually perpendicular}$
$\displaystyle \therefore \overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot\overrightarrow{c}=\overrightarrow{c}\cdot\overrightarrow{a}=0$
$\displaystyle \text{Now, }|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^{2}=(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})$
$\displaystyle =|\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}+|\overrightarrow{c}|^{2}=3a^{2}$
$\displaystyle \text{Let }(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\text{ make angles }\alpha,\beta,\gamma\text{ with }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$
$\displaystyle (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot\overrightarrow{a}=|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{a}|\cos\alpha$
$\displaystyle =(\sqrt{3}a)(a)\cos\alpha=\sqrt{3}a^{2}\cos\alpha$
$\displaystyle \text{But LHS }=\overrightarrow{a}\cdot\overrightarrow{a}=a^{2}$
$\displaystyle \Rightarrow a^{2}=\sqrt{3}a^{2}\cos\alpha$
$\displaystyle \Rightarrow \cos\alpha=\frac{1}{\sqrt{3}}$
$\displaystyle \text{Similarly, }\cos\beta=\frac{1}{\sqrt{3}},\ \cos\gamma=\frac{1}{\sqrt{3}}$
$\displaystyle \therefore \alpha=\beta=\gamma=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$\\$

$\displaystyle \textbf{Question 18: } \text{Find a unit vector perpendicular to each of the vectors}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b} \text{ and } \overrightarrow{a}-\overrightarrow{b}, \text{ where } \overrightarrow{a}=3\overrightarrow{i}+2\overrightarrow{j}+2\overrightarrow{k} \text{ and}$
$\displaystyle \overrightarrow{b}=\overrightarrow{i}+2\overrightarrow{j}-2\overrightarrow{k}. \quad \text{ISC 2016}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=3\widehat{i}+2\widehat{j}+2\widehat{k},\ \overrightarrow{b}=\widehat{i}+2\widehat{j}-2\widehat{k}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}=4\widehat{i}+4\widehat{j}=\overrightarrow{c}$
$\displaystyle \overrightarrow{a}-\overrightarrow{b}=2\widehat{i}+4\widehat{k}=\overrightarrow{d}$
$\displaystyle \overrightarrow{c}\times\overrightarrow{d}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\4&4&0\\2&0&4\end{vmatrix}$
$\displaystyle =\widehat{i}(16-0)-\widehat{j}(16-0)+\widehat{k}(0-8)$
$\displaystyle =16\widehat{i}-16\widehat{j}-8\widehat{k}$
$\displaystyle \text{Required unit vector }=\frac{16\widehat{i}-16\widehat{j}-8\widehat{k}}{\sqrt{16^{2}+(-16)^{2}+(-8)^{2}}}$
$\displaystyle =\frac{16\widehat{i}-16\widehat{j}-8\widehat{k}}{24}$
$\displaystyle =\pm\left(\frac{2}{3}\widehat{i}-\frac{2}{3}\widehat{j}-\frac{1}{3}\widehat{k}\right)$
$\\$

$\displaystyle \textbf{Question 19: } \text{Using vectors, prove that angle in a semi-circle is a} \text{right angle.} \quad \text{ISC 2015}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }O\text{ be centre of semicircle and }AB\text{ diameter}$
$\displaystyle \text{Let }\overrightarrow{OA}=-\overrightarrow{OB}=\overrightarrow{a},\ \overrightarrow{OC}=\overrightarrow{b}$ $\displaystyle \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\overrightarrow{b}+\overrightarrow{a}$
$\displaystyle \overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=\overrightarrow{b}-\overrightarrow{a}$
$\displaystyle \overrightarrow{AC}\cdot\overrightarrow{BC}=(\overrightarrow{b}+\overrightarrow{a})\cdot(\overrightarrow{b}-\overrightarrow{a})$
$\displaystyle =|\overrightarrow{b}|^{2}-|\overrightarrow{a}|^{2}=0\ (\text{since }|\overrightarrow{a}|=|\overrightarrow{b}|)$
$\displaystyle \therefore \overrightarrow{AC}\perp\overrightarrow{BC}$
$\displaystyle \Rightarrow \angle ACB=90^\circ$
$\displaystyle \overrightarrow{AC}\cdot\overrightarrow{BC}=(\overrightarrow{b}+\overrightarrow{a})\cdot(\overrightarrow{b}-\overrightarrow{a})$
$\displaystyle =\overrightarrow{b}\cdot\overrightarrow{b}-\overrightarrow{b}\cdot\overrightarrow{a}+\overrightarrow{a}\cdot\overrightarrow{b}-\overrightarrow{a}\cdot\overrightarrow{a}$
$\displaystyle =b^{2}-a^{2}=r^{2}-r^{2}=0$
$\displaystyle \therefore \overrightarrow{AC}\perp\overrightarrow{BC}$
$\displaystyle \therefore \angle ACB=90^\circ$
$\displaystyle \text{Hence, the angle in a semi-circle is a right angle.}$
$\\$

$\displaystyle \textbf{Question 20: } \text{In a } \triangle ABC, \text{ using vectors, prove that} c^{2}=a^{2}+b^{2}-2ab\cos C. \quad \text{ISC 2014}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle ABC,\ \overrightarrow{BC}=\overrightarrow{a},\ \overrightarrow{CA}=\overrightarrow{b}\text{ and }\overrightarrow{AB}=\overrightarrow{c}$ $\displaystyle |\overrightarrow{BC}|=a,\ |\overrightarrow{CA}|=b\text{ and }|\overrightarrow{AB}|=c$
$\displaystyle \text{By triangle law of addition,}$
$\displaystyle \overrightarrow{BA}=\overrightarrow{BC}+\overrightarrow{CA}$
$\displaystyle \Rightarrow -\overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}$
$\displaystyle \Rightarrow \overrightarrow{c}=-(\overrightarrow{a}+\overrightarrow{b})$
$\displaystyle \text{Now, }c^{2}=\overrightarrow{c}\cdot\overrightarrow{c}=(\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})$
$\displaystyle =\overrightarrow{a}\cdot\overrightarrow{a}+\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{b}$
$\displaystyle =a^{2}+b^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle =a^{2}+b^{2}+2ab\cos(\pi-C)$
$\displaystyle \text{But }\cos(\pi-C)=-\cos C$
$\displaystyle \therefore c^{2}=a^{2}+b^{2}-2ab\cos C$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 21: } \text{If } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are unit vector and } \theta \text{ is the angle between}$
$\displaystyle \text{them, then show that } |\overrightarrow{a}-\overrightarrow{b}|=2\sin\frac{\theta}{2}. \quad \text{ISC 2013}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}|=1,\ |\overrightarrow{b}|=1\text{ and }\theta\text{ is the angle between them}$
$\displaystyle \text{Now, }|\overrightarrow{a}-\overrightarrow{b}|^{2}=(\overrightarrow{a}-\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})$
$\displaystyle =a^{2}+b^{2}-2\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle =1+1-2|\overrightarrow{a}||\overrightarrow{b}|\cos\theta$
$\displaystyle =2-2\cos\theta=2(1-\cos\theta)$
$\displaystyle =2\cdot 2\sin^{2}\frac{\theta}{2}=4\sin^{2}\frac{\theta}{2}$
$\displaystyle \therefore |\overrightarrow{a}-\overrightarrow{b}|=2\sin\frac{\theta}{2}$
$\\$

$\displaystyle \textbf{Question 22: } \text{In any } \triangle ABC, \text{ prove by vector method}$
$\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}. \quad \text{ISC 2012}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle ABC,\ \overrightarrow{BC}=\overrightarrow{a},\ \overrightarrow{CA}=\overrightarrow{b}\text{ and }\overrightarrow{AB}=\overrightarrow{c}$ $\displaystyle |\overrightarrow{BC}|=a,\ |\overrightarrow{CA}|=b\text{ and }|\overrightarrow{AB}|=c$
$\displaystyle \text{By triangle law of addition,}$
$\displaystyle \overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}=0$
$\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\displaystyle \text{Taking cross product with }\overrightarrow{a},$
$\displaystyle \overrightarrow{a}\times(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=0$
$\displaystyle \Rightarrow \overrightarrow{a}\times\overrightarrow{a}+\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{a}\times\overrightarrow{c}=0$
$\displaystyle \Rightarrow \overrightarrow{a}\times\overrightarrow{b}=-(\overrightarrow{a}\times\overrightarrow{c})=\overrightarrow{c}\times\overrightarrow{a}\quad ...(i)$
$\displaystyle \text{Similarly, } \overrightarrow{c}\times\overrightarrow{a}=\overrightarrow{b}\times\overrightarrow{c}\quad ...(ii)$
$\displaystyle \text{From (i) and (ii), } \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{c}\times\overrightarrow{a}$
$\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{b}\times\overrightarrow{c}|=|\overrightarrow{c}\times\overrightarrow{a}|$
$\displaystyle \Rightarrow ab\sin(\pi-C)=bc\sin(\pi-A)=ca\sin(\pi-B)$
$\displaystyle \Rightarrow ab\sin C=bc\sin A=ca\sin B$
$\displaystyle \Rightarrow \frac{\sin C}{c}=\frac{\sin A}{a}=\frac{\sin B}{b}$
$\displaystyle \Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 23: } \text{If } D,E \text{ and } F \text{ are the mid-points of the sides of a } \triangle ABC, \text{ prove by} \\ \text{vector method that area of } \triangle DEF=\frac{1}{4}(\text{area of } \triangle ABC). \quad \text{ISC 2011, 01}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the position vectors of }A,B,C\text{ be }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ respectively}$
$\displaystyle \text{Given, }D,E,F\text{ are mid-points of sides of }\triangle ABC$ $\displaystyle \therefore \text{Position vectors of }D,E,F\text{ are } \\ \frac{\overrightarrow{a}+\overrightarrow{b}}{2},\frac{\overrightarrow{a}+\overrightarrow{c}}{2},\frac{\overrightarrow{b}+\overrightarrow{c}}{2}\text{ respectively}$
$\displaystyle \text{Now, }\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a},\quad \overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{a}$
$\displaystyle \text{Then, }\overrightarrow{EF}=\frac{\overrightarrow{b}+\overrightarrow{c}}{2}-\frac{\overrightarrow{a}+\overrightarrow{c}}{2}=\frac{1}{2}(\overrightarrow{b}-\overrightarrow{a})$
$\displaystyle \text{and }\overrightarrow{DF}=\frac{\overrightarrow{b}+\overrightarrow{c}}{2}-\frac{\overrightarrow{a}+\overrightarrow{b}}{2}=\frac{1}{2}(\overrightarrow{c}-\overrightarrow{a})$
$\displaystyle \text{Now, Area of }\triangle ABC=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|$
$\displaystyle =\frac{1}{2}|(\overrightarrow{b}-\overrightarrow{a})\times(\overrightarrow{c}-\overrightarrow{a})|\quad ...(i)$
$\displaystyle \text{Area of }\triangle DEF=\frac{1}{2}|\overrightarrow{EF}\times\overrightarrow{DF}|$
$\displaystyle =\frac{1}{2}\left|\frac{1}{2}(\overrightarrow{b}-\overrightarrow{a})\times\frac{1}{2}(\overrightarrow{c}-\overrightarrow{a})\right|$
$\displaystyle =\frac{1}{4}\times\frac{1}{2}|(\overrightarrow{b}-\overrightarrow{a})\times(\overrightarrow{c}-\overrightarrow{a})|$
$\displaystyle =\frac{1}{4}\times \text{Area of }\triangle ABC\quad [\text{from Eq. (i)}]$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 24: } \text{In any } \triangle ABC, \text{ prove by vector method}$
$\displaystyle \cos B=\frac{c^{2}+a^{2}-b^{2}}{2ca}. \quad \text{ISC 2010}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }ABC\text{ be a triangle with }\overrightarrow{BC}=\overrightarrow{a},\overrightarrow{CA}=\overrightarrow{b}\text{ and }\overrightarrow{AB}=\overrightarrow{c}$ $\displaystyle \therefore |\overrightarrow{BC}|=a,\quad |\overrightarrow{CA}|=b,\quad |\overrightarrow{AB}|=c$
$\displaystyle \text{By triangle law of addition, }\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}$
$\displaystyle \Rightarrow \overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{0}$
$\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$
$\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{c}=-\overrightarrow{b}$
$\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{c})\cdot(\overrightarrow{a}+\overrightarrow{c})=(-\overrightarrow{b})\cdot(-\overrightarrow{b})$
$\displaystyle \Rightarrow |\overrightarrow{a}|^{2}+|\overrightarrow{c}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{c}=|\overrightarrow{b}|^{2}$
$\displaystyle \Rightarrow a^{2}+c^{2}+2|\overrightarrow{a}||\overrightarrow{c}|\cos(\pi-B)=b^{2}$
$\displaystyle \Rightarrow a^{2}+c^{2}-2ac\cos B=b^{2}$
$\displaystyle \Rightarrow 2ac\cos B=a^{2}+c^{2}-b^{2}$
$\displaystyle \therefore \cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 25: } \text{Find the area of a parallelogram whose diagonals are determined by the }$
$\displaystyle \text{vectors } \overrightarrow{a}=3\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k} \text{ and} \overrightarrow{b}=\overrightarrow{i}-3\overrightarrow{j}+4\overrightarrow{k}. \quad \text{ISC 2009}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, diagonals of a parallelogram are}$
$\displaystyle \overrightarrow{a}=3\widehat{i}+\widehat{j}-2\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}-3\widehat{j}+4\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&1&-2\\1&-3&4\end{vmatrix}$
$\displaystyle =[4-6]\widehat{i}-[12+2]\widehat{j}+[-9-1]\widehat{k}$
$\displaystyle =-2\widehat{i}-14\widehat{j}-10\widehat{k}$
$\displaystyle \therefore \text{Area of parallelogram}=\frac{1}{2}|\overrightarrow{a}\times\overrightarrow{b}|$
$\displaystyle =\frac{1}{2}|-2\widehat{i}-14\widehat{j}-10\widehat{k}|$
$\displaystyle =\frac{1}{2}\sqrt{(-2)^{2}+(-14)^{2}+(-10)^{2}}$
$\displaystyle =\frac{1}{2}\sqrt{4+196+100}$
$\displaystyle =\frac{1}{2}\sqrt{300}=\frac{10\sqrt{3}}{2}=5\sqrt{3}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 26: } \text{(i) Prove by vector method that the diameter of a circle will subtend a }$
$\displaystyle \text{right angle at a point on its } \text{circumference.}$
$\displaystyle \text{(ii) If } \overrightarrow{a},\overrightarrow{b} \text{ and } \overrightarrow{c} \text{ represent the position vectors of the} \text{points with coordinates} \\ (2,-10,2), (3,1,2) \text{ and} (2,1,3), \text{ respectively, find the value of} \overrightarrow{a}\times(\overrightarrow{b}\times\overrightarrow{c}). \quad \text{ISC 2009}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, } AB \text{ is the diameter of the circle.}$ $\displaystyle \sin \theta = \frac{8}{2 \times 5} = \frac{4}{5}$
$\displaystyle \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{4}{5}\right)^2}$
$\displaystyle = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \frac{3}{5}$
$\displaystyle \text{In } \triangle AOC,\ \overrightarrow{CA} = \overrightarrow{CO} + \overrightarrow{OA} \quad \text{(i)}$
$\displaystyle \text{[triangle law of addition]}$
$\displaystyle \text{and in } \triangle BOC,\ \overrightarrow{CB} = \overrightarrow{CO} + \overrightarrow{OB} \quad \text{(ii)}$
$\displaystyle \text{[triangle law of addition]}$
$\displaystyle \text{Now, } \overrightarrow{CA} \cdot \overrightarrow{CB} = (\overrightarrow{CO} + \overrightarrow{OA}) \cdot (\overrightarrow{CO} + \overrightarrow{OB})$
$\displaystyle = \overrightarrow{CO} \cdot \overrightarrow{CO} + \overrightarrow{CO} \cdot \overrightarrow{OB} + \overrightarrow{OA} \cdot \overrightarrow{CO} + \overrightarrow{OA} \cdot \overrightarrow{OB}$
$\displaystyle = CO^2 + \overrightarrow{CO} \cdot \overrightarrow{OB} - \overrightarrow{OB} \cdot \overrightarrow{CO} - \overrightarrow{OA} \cdot \overrightarrow{OA}$
$\displaystyle [\because\ \overrightarrow{OB} = -\overrightarrow{OA}]$
$\displaystyle = CO^2 - OA^2 = 0 \quad [\because\ CO = OA = \text{radius}]$
$\displaystyle \therefore \overrightarrow{CA} \perp \overrightarrow{CB} \Rightarrow \angle ACB = 90^\circ$
$\displaystyle \text{Hence, diameter of a circle will subtend a right angle at a point on its circumference.}$
$\displaystyle \text{(ii) Given, } \overrightarrow{a} = 2\widehat{i} - 10\widehat{j} + 2\widehat{k},\ \overrightarrow{b} = 3\widehat{i} + \widehat{j} + 2\widehat{k},\ \overrightarrow{c} = 2\widehat{i} + \widehat{j} + 3\widehat{k}$
$\displaystyle \text{Now, } \overrightarrow{b} \times \overrightarrow{c} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 3 & 1 & 2 \\ 2 & 1 & 3 \end{vmatrix}$
$\displaystyle = (3 - 2)\widehat{i} - (9 - 4)\widehat{j} + (3 - 2)\widehat{k}$
$\displaystyle = \widehat{i} - 5\widehat{j} + \widehat{k}$
$\displaystyle \therefore \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 2 & -10 & 2 \\ 1 & -5 & 1 \end{vmatrix}$
$\displaystyle = (-10 + 10)\widehat{i} - (2 - 2)\widehat{j} + (-10 + 10)\widehat{k}$
$\displaystyle = 0\widehat{i} + 0\widehat{j} + 0\widehat{k} = \overrightarrow{0}$
$\\$

$\displaystyle \textbf{Question 27: } \text{Find } \overrightarrow{a}\cdot\overrightarrow{b}, \text{ if } |\overrightarrow{a}|=2, |\overrightarrow{b}|=5 \text{ and } |\overrightarrow{a}\times\overrightarrow{b}|=8.$
$\displaystyle \quad \text{ISC 2008}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } |\overrightarrow{a} \times \overrightarrow{b}| = 8,\ |\overrightarrow{a}| = 2,\ |\overrightarrow{b}| = 5$
$\displaystyle \overrightarrow{a} \times \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}| \sin \theta \ \widehat{n}$
$\displaystyle |\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}||\overrightarrow{b}| \sin \theta$
$\displaystyle \sin \theta = \frac{|\overrightarrow{a} \times \overrightarrow{b}|}{|\overrightarrow{a}||\overrightarrow{b}|} = \frac{8}{2 \times 5} = \frac{4}{5}$
$\displaystyle \cos \theta = \sqrt{1 - \sin^2 \theta} = \frac{3}{5}$
$\displaystyle \overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}| \cos \theta = 2 \times 5 \times \frac{3}{5} = 6$
$\\$

$\displaystyle \textbf{Question 28: } \text{Given } \overrightarrow{a}=\overrightarrow{i}-2\overrightarrow{j}+\overrightarrow{k}, \overrightarrow{b}=2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k} \text{ and}$
$\displaystyle \overrightarrow{c}=\overrightarrow{i}+2\overrightarrow{j}-\overrightarrow{k}. \text{ Then, find } \overrightarrow{a}\times(\overrightarrow{b}\times\overrightarrow{c}). \quad \text{ISC 2008, 02}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \overrightarrow{a} = \widehat{i} - 2\widehat{j} + \widehat{k},\ \overrightarrow{b} = 2\widehat{i} + \widehat{j} + \widehat{k},\ \overrightarrow{c} = \widehat{i} + 2\widehat{j} - \widehat{k}$
$\displaystyle \text{Now, } \overrightarrow{b} \times \overrightarrow{c} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 2 & 1 & 1 \\ 1 & 2 & -1 \end{vmatrix}$
$\displaystyle = (-1 - 2)\widehat{i} - (-2 - 1)\widehat{j} + (4 - 1)\widehat{k}$
$\displaystyle = -3\widehat{i} + 3\widehat{j} + 3\widehat{k}$
$\displaystyle \text{Now, } \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & -2 & 1 \\ -3 & 3 & 3 \end{vmatrix}$
$\displaystyle = (-6 - 3)\widehat{i} - (3 + 3)\widehat{j} + (3 - 6)\widehat{k}$
$\displaystyle = -9\widehat{i} - 6\widehat{j} - 3\widehat{k}$
$\displaystyle \text{which is the required vector.}$
$\\$

$\displaystyle \textbf{Question 29: } \text{Find a unit vector perpendicular to the vectors}$
$\displaystyle 4\overrightarrow{i}+3\overrightarrow{j}+\overrightarrow{k} \text{ and } 2\overrightarrow{i}-\overrightarrow{j}+2\overrightarrow{k}. \text{ Determine the sine of the}$
$\displaystyle \text{angle between these two vectors.} \quad \text{ISC 2007}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } \overrightarrow{a} = 4\widehat{i} + 3\widehat{j} + \widehat{k},\ \overrightarrow{b} = 2\widehat{i} - \widehat{j} + 2\widehat{k}$
$\displaystyle \text{Now, } \overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 4 & 3 & 1 \\ 2 & -1 & 2 \end{vmatrix}$
$\displaystyle = (6 + 1)\widehat{i} - (8 - 2)\widehat{j} + (-4 - 6)\widehat{k}$
$\displaystyle = 7\widehat{i} - 6\widehat{j} - 10\widehat{k}$
$\displaystyle \text{Now, unit vector perpendicular to } \overrightarrow{a} \text{ and } \overrightarrow{b}$
$\displaystyle \widehat{n} = \frac{\overrightarrow{a} \times \overrightarrow{b}}{|\overrightarrow{a} \times \overrightarrow{b}|} = \frac{7\widehat{i} - 6\widehat{j} - 10\widehat{k}}{\sqrt{7^2 + (-6)^2 + (-10)^2}}$
$\displaystyle = \frac{7\widehat{i} - 6\widehat{j} - 10\widehat{k}}{\sqrt{49 + 36 + 100}} = \frac{7\widehat{i} - 6\widehat{j} - 10\widehat{k}}{\sqrt{185}}$
$\displaystyle \text{Now, } |\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}||\overrightarrow{b}| \sin \theta$
$\displaystyle \sqrt{185} = \sqrt{16+9+1}\sqrt{4+1+4}\sin \theta$
$\displaystyle \sin \theta = \frac{\sqrt{185}}{\sqrt{26}\sqrt{9}} = \frac{\sqrt{185}}{3\sqrt{26}}$
$\\$

$\displaystyle \textbf{Question 30: } \text{The vectors } -2\overrightarrow{i}+4\overrightarrow{j}+4\overrightarrow{k} \text{ and } -4\overrightarrow{i}-2\overrightarrow{k} \text{ represent the}$
$\displaystyle \text{diagonals } BD \text{ and } AC \text{ of a parallelogram } ABCD. \text{ Then,}$
$\displaystyle \text{find the area of the parallelogram.} \quad \text{ISC 2006}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given diagonals of parallelogram are}$
$\displaystyle \overrightarrow{BD} = -2\widehat{i} + 4\widehat{j} + 4\widehat{k}$
$\displaystyle \text{and } \overrightarrow{AC} = -4\widehat{i} - 2\widehat{k}$
$\displaystyle \text{Now, area of parallelogram } = \frac{1}{2}|\overrightarrow{BD} \times \overrightarrow{AC}|$
$\displaystyle \overrightarrow{BD} \times \overrightarrow{AC} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ -2 & 4 & 4 \\ -4 & 0 & -2 \end{vmatrix}$
$\displaystyle = \widehat{i}(-8 - 0) - \widehat{j}(4 - (-16)) + \widehat{k}(0 + 16)$
$\displaystyle = -8\widehat{i} - 20\widehat{j} + 16\widehat{k}$
$\displaystyle |\overrightarrow{BD} \times \overrightarrow{AC}| = \sqrt{(-8)^2 + (-20)^2 + 16^2} = \sqrt{720}$
$\displaystyle \text{Area of parallelogram } = \frac{1}{2}\sqrt{720} = \frac{1}{2} \times 12\sqrt{5} = 6\sqrt{5} \text{ sq units}$
$\\$

$\displaystyle \textbf{Question 31: } \text{If } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are unit vectors such that } 2\overrightarrow{a}-4\overrightarrow{b} \text{ and}$
$\displaystyle 10\overrightarrow{a}+8\overrightarrow{b} \text{ are perpendicular to each other. Then, find}$
$\displaystyle \text{the angle between the vectors } \overrightarrow{a} \text{ and } \overrightarrow{b}. \quad \text{ISC 2005}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } 2\overrightarrow{a} - 4\overrightarrow{b} \text{ and } 10\overrightarrow{a} + 8\overrightarrow{b} \text{ are perpendicular}$
$\displaystyle (2\overrightarrow{a} - 4\overrightarrow{b}) \cdot (10\overrightarrow{a} + 8\overrightarrow{b}) = 0$
$\displaystyle 20(\overrightarrow{a} \cdot \overrightarrow{a}) + 16(\overrightarrow{a} \cdot \overrightarrow{b}) - 40(\overrightarrow{b} \cdot \overrightarrow{a}) - 32(\overrightarrow{b} \cdot \overrightarrow{b}) = 0$
$\displaystyle 20|\overrightarrow{a}|^2 - 24(\overrightarrow{a} \cdot \overrightarrow{b}) - 32|\overrightarrow{b}|^2 = 0$
$\displaystyle 20 - 24(\overrightarrow{a} \cdot \overrightarrow{b}) - 32 = 0 \quad [\because\ |\overrightarrow{a}| = |\overrightarrow{b}| = 1]$
$\displaystyle -24(\overrightarrow{a} \cdot \overrightarrow{b}) - 12 = 0$
$\displaystyle 24|\overrightarrow{a}||\overrightarrow{b}|\cos \theta = -12$
$\displaystyle 24\cos \theta = -12$
$\displaystyle \cos \theta = -\frac{1}{2}$
$\displaystyle \theta = \frac{2\pi}{3}$
$\\$

$\displaystyle \textbf{Question 32: } \text{Show that the sum of the squares of the sides of a}$
$\displaystyle \text{parallelogram is equal to the sum of the squares of the}$
$\displaystyle \text{diagonals of the parallelogram.} \quad \text{ISC 2004, 01}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } ABCD \text{ be a parallelogram}$ $\displaystyle \text{Using triangle law property, } \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} \quad \text{(i)}$
$\displaystyle \overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} \quad \text{(ii)}$
$\displaystyle \text{On squaring and adding, } (AC)^2 + (BD)^2 = (AB + BC)^2 + (BA + AD)^2$
$\displaystyle = AB^2 + BC^2 + 2|AB||BC|\cos(\angle ABC) + BA^2 + AD^2 + 2|BA||AD|\cos(\angle BAD)$
$\displaystyle = AB^2 + BC^2 + CD^2 + AD^2 + 2|AB||BC|[\cos(\angle ABC) - \cos(\angle ABC)]$
$\displaystyle = AB^2 + BC^2 + CD^2 + AD^2$
$\displaystyle \therefore AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + AD^2$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 33: } \text{Find the unit vector perpendicular to the two vectors}$
$\displaystyle \overrightarrow{i}+2\overrightarrow{j}-\overrightarrow{k} \text{ and } 2\overrightarrow{i}+3\overrightarrow{j}+\overrightarrow{k}. \quad \text{ISC 2004}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{A}=\widehat{i}+2\widehat{j}-\widehat{k},\ \overrightarrow{B}=2\widehat{i}+3\widehat{j}+\widehat{k}$
$\displaystyle \text{Unit vector perpendicular to both is } \widehat{n}=\frac{\overrightarrow{A}\times\overrightarrow{B}}{|\overrightarrow{A}\times\overrightarrow{B}|}$
$\displaystyle \overrightarrow{A}\times\overrightarrow{B}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&-1\\2&3&1\end{vmatrix}$
$\displaystyle =\widehat{i}(2\cdot1-(-1)\cdot3)-\widehat{j}(1\cdot1-(-1)\cdot2)+\widehat{k}(1\cdot3-2\cdot2)$
$\displaystyle =5\widehat{i}-3\widehat{j}-\widehat{k}$
$\displaystyle |\overrightarrow{A}\times\overrightarrow{B}|=\sqrt{25+9+1}=\sqrt{35}$
$\displaystyle \therefore \widehat{n}=\frac{5\widehat{i}-3\widehat{j}-\widehat{k}}{\sqrt{35}}$
$\\$

$\displaystyle \textbf{Question 34: } \text{If } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are two vectors such that } |\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|,$
$\displaystyle \text{then show that vector } (2\overrightarrow{a}+\overrightarrow{b}) \text{ is perpendicular to } \overrightarrow{b}. \quad \text{ISC 2004}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|$
$\displaystyle \text{Suppose }(2\overrightarrow{a}+\overrightarrow{b})\cdot\overrightarrow{b}\neq 0$
$\displaystyle \Rightarrow 2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^{2}\neq 0$
$\displaystyle \text{Adding }|\overrightarrow{a}|^{2}\text{ to both sides,}$
$\displaystyle |\overrightarrow{a}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^{2}\neq |\overrightarrow{a}|^{2}$
$\displaystyle \Rightarrow |\overrightarrow{a}+\overrightarrow{b}|^{2}\neq |\overrightarrow{a}|^{2}$
$\displaystyle \Rightarrow |\overrightarrow{a}+\overrightarrow{b}|\neq |\overrightarrow{a}|\quad \text{(contradiction)}$
$\displaystyle \therefore (2\overrightarrow{a}+\overrightarrow{b})\cdot\overrightarrow{b}=0$
$\displaystyle \Rightarrow (2\overrightarrow{a}+\overrightarrow{b})\perp \overrightarrow{b}$
$\\$

$\displaystyle \textbf{Question 35: } \text{Using vectors, show that the perpendiculars from the}$
$\displaystyle \text{vertices to the opposite sides of the } \triangle ABC \text{ are concurrent.} \quad \text{ISC 2003}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{h}\text{ be position vectors of }A,B,C,H$ $\displaystyle \text{Since }AH\perp BC$
$\displaystyle \Rightarrow (\overrightarrow{h}-\overrightarrow{a})\cdot(\overrightarrow{c}-\overrightarrow{b})=0$
$\displaystyle \Rightarrow \overrightarrow{h}\cdot(\overrightarrow{c}-\overrightarrow{b})-\overrightarrow{a}\cdot(\overrightarrow{c}-\overrightarrow{b})=0$
$\displaystyle \Rightarrow \overrightarrow{h}\cdot\overrightarrow{c}-\overrightarrow{h}\cdot\overrightarrow{b}-\overrightarrow{a}\cdot\overrightarrow{c}+\overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \text{Again, since }BH\perp CA$
$\displaystyle \therefore (\overrightarrow{h}-\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{c})=0$
$\displaystyle \Rightarrow \overrightarrow{h}\cdot(\overrightarrow{a}-\overrightarrow{c})-\overrightarrow{b}\cdot(\overrightarrow{a}-\overrightarrow{c})=0$
$\displaystyle \Rightarrow \overrightarrow{h}\cdot\overrightarrow{a}-\overrightarrow{h}\cdot\overrightarrow{c}-\overrightarrow{b}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{c}=0 \quad ...(ii)$
$\displaystyle \text{On adding Eqs. (i) and (ii), we get}$
$\displaystyle \overrightarrow{h}\cdot\overrightarrow{a}-\overrightarrow{h}\cdot\overrightarrow{b}-\overrightarrow{a}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{b}+\overrightarrow{h}\cdot\overrightarrow{a}-\overrightarrow{h}\cdot\overrightarrow{c}-\overrightarrow{b}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{c}=0$
$\displaystyle \Rightarrow (\overrightarrow{h}-\overrightarrow{c})\cdot(\overrightarrow{a}-\overrightarrow{b})=0$
$\displaystyle \therefore CH\perp BA$
$\displaystyle \text{This proves that perpendiculars drawn from the vertices of a triangle to opposite} \\ \text{sides are concurrent.}$
$\\$

$\displaystyle \textbf{Question 36: } \text{The position vectors of the vertices of a triangle are given as }$
$\displaystyle 2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}, \overrightarrow{i}-3\overrightarrow{j}-5\overrightarrow{k} \text{ and } 3\overrightarrow{i}-4\overrightarrow{j}-4\overrightarrow{k}.$
$\displaystyle \text{Prove that it is a right angled triangle.} \quad \text{ISC 2003, 1999}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ be the position vectors of }A,B,C\text{ respectively}$
$\displaystyle \text{Let }\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{b}=\widehat{i}-3\widehat{j}-5\widehat{k}$
$\displaystyle \text{and }\overrightarrow{c}=3\widehat{i}-4\widehat{j}-4\widehat{k}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}=(\widehat{i}-3\widehat{j}-5\widehat{k})-(2\widehat{i}-\widehat{j}+\widehat{k})$
$\displaystyle =-\widehat{i}-2\widehat{j}-6\widehat{k}$
$\displaystyle |\overrightarrow{AB}|=\sqrt{(-1)^{2}+(-2)^{2}+(-6)^{2}}$
$\displaystyle =\sqrt{1+4+36}=\sqrt{41}$
$\displaystyle \overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b}=(3\widehat{i}-4\widehat{j}-4\widehat{k})-(\widehat{i}-3\widehat{j}-5\widehat{k})$
$\displaystyle =2\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle |\overrightarrow{BC}|=\sqrt{2^{2}+(-1)^{2}+1^{2}}$
$\displaystyle =\sqrt{4+1+1}=\sqrt{6}$
$\displaystyle \overrightarrow{CA}=\overrightarrow{a}-\overrightarrow{c}=(2\widehat{i}-\widehat{j}+\widehat{k})-(3\widehat{i}-4\widehat{j}-4\widehat{k})$
$\displaystyle =-\widehat{i}+3\widehat{j}+5\widehat{k}$
$\displaystyle |\overrightarrow{CA}|=\sqrt{(-1)^{2}+3^{2}+5^{2}}$
$\displaystyle =\sqrt{1+9+25}=\sqrt{35}$
$\displaystyle \text{Now, }|\overrightarrow{AB}|^{2}=41$
$\displaystyle \text{and }|\overrightarrow{BC}|^{2}+|\overrightarrow{CA}|^{2}=6+35=41$
$\displaystyle \therefore |\overrightarrow{AB}|^{2}=|\overrightarrow{BC}|^{2}+|\overrightarrow{CA}|^{2}$
$\displaystyle \text{Hence, }ABC\text{ is a right-angled triangle.}$
$\\$

$\displaystyle \textbf{Question 37: } \text{Using vectors, show that the medians of a triangle meet at a point.}$
$\displaystyle \quad \text{ISC 2002}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }ABC\text{ be a triangle and }D,E,F\text{ be mid-points of }BC,AC,AB\text{ respectively}$ $\displaystyle \text{Let }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ be position vectors of }A,B,C\text{ respectively}$
$\displaystyle \text{Then, position vectors of }D,E,F\text{ are }\frac{\overrightarrow{b}+\overrightarrow{c}}{2},\frac{\overrightarrow{a}+\overrightarrow{c}}{2},\frac{\overrightarrow{a}+\overrightarrow{b}}{2}\text{ respectively}$
$\displaystyle \text{Position vector of point }G\text{ dividing }AD\text{ in ratio }2:1\text{ is}$
$\displaystyle \frac{1\cdot\overrightarrow{a}+2\cdot\frac{\overrightarrow{b}+\overrightarrow{c}}{2}}{1+2}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}$
$\displaystyle \text{Similarly, position vectors of point }G\text{ dividing }BE\text{ and }CF\text{ in ratio }2:1$
$\displaystyle \text{are each equal to }\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}$
$\displaystyle \text{Thus, medians of a triangle are concurrent and centroid has position vector }\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}$
$\\$

$\displaystyle \textbf{Question 38: } \text{The vectors } \overrightarrow{a}=3\overrightarrow{i}+x\overrightarrow{j}-\overrightarrow{k} \text{ and } \overrightarrow{b}=2\overrightarrow{i}+\overrightarrow{j}+y\overrightarrow{k} \text{ are}$
$\displaystyle \text{mutually perpendicular. Given that } |\overrightarrow{a}|=|\overrightarrow{b}|. \text{ Then, find}$
$\displaystyle \text{the values of } x \text{ and } y. \quad \text{ISC 2002}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \overrightarrow{a} = 3\widehat{i} + x\widehat{j} - \widehat{k},\ \overrightarrow{b} = 2\widehat{i} + \widehat{j} + y\widehat{k}$
$\displaystyle |\overrightarrow{a}| = \sqrt{9 + x^2 + 1}$
$\displaystyle |\overrightarrow{b}| = \sqrt{4 + 1 + y^2}$
$\displaystyle \text{Given } |\overrightarrow{a}| = |\overrightarrow{b}|$
$\displaystyle 9 + x^2 + 1 = 4 + 1 + y^2$
$\displaystyle x^2 - y^2 = -5$
$\displaystyle x^2 - y^2 = 5 - 10$
$\displaystyle x^2 - y^2 = -5 \quad \text{...(i)}$
$\displaystyle \text{Also, it is given that } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are mutually perpendicular.}$
$\displaystyle \overrightarrow{a} \cdot \overrightarrow{b} = 0$
$\displaystyle (3\widehat{i} + x\widehat{j} - \widehat{k}) \cdot (2\widehat{i} + \widehat{j} + y\widehat{k}) = 0$
$\displaystyle 6 + x - y = 0$
$\displaystyle y - x = 6 \quad \text{...(ii)}$
$\displaystyle \text{On putting the value of } y \text{ from Eq. (ii) in Eq. (i), we get}$
$\displaystyle x^2 - (6 + x)^2 = -5$
$\displaystyle x^2 - 36 - x^2 - 12x = -5$
$\displaystyle -36 - 12x = -5$
$\displaystyle 12x = -31$
$\displaystyle x = -\frac{31}{12}$
$\displaystyle \text{Put } x = -\frac{31}{12} \text{ in Eq. (ii), we get}$
$\displaystyle y - \left(-\frac{31}{12}\right) = 6$
$\displaystyle y + \frac{31}{12} = 6$
$\displaystyle y = 6 - \frac{31}{12}$
$\displaystyle y = \frac{72 - 31}{12}$
$\displaystyle y = \frac{41}{12}$
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$\displaystyle \textbf{Question 39: } \text{If } \overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \text{ and } \overrightarrow{d} \text{ are the position vectors of four points}$
$\displaystyle A,B,C \text{ and } D \text{ respectively and if } \overrightarrow{b}-\overrightarrow{a}=\overrightarrow{c}-\overrightarrow{d}, \text{ show that } ABCD \\ \text{is a parallelogram.} \quad \text{ISC 2000}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d}\text{ are position vectors of four points }A,B,C,D\text{ respectively}$ $\displaystyle \text{Now, }\overrightarrow{b}-\overrightarrow{a}=\overrightarrow{AB}\text{ and }\overrightarrow{c}-\overrightarrow{d}=\overrightarrow{DC}$
$\displaystyle \text{Also, given }\overrightarrow{b}-\overrightarrow{a}=\overrightarrow{c}-\overrightarrow{d}$
$\displaystyle \therefore \overrightarrow{AB}=\overrightarrow{DC}$
$\displaystyle \Rightarrow |\overrightarrow{AB}|=|\overrightarrow{DC}|$
$\displaystyle \text{and }\overrightarrow{AB}\parallel\overrightarrow{DC}$
$\displaystyle \text{Hence, }ABCD\text{ is a parallelogram.}$
$\\$

$\displaystyle \textbf{Question 40: } \text{Prove by the vector method that the middle point of}$
$\displaystyle \text{the hypotenuse of a right angled triangle is at}$
$\displaystyle \text{equidistant from the vertices of the triangle.} \quad \text{ISC 2000}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } ABC \text{ be a right angled triangle. Taking } B \text{ as origin. Let the position vectors of } \\ A \text{ and } C \text{ be respectively } \overrightarrow{a} \text{ and } \overrightarrow{b}.$ $\displaystyle \overrightarrow{BA} = \overrightarrow{a} \text{ and } \overrightarrow{BC} = \overrightarrow{b}$
$\displaystyle \text{Let } D \text{ be the mid-point of hypotenuse } AC.$
$\displaystyle |\overrightarrow{DA}| = |\overrightarrow{DC}|$
$\displaystyle \text{Now, } \overrightarrow{BA} = \overrightarrow{BD} + \overrightarrow{DA} \text{ and } \overrightarrow{BC} = \overrightarrow{BD} + \overrightarrow{DC}$
$\displaystyle \therefore \overrightarrow{BA} \cdot \overrightarrow{BC} = (\overrightarrow{BD} + \overrightarrow{DA}) \cdot (\overrightarrow{BD} + \overrightarrow{DC})$
$\displaystyle = \overrightarrow{BD} \cdot \overrightarrow{BD} + \overrightarrow{BD} \cdot \overrightarrow{DC} + \overrightarrow{DA} \cdot \overrightarrow{BD} + \overrightarrow{DA} \cdot \overrightarrow{DC}$
$\displaystyle \text{But } \overrightarrow{DA} = -\overrightarrow{DC}$
$\displaystyle \text{[since, } \overrightarrow{DA} \text{ and } \overrightarrow{DC} \text{ are equal in magnitude but opposite in direction]}$
$\displaystyle \therefore \overrightarrow{BA} \cdot \overrightarrow{BC} = |\overrightarrow{BD}|^2 + \overrightarrow{BD} \cdot \overrightarrow{DC} - \overrightarrow{BD} \cdot \overrightarrow{DC} - |\overrightarrow{DA}|^2$
$\displaystyle = BD^2 - DA^2$
$\displaystyle \text{Also, } \overrightarrow{BA} \text{ and } \overrightarrow{BC} \text{ are perpendicular to each other.}$
$\displaystyle \therefore \overrightarrow{BA} \cdot \overrightarrow{BC} = 0$
$\displaystyle \Rightarrow BD^2 - DA^2 = 0$
$\displaystyle \Rightarrow BD = DA$
$\displaystyle \text{Also, } DA = DC \quad \text{[taking square root]}$
$\displaystyle \therefore BD = DA = DC$
$\displaystyle \text{Hence, } D \text{ is equidistant from } A, B \text{ and } C.$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 41: } \text{The given figure shows an air plant holder which is in the shape of }$
$\displaystyle \text{tetrahedron. Let } A(1,1,1), B(2,1,3), C(3,2,2) \text{ and } D(3,3,4) \text{are the vertices of }$
$\displaystyle \text{air plant holder. Based on the above } \text{information, answer the following questions.}$ $\displaystyle \text{(i) The vector } \overrightarrow{AB} \text{ is}$
$\displaystyle \text{(a) } -\overrightarrow{i}-2\overrightarrow{k} \quad \text{(b) } 2\overrightarrow{i}+\overrightarrow{k} \quad \text{(c) } \overrightarrow{i}+2\overrightarrow{k} \quad \text{(d) } -2\overrightarrow{i}-\overrightarrow{k}$
$\displaystyle \text{(ii) The vector } \overrightarrow{AC} \text{ is}$
$\displaystyle \text{(a) } 2\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k} \quad \text{(b) } 2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k} \quad \text{(c) } -2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k} \quad \text{(d) } \overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k}$
$\displaystyle \text{(iii) Area of } \triangle ABC \text{ is}$
$\displaystyle \text{(a) } \frac{\sqrt{11}}{2} \text{ sq units} \quad \text{(b) } \frac{\sqrt{14}}{2} \text{ sq units} \quad \text{(c) } \frac{\sqrt{13}}{2} \text{ sq units} \quad \text{(d) } \frac{\sqrt{17}}{2} \text{ sq units}$
$\displaystyle \text{(iv) The unit vector along } \overrightarrow{AB} \text{ is}$
$\displaystyle \text{(a) } \frac{-2\overrightarrow{i}-\overrightarrow{k}}{\sqrt{5}} \quad \text{(b) } \frac{-\overrightarrow{i}-2\overrightarrow{k}}{\sqrt{5}} \quad \text{(c) } \frac{2\overrightarrow{i}+\overrightarrow{k}}{\sqrt{5}} \quad \text{(d) } \frac{\overrightarrow{i}+2\overrightarrow{k}}{\sqrt{5}} \quad \text{ISC 2000}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) (c) }\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$\displaystyle =(2\widehat{i}+\widehat{j}+3\widehat{k})-(\widehat{i}+\widehat{j}+\widehat{k})=\widehat{i}+2\widehat{k}$
$\displaystyle \text{(ii) (b) }\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}$
$\displaystyle =(3\widehat{i}+2\widehat{j}+2\widehat{k})-(\widehat{i}+\widehat{j}+\widehat{k})=2\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{(iii) (b) Area of }\triangle ABC=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|$
$\displaystyle =\frac{1}{2}\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&0&2\\2&1&1\end{vmatrix}$
$\displaystyle =\frac{1}{2}(\widehat{i}(0-2)-\widehat{j}(1-4)+\widehat{k}(1-0))$
$\displaystyle =\frac{1}{2}(-2\widehat{i}+3\widehat{j}+\widehat{k})$
$\displaystyle \therefore \frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|=\frac{1}{2}\sqrt{4+9+1}=\frac{\sqrt{14}}{2}\text{ sq units}$
$\displaystyle \text{(iv) (d) Unit vector along }\overrightarrow{AB}=\frac{\overrightarrow{AB}}{|\overrightarrow{AB}|}$
$\displaystyle =\frac{\widehat{i}+2\widehat{k}}{\sqrt{1^{2}+2^{2}}}=\frac{\widehat{i}+2\widehat{k}}{\sqrt{5}}$
$\\$