Class 12: Vectors (Test Problems)

$\displaystyle \textbf{Question 1. }\text{If }m_1,m_2,m_3\text{ and }m_4\text{ are respectively the magnitudes}$
$\displaystyle \text{of the vectors }\overrightarrow{a_1}=2\widehat{i}-\widehat{j}+\widehat{k},\ \overrightarrow{a_2}=3\widehat{i}-4\widehat{j}-4\widehat{k}, \overrightarrow{a_3}=\widehat{i}+\widehat{j}-\widehat{k}\text{ and }$
$\displaystyle \overrightarrow{a_4}=-\widehat{i}+3\widehat{j}+\widehat{k},\text{ then the correct } \text{order of }m_1,m_2,m_3\text{ and }m_4\text{ is}$
$\displaystyle \text{(a) }m_3<m_1<m_4<m_2$
$\displaystyle \text{(b) }m_3<m_1<m_2<m_4$
$\displaystyle \text{(c) }m_3<m_4<m_1<m_2$
$\displaystyle \text{(d) }m_3<m_4<m_2<m_1$
$\displaystyle \text{Answer:}$
$\displaystyle m_1=\left|\overrightarrow{a_1}\right|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}$
$\displaystyle m_2=\left|\overrightarrow{a_2}\right|=\sqrt{3^2+(-4)^2+(-4)^2}=\sqrt{41}$
$\displaystyle m_3=\left|\overrightarrow{a_3}\right|=\sqrt{1^2+1^2+(-1)^2}=\sqrt{3}$
$\displaystyle m_4=\left|\overrightarrow{a_4}\right|=\sqrt{(-1)^2+3^2+1^2}=\sqrt{11}$
$\displaystyle \text{Since }\sqrt{3}<\sqrt{6}<\sqrt{11}<\sqrt{41}$
$\displaystyle \therefore m_3<m_1<m_4<m_2$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 2. }\text{If }\overrightarrow{a}\text{ is a non-zero vector of magnitude }a\text{ and }\lambda\text{ is a non-zero scalar, then }$
$\displaystyle \lambda\overrightarrow{a}\text{ is a unit vector if}$
$\displaystyle \text{(a) }\lambda=1\qquad\text{(b) }\lambda=-1\qquad\text{(c) }a=|\lambda|\qquad\text{(d) }a=\frac{1}{|\lambda|}$
$\displaystyle \text{Answer:}$
$\displaystyle |\lambda\overrightarrow{a}|=|\lambda|\,|\overrightarrow{a}|$
$\displaystyle =|\lambda|\,a$
$\displaystyle \text{Since }\lambda\overrightarrow{a}\text{ is a unit vector,}$
$\displaystyle |\lambda|\,a=1$
$\displaystyle \therefore a=\frac{1}{|\lambda|}$
$\displaystyle \therefore \text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 3. }\text{If }|\overrightarrow{a}|=4\text{ and }-3\leq\lambda\leq2,\text{ then the range of }|\lambda\overrightarrow{a}|\text{ is}$
$\displaystyle \text{(a) }[0,8]\qquad\text{(b) }[-12,8]\qquad\text{(c) }[0,12]\qquad\text{(d) }[8,12]$
$\displaystyle \text{Answer:}$
$\displaystyle |\lambda\overrightarrow{a}|=|\lambda|\,|\overrightarrow{a}|$
$\displaystyle =4|\lambda|$
$\displaystyle \text{Given }-3\leq\lambda\leq2$
$\displaystyle \therefore 0\leq|\lambda|\leq3$
$\displaystyle \therefore 0\leq4|\lambda|\leq12$
$\displaystyle \therefore |\lambda\overrightarrow{a}|\in[0,12]$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 4. }\text{The position vector of the point which divides the join of points }$
$\displaystyle 2\overrightarrow{a}-3\overrightarrow{b}\text{ and } \overrightarrow{a}+\overrightarrow{b}\text{ in the ratio }3:1,\text{ is}$
$\displaystyle \text{(a) }\frac{3\overrightarrow{a}-2\overrightarrow{b}}{2}\qquad\text{(b) }\frac{7\overrightarrow{a}-8\overrightarrow{b}}{4}$
$\displaystyle \text{(c) }\frac{3\overrightarrow{a}}{4}\qquad\text{(d) }\frac{5\overrightarrow{a}}{4}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P=2\overrightarrow{a}-3\overrightarrow{b}\text{ and }Q=\overrightarrow{a}+\overrightarrow{b}$
$\displaystyle \text{If }R\text{ divides }PQ\text{ internally in the ratio }3:1,$
$\displaystyle \overrightarrow{OR}=\frac{3\overrightarrow{Q}+1\overrightarrow{P}}{3+1}$
$\displaystyle =\frac{3(\overrightarrow{a}+\overrightarrow{b})+(2\overrightarrow{a}-3\overrightarrow{b})}{4}$
$\displaystyle =\frac{5\overrightarrow{a}}{4}$
$\displaystyle \therefore \text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 5. }\text{The projection vector of }\overrightarrow{a}\text{ on }\overrightarrow{b}\text{ is}$
$\displaystyle \text{(a) }\left(\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|^{2}}\right)\overrightarrow{b}\qquad\text{(b) }\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}$
$\displaystyle \text{(c) }\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|}\qquad\text{(d) }\left(\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|^{2}}\right)\overrightarrow{b}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Projection vector of }\overrightarrow{a}\text{ on }\overrightarrow{b}$
$\displaystyle =\left(\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|^{2}}\right)\overrightarrow{b}$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 6. }\text{If }\overrightarrow{a}\cdot\overrightarrow{b}=\frac{1}{2}|\overrightarrow{a}|\ |\overrightarrow{b}|,\text{ then the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is}$
$\displaystyle \text{(a) }0^\circ \qquad \text{(b) }30^\circ \qquad \text{(c) }60^\circ \qquad \text{(d) }90^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos\theta$
$\displaystyle \therefore |\overrightarrow{a}||\overrightarrow{b}|\cos\theta=\frac{1}{2}|\overrightarrow{a}||\overrightarrow{b}|$
$\displaystyle \therefore \cos\theta=\frac{1}{2}$
$\displaystyle \therefore \theta=60^\circ$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 7. }\text{If }\lambda(3\widehat{i}+2\widehat{j}-6\widehat{k})\text{ is a unit vector, then the value of }\lambda\text{ is}$
$\displaystyle \text{(a) }\pm\frac{1}{7} \qquad \text{(b) }\pm7 \qquad \text{(c) }\pm\sqrt{43} \qquad \text{(d) }\pm\frac{1}{\sqrt{49}}$
$\displaystyle \text{Answer:}$
$\displaystyle \left|\lambda (3\widehat{i}+2\widehat{j}-6\widehat{k})\right|=1$
$\displaystyle |\lambda|\sqrt{3^2+2^2+(-6)^2}=1$
$\displaystyle |\lambda|\sqrt{49}=1$
$\displaystyle 7|\lambda|=1$
$\displaystyle |\lambda|=\frac{1}{7}$
$\displaystyle \therefore \lambda=\pm\frac{1}{7}$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 8. }\text{The figure formed by four points }\widehat{i}+\widehat{j}+\widehat{k}, 2\widehat{i}+3\widehat{j},\$
$\displaystyle 3\widehat{i}+5\widehat{j}-2\widehat{k}\text{ and }\widehat{k}-\widehat{j}\text{ is a}$
$\displaystyle \text{(a) parallelogram}\qquad \text{(b) rectangle}$
$\displaystyle \text{(c) trapezium}\qquad \text{(d) square}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=(1,1,1),\ B=(2,3,0),\ C=(3,5,-2),\ D=(0,-1,1)$
$\displaystyle \overrightarrow{AB}=B-A=(1,2,-1)$
$\displaystyle \overrightarrow{CD}=D-C=(-3,-6,3)=-3(1,2,-1)$
$\displaystyle \therefore \overrightarrow{AB}\parallel \overrightarrow{CD}$
$\displaystyle \overrightarrow{BC}=C-B=(1,2,-2)$
$\displaystyle \overrightarrow{AD}=D-A=(-1,-2,0)$
$\displaystyle \therefore \overrightarrow{BC}\text{ is not parallel to }\overrightarrow{AD}$
$\displaystyle \text{Since only one pair of opposite sides is parallel, the figure is a trapezium.}$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 9. }\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are two unit vectors inclined at an angle}$
$\displaystyle \frac{\pi}{3},\text{ then the value of }|\overrightarrow{a}+\overrightarrow{b}|\text{ is}$
$\displaystyle \text{(a) equal to }1\qquad \text{(b) greater than }1$
$\displaystyle \text{(c) equal to }0\qquad \text{(d) less than }1$
$\displaystyle \text{Answer:}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{a}||\overrightarrow{b}|\cos\frac{\pi}{3}$
$\displaystyle =1^2+1^2+2(1)(1)\cdot\frac{1}{2}$
$\displaystyle =3$
$\displaystyle \therefore |\overrightarrow{a}+\overrightarrow{b}|=\sqrt{3}$
$\displaystyle \therefore |\overrightarrow{a}+\overrightarrow{b}|>1$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 10. }\text{Let }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\ \overrightarrow{b}=4\widehat{i}-2\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{c}=\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \text{and find a vector of magnitude }6\text{ units which is parallel to the vector } \\ 2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}.$
$\displaystyle \text{Answer:}$
$\displaystyle 2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}$
$\displaystyle =2(\widehat{i}+\widehat{j}+\widehat{k})-(4\widehat{i}-2\widehat{j}+3\widehat{k})+3(\widehat{i}-2\widehat{j}+\widehat{k})$
$\displaystyle =(2\widehat{i}+2\widehat{j}+2\widehat{k})+(-4\widehat{i}+2\widehat{j}-3\widehat{k})+(3\widehat{i}-6\widehat{j}+3\widehat{k})$
$\displaystyle =\widehat{i}-2\widehat{j}+2\widehat{k}$
$\displaystyle \text{Magnitude of }\widehat{i}-2\widehat{j}+2\widehat{k}$
$\displaystyle =\sqrt{1^{2}+(-2)^{2}+2^{2}}=\sqrt{9}=3$
$\displaystyle \text{Unit vector parallel to it }=\frac{1}{3}(\widehat{i}-2\widehat{j}+2\widehat{k})$
$\displaystyle \text{Required vector of magnitude }6$
$\displaystyle =6\cdot\frac{1}{3}(\widehat{i}-2\widehat{j}+2\widehat{k})$
$\displaystyle =2\widehat{i}-4\widehat{j}+4\widehat{k}$
$\\$

$\displaystyle \textbf{Question 11. }\text{A vector }\overrightarrow{r}\text{ is inclined at equal angles to the three axes.}$
$\displaystyle \text{If the magnitude of }\overrightarrow{r}\text{ is }2\sqrt{3}\text{ units, then find the value of }\overrightarrow{r}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the direction cosines of }\overrightarrow{r}\text{ be }l,m,n$
$\displaystyle \text{Since }\overrightarrow{r}\text{ is equally inclined to the three axes,}$
$\displaystyle l=m=n$
$\displaystyle \text{Also, }l^{2}+m^{2}+n^{2}=1$
$\displaystyle \therefore 3l^{2}=1$
$\displaystyle \therefore l=m=n=\frac{1}{\sqrt{3}}$
$\displaystyle \therefore \overrightarrow{r}=|\overrightarrow{r}|\left(\frac{\widehat{i}}{\sqrt{3}}+\frac{\widehat{j}}{\sqrt{3}}+\frac{\widehat{k}}{\sqrt{3}}\right)$
$\displaystyle =2\sqrt{3}\left(\frac{\widehat{i}+\widehat{j}+\widehat{k}}{\sqrt{3}}\right)$
$\displaystyle =2(\widehat{i}+\widehat{j}+\widehat{k})$
$\displaystyle \therefore \overrightarrow{r}=2\widehat{i}+2\widehat{j}+2\widehat{k}$
$\\$

$\displaystyle \textbf{Question 12: }\text{If a unit vector }\overrightarrow{a}\text{ makes angle }\frac{\pi}{4}\text{ with }\widehat{i},\ \frac{\pi}{3}\text{ with }\widehat{j}\text{ and an acute}$
$\displaystyle \text{angle }\theta\text{ with }\widehat{k},\text{ then find the components of }\overrightarrow{a}\text{ and the angle }\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the direction cosines of }\overrightarrow{a}\text{ be }l,m,n$
$\displaystyle l=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}},\qquad m=\cos\frac{\pi}{3}=\frac{1}{2}$
$\displaystyle \text{Since }\overrightarrow{a}\text{ is a unit vector,}$
$\displaystyle l^{2}+m^{2}+n^{2}=1$
$\displaystyle \therefore \frac{1}{2}+\frac{1}{4}+n^{2}=1$
$\displaystyle \therefore n^{2}=\frac{1}{4}$
$\displaystyle \therefore n=\frac{1}{2}\text{ (since }\theta\text{ is acute)}$
$\displaystyle \therefore \cos\theta=\frac{1}{2}$
$\displaystyle \therefore \theta=\frac{\pi}{3}$
$\displaystyle \text{Hence, }$
$\displaystyle \overrightarrow{a}=\frac{1}{\sqrt{2}}\widehat{i}+\frac{1}{2}\widehat{j}+\frac{1}{2}\widehat{k}$
$\displaystyle \text{and }\theta=\frac{\pi}{3}$
$\\$

$\displaystyle \textbf{Question 13. }\text{Find the position vector of a point }C\text{ which divides the line segment joining }$
$\displaystyle A\text{ and }B \text{whose position vectors are }2\overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{a}-3\overrightarrow{b},\text{ externally in the ratio }$
$\displaystyle 1:2. \text{ Also, show that }A\text{ is the mid-point of the line segment }BC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{OA}=2\overrightarrow{a}+\overrightarrow{b},\qquad \overrightarrow{OB}=\overrightarrow{a}-3\overrightarrow{b}$
$\displaystyle \text{Since }C\text{ divides }AB\text{ externally in the ratio }1:2,$
$\displaystyle \overrightarrow{OC}=\frac{1\cdot\overrightarrow{OB}-2\cdot\overrightarrow{OA}}{1-2}$
$\displaystyle =\frac{(\overrightarrow{a}-3\overrightarrow{b})-2(2\overrightarrow{a}+\overrightarrow{b})}{-1}$
$\displaystyle =\frac{\overrightarrow{a}-3\overrightarrow{b}-4\overrightarrow{a}-2\overrightarrow{b}}{-1}$
$\displaystyle =3\overrightarrow{a}+5\overrightarrow{b}$
$\displaystyle \therefore \text{Position vector of }C\text{ is }3\overrightarrow{a}+5\overrightarrow{b}$
$\displaystyle \text{Now, midpoint of }BC$
$\displaystyle =\frac{\overrightarrow{OB}+\overrightarrow{OC}}{2}$
$\displaystyle =\frac{(\overrightarrow{a}-3\overrightarrow{b})+(3\overrightarrow{a}+5\overrightarrow{b})}{2}$
$\displaystyle =\frac{4\overrightarrow{a}+2\overrightarrow{b}}{2}$
$\displaystyle =2\overrightarrow{a}+\overrightarrow{b}$
$\displaystyle =\overrightarrow{OA}$
$\displaystyle \therefore A\text{ is the midpoint of }BC$
$\\$

$\displaystyle \textbf{Question 14. }\text{Points }L,M\text{ and }N\text{ divide the sides }BC,\ CA\text{ and }AB\text{ of }\triangle ABC$
$\displaystyle \text{in the ratio }1:4,\ 3:2\text{ and }3:7,\text{ respectively. Prove that } \overrightarrow{AL}+\overrightarrow{BM}+\overrightarrow{CN}$
$\displaystyle \text{ is a vector parallel to }\overrightarrow{CK},\text{ where }K \text{ divides }AB\text{ in the ratio }1:3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the position vectors of }A,B,C\text{ be }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ respectively.}$
$\displaystyle \text{Since }L\text{ divides }BC\text{ in the ratio }1:4,$
$\displaystyle \overrightarrow{OL}=\frac{4\overrightarrow{b}+\overrightarrow{c}}{5}$
$\displaystyle \therefore \overrightarrow{AL}=\overrightarrow{OL}-\overrightarrow{OA}=\frac{4\overrightarrow{b}+\overrightarrow{c}-5\overrightarrow{a}}{5}$
$\displaystyle \text{Since }M\text{ divides }CA\text{ in the ratio }3:2,$
$\displaystyle \overrightarrow{OM}=\frac{2\overrightarrow{c}+3\overrightarrow{a}}{5}$
$\displaystyle \therefore \overrightarrow{BM}=\overrightarrow{OM}-\overrightarrow{OB}=\frac{2\overrightarrow{c}+3\overrightarrow{a}-5\overrightarrow{b}}{5}$
$\displaystyle \text{Since }N\text{ divides }AB\text{ in the ratio }3:7,$
$\displaystyle \overrightarrow{ON}=\frac{7\overrightarrow{a}+3\overrightarrow{b}}{10}$
$\displaystyle \therefore \overrightarrow{CN}=\overrightarrow{ON}-\overrightarrow{OC}=\frac{7\overrightarrow{a}+3\overrightarrow{b}-10\overrightarrow{c}}{10}$
$\displaystyle \overrightarrow{AL}+\overrightarrow{BM}+\overrightarrow{CN}$
$\displaystyle =\frac{4\overrightarrow{b}+\overrightarrow{c}-5\overrightarrow{a}}{5}+\frac{2\overrightarrow{c}+3\overrightarrow{a}-5\overrightarrow{b}}{5}+\frac{7\overrightarrow{a}+3\overrightarrow{b}-10\overrightarrow{c}}{10}$
$\displaystyle =\frac{-4\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}}{10}$
$\displaystyle =\frac{3}{10}\left(\overrightarrow{b}-\frac{4}{3}\overrightarrow{a}+\frac{1}{3}\overrightarrow{c}\right)$
$\displaystyle \text{Now }K\text{ divides }AB\text{ in the ratio }1:3$
$\displaystyle \therefore \overrightarrow{OK}=\frac{3\overrightarrow{a}+\overrightarrow{b}}{4}$
$\displaystyle \therefore \overrightarrow{CK}=\overrightarrow{OK}-\overrightarrow{OC}$
$\displaystyle =\frac{3\overrightarrow{a}+\overrightarrow{b}-4\overrightarrow{c}}{4}$
$\displaystyle \overrightarrow{AL}+\overrightarrow{BM}+\overrightarrow{CN}$
$\displaystyle =\frac{2}{5}\,\overrightarrow{CK}$
$\displaystyle \therefore \overrightarrow{AL}+\overrightarrow{BM}+\overrightarrow{CN}\text{ is parallel to }\overrightarrow{CK}$
$\\$

$\displaystyle \textbf{Question 15. }\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are two unit vectors, then find the angle}$
$\displaystyle \text{between }\overrightarrow{a}\text{ and }\overrightarrow{b},\text{ given that }\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\text{ is a unit vector.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ be }\theta$
$\displaystyle |\overrightarrow{a}|=|\overrightarrow{b}|=1$
$\displaystyle \text{Also, }\left|\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right|=1$
$\displaystyle \therefore \left|\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right|^2=1$
$\displaystyle \therefore 3|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2\sqrt{3}(\overrightarrow{a}\cdot\overrightarrow{b})=1$
$\displaystyle \therefore 3+1-2\sqrt{3}\cos\theta=1$
$\displaystyle \therefore 2\sqrt{3}\cos\theta=3$
$\displaystyle \therefore \cos\theta=\frac{\sqrt{3}}{2}$
$\displaystyle \therefore \theta=30^\circ$
$\\$

$\displaystyle \textbf{Question 16. }\text{If }\overrightarrow{a}=5\widehat{i}-\widehat{j}-3\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}+3\widehat{j}-5\widehat{k},\text{ then show that}$
$\displaystyle \text{the vectors }(\overrightarrow{a}+\overrightarrow{b})\text{ and }(\overrightarrow{a}-\overrightarrow{b})\text{ are perpendicular.}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}=(5+1)\widehat{i}+(-1+3)\widehat{j}+(-3-5)\widehat{k}$
$\displaystyle =6\widehat{i}+2\widehat{j}-8\widehat{k}$
$\displaystyle \overrightarrow{a}-\overrightarrow{b}=(5-1)\widehat{i}+(-1-3)\widehat{j}+(-3+5)\widehat{k}$
$\displaystyle =4\widehat{i}-4\widehat{j}+2\widehat{k}$
$\displaystyle \text{Now, }(\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})$
$\displaystyle =(6)(4)+(2)(-4)+(-8)(2)$
$\displaystyle =24-8-16$
$\displaystyle =0$
$\displaystyle \therefore (\overrightarrow{a}+\overrightarrow{b})\perp(\overrightarrow{a}-\overrightarrow{b})$
$\\$

$\displaystyle \textbf{Question 17. }\text{If }\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{b}=3\widehat{i}+5\widehat{j}-2\widehat{k},\text{ then find } |\overrightarrow{a}\times\overrightarrow{b}|.$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&1&3\\3&5&-2\end{vmatrix}$
$\displaystyle =\widehat{i}(-2-15)-\widehat{j}(-4-9)+\widehat{k}(10-3)$
$\displaystyle =-17\widehat{i}+13\widehat{j}+7\widehat{k}$
$\displaystyle \therefore |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{(-17)^2+13^2+7^2}$
$\displaystyle =\sqrt{289+169+49}=\sqrt{507}=13\sqrt{3}$
$\\$

$\displaystyle \textbf{Question 18. }\text{If vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are such that }|\overrightarrow{a}|=3,\ |\overrightarrow{b}|=\frac{2}{3}$
$\displaystyle \text{and }\overrightarrow{a}\times\overrightarrow{b}\text{ is a unit vector, then find the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}.$
$\displaystyle \text{Answer:}$
$\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta$
$\displaystyle \text{Since }\overrightarrow{a}\times\overrightarrow{b}\text{ is a unit vector, }|\overrightarrow{a}\times\overrightarrow{b}|=1$
$\displaystyle \therefore 1=3\cdot\frac{2}{3}\sin\theta$
$\displaystyle \therefore 1=2\sin\theta$
$\displaystyle \therefore \sin\theta=\frac{1}{2}$
$\displaystyle \therefore \theta=30^\circ$
$\\$

$\displaystyle \textbf{Question 19. }\text{Find the angle between two vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ with}$
$\displaystyle \text{magnitudes }2\text{ and }1\text{ respectively, such that }\overrightarrow{a}\cdot\overrightarrow{b}=\sqrt{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos\theta$
$\displaystyle \therefore \sqrt{3}=2\cdot1\cdot\cos\theta$
$\displaystyle \therefore \cos\theta=\frac{\sqrt{3}}{2}$
$\displaystyle \therefore \theta=30^\circ$
$\displaystyle \therefore \text{The angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }30^\circ.$
$\\$

$\displaystyle \textbf{Question 20. }\text{If a unit vector }\overrightarrow{a}\text{ makes angle }\frac{\pi}{4}\text{ with }\widehat{i},\ \frac{\pi}{3}\text{ with }\widehat{j}\text{ and}$
$\displaystyle \text{an acute angle }\theta\text{ with }\widehat{k},\text{ then find the components of }\overrightarrow{a}\text{ and angle }\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the direction cosines of }\overrightarrow{a}\text{ be }l,m,n$
$\displaystyle l=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}},\qquad m=\cos\frac{\pi}{3}=\frac{1}{2}$
$\displaystyle \text{Since }\overrightarrow{a}\text{ is a unit vector,}$
$\displaystyle l^2+m^2+n^2=1$
$\displaystyle \therefore \frac{1}{2}+\frac{1}{4}+n^2=1$
$\displaystyle \therefore n^2=\frac{1}{4}$
$\displaystyle \therefore n=\pm\frac{1}{2}$
$\displaystyle \text{Since }\theta\text{ is acute, }n=\cos\theta>0$
$\displaystyle \therefore n=\frac{1}{2}$
$\displaystyle \therefore \cos\theta=\frac{1}{2}$
$\displaystyle \therefore \theta=\frac{\pi}{3}$
$\displaystyle \therefore \overrightarrow{a}=\frac{1}{\sqrt{2}}\widehat{i}+\frac{1}{2}\widehat{j}+\frac{1}{2}\widehat{k}$
$\\$

$\displaystyle \textbf{Question 21. }\text{If }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are unit vectors such that } \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0,$
$\displaystyle \text{ then find the value of } \overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}.$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\displaystyle \therefore |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=0$
$\displaystyle \therefore |\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \therefore 1+1+1+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \therefore 3+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \therefore \overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}=-\frac{3}{2}$
$\\$

$\displaystyle \textbf{Question 22. }\text{Find the value of }p\text{ for which two vectors } \overrightarrow{a}=3\widehat{i}+2\widehat{j}+9\widehat{k}\text{ and }$
$\displaystyle \overrightarrow{b}=\widehat{i}+p\widehat{j}+3\widehat{k}\text{ are}$
$\displaystyle \text{(i) perpendicular}\qquad \text{(ii) parallel}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=3\widehat{i}+2\widehat{j}+9\widehat{k},\qquad \overrightarrow{b}=\widehat{i}+p\widehat{j}+3\widehat{k}$
$\displaystyle \text{(i) For perpendicular vectors, }\overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \therefore (3)(1)+(2)(p)+(9)(3)=0$
$\displaystyle \therefore 3+2p+27=0$
$\displaystyle \therefore 2p+30=0$
$\displaystyle \therefore p=-15$
$\displaystyle \text{Hence, }p=-15$
$\displaystyle \text{(ii) For parallel vectors, corresponding components are proportional}$
$\displaystyle \therefore \frac{3}{1}=\frac{2}{p}=\frac{9}{3}$
$\displaystyle \therefore 3=\frac{2}{p}=3$
$\displaystyle \therefore \frac{2}{p}=3$
$\displaystyle \therefore p=\frac{2}{3}$
$\displaystyle \text{Hence, }p=\frac{2}{3}$
$\\$

$\displaystyle \textbf{Question 23. }\text{If }\overrightarrow{a}=7\widehat{i}+\widehat{j}-4\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+6\widehat{j}+3\widehat{k},\text{ then find the}$
$\displaystyle \text{projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=7(2)+1(6)+(-4)(3)$
$\displaystyle =14+6-12=8$
$\displaystyle |\overrightarrow{b}|=\sqrt{2^2+6^2+3^2}$
$\displaystyle =\sqrt{4+36+9}=\sqrt{49}=7$
$\displaystyle \therefore \text{Projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{8}{7}$
$\\$

$\displaystyle \textbf{Question 24. }\text{If the dot products of a vector with vectors }3\widehat{i}-5\widehat{k},$
$\displaystyle 2\widehat{i}+7\widehat{j}\text{ and }\widehat{i}+\widehat{j}+\widehat{k}\text{ are respectively }-1,6\text{ and }5,\text{ then find the vector.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required vector be }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$
$\displaystyle \text{Given, }\overrightarrow{r}\cdot(3\widehat{i}-5\widehat{k})=-1$
$\displaystyle \therefore 3x-5z=-1\qquad (1)$
$\displaystyle \overrightarrow{r}\cdot(2\widehat{i}+7\widehat{j})=6$
$\displaystyle \therefore 2x+7y=6\qquad (2)$
$\displaystyle \overrightarrow{r}\cdot(\widehat{i}+\widehat{j}+\widehat{k})=5$
$\displaystyle \therefore x+y+z=5\qquad (3)$
$\displaystyle \text{From (3), }y=5-x-z$
$\displaystyle \text{Substituting in (2),}$
$\displaystyle 2x+7(5-x-z)=6$
$\displaystyle \therefore 2x+35-7x-7z=6$
$\displaystyle \therefore -5x-7z=-29$
$\displaystyle \therefore 5x+7z=29\qquad (4)$
$\displaystyle \text{Now, solving (1) and (4),}$
$\displaystyle 3x-5z=-1\qquad (1)$
$\displaystyle 5x+7z=29\qquad (4)$
$\displaystyle \text{Multiplying (1) by }5\text{ and (4) by }3,$
$\displaystyle 15x-25z=-5$
$\displaystyle 15x+21z=87$
$\displaystyle \text{Subtracting, }46z=92$
$\displaystyle \therefore z=2$
$\displaystyle \text{Substituting }z=2\text{ in (1),}$
$\displaystyle 3x-5(2)=-1$
$\displaystyle \therefore 3x-10=-1$
$\displaystyle \therefore 3x=9$
$\displaystyle \therefore x=3$
$\displaystyle \text{Substituting }x=3,\ z=2\text{ in (3),}$
$\displaystyle 3+y+2=5$
$\displaystyle \therefore y=0$
$\displaystyle \therefore \overrightarrow{r}=3\widehat{i}+0\widehat{j}+2\widehat{k}$
$\displaystyle \therefore \overrightarrow{r}=3\widehat{i}+2\widehat{k}$
$\\$

$\displaystyle \textbf{Question 25. }\text{If }\overrightarrow{a}=2\widehat{i}+\widehat{j}-3\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}-2\widehat{j}+\widehat{k},\text{ then find a vector}$
$\displaystyle \text{of magnitude }5\text{ perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{A vector perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }\overrightarrow{a}\times\overrightarrow{b}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&1&-3\\1&-2&1\end{vmatrix}$
$\displaystyle =\widehat{i}(1\cdot1-(-3)(-2))-\widehat{j}(2\cdot1-(-3)(1))+\widehat{k}(2\cdot(-2)-1\cdot1)$
$\displaystyle =-5\widehat{i}-5\widehat{j}-5\widehat{k}$
$\displaystyle \therefore |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{(-5)^2+(-5)^2+(-5)^2}=5\sqrt{3}$
$\displaystyle \text{Unit vector in the direction of }\overrightarrow{a}\times\overrightarrow{b}=\frac{-\widehat{i}-\widehat{j}-\widehat{k}}{\sqrt{3}}$
$\displaystyle \therefore \text{Required vector}=5\left(\frac{-\widehat{i}-\widehat{j}-\widehat{k}}{\sqrt{3}}\right)$
$\displaystyle =-\frac{5}{\sqrt{3}}(\widehat{i}+\widehat{j}+\widehat{k})$
$\displaystyle \therefore \text{A vector of magnitude }5\text{ perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is}$
$\displaystyle -\frac{5}{\sqrt{3}}(\widehat{i}+\widehat{j}+\widehat{k})$
$\\$

$\displaystyle \textbf{Question 26. }\text{If }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=\widehat{j}-\widehat{k},\text{ then find a vector }\overrightarrow{c},$
$\displaystyle \text{such that }\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b}\text{ and }\overrightarrow{a}\cdot\overrightarrow{c}=3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{c}=x\widehat{i}+y\widehat{j}+z\widehat{k}$
$\displaystyle \text{Then }(\widehat{i}+\widehat{j}+\widehat{k})\times(x\widehat{i}+y\widehat{j}+z\widehat{k})=\widehat{j}-\widehat{k}$
$\displaystyle \begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&1&1\\x&y&z\end{vmatrix}=\widehat{j}-\widehat{k}$
$\displaystyle (z-y)\widehat{i}-(z-x)\widehat{j}+(y-x)\widehat{k}=\widehat{j}-\widehat{k}$
$\displaystyle \therefore z-y=0$
$\displaystyle \therefore -(z-x)=1$
$\displaystyle \therefore y-x=-1$
$\displaystyle \therefore y=z,\quad x-z=1,\quad y=x-1$
$\displaystyle \text{Hence }y=z=x-1$
$\displaystyle \text{Also, }\overrightarrow{a}\cdot\overrightarrow{c}=3$
$\displaystyle \therefore x+y+z=3$
$\displaystyle \therefore x+(x-1)+(x-1)=3$
$\displaystyle \therefore 3x-2=3$
$\displaystyle \therefore x=\frac{5}{3}$
$\displaystyle \therefore y=z=\frac{2}{3}$
$\displaystyle \therefore \overrightarrow{c}=\frac{5}{3}\widehat{i}+\frac{2}{3}\widehat{j}+\frac{2}{3}\widehat{k}$
$\\$

$\displaystyle \textbf{Question 27. }\text{Using vectors, find the area of triangle with vertices }A(1,1,2), \\ B(2,3,5)\text{ and }C(1,5,5).$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{AB}=(2-1)\widehat{i}+(3-1)\widehat{j}+(5-2)\widehat{k}$
$\displaystyle =\widehat{i}+2\widehat{j}+3\widehat{k}$
$\displaystyle \overrightarrow{AC}=(1-1)\widehat{i}+(5-1)\widehat{j}+(5-2)\widehat{k}$
$\displaystyle =4\widehat{j}+3\widehat{k}$
$\displaystyle \overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&3\\0&4&3\end{vmatrix}$
$\displaystyle =-6\widehat{i}-3\widehat{j}+4\widehat{k}$
$\displaystyle |\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{(-6)^{2}+(-3)^{2}+4^{2}}=\sqrt{61}$
$\displaystyle \therefore \text{Area of triangle }ABC=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|$
$\displaystyle =\frac{\sqrt{61}}{2}$
$\\$

$\displaystyle \textbf{Question 28. }\text{If }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are three vectors, such that}$
$\displaystyle |\overrightarrow{a}|=5,\ |\overrightarrow{b}|=12,\ |\overrightarrow{c}|=13\text{ and }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0},\text{ then find the}$
$\displaystyle \text{value of }\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}.$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$
$\displaystyle \therefore \overrightarrow{c}=-(\overrightarrow{a}+\overrightarrow{b})$
$\displaystyle |\overrightarrow{c}|^{2}=|\overrightarrow{a}+\overrightarrow{b}|^{2}$
$\displaystyle 13^{2}=5^{2}+12^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle 169=25+144+2\overrightarrow{a}\cdot\overrightarrow{b}$
$\displaystyle \therefore \overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \text{Now, }(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})^{2}=0$
$\displaystyle |\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}+|\overrightarrow{c}|^{2}+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle 25+144+169+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle 338+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0$
$\displaystyle \therefore \overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}=-169$
$\\$

$\displaystyle \textbf{Question 29. }\text{The diagonals }AC\text{ and }BD\text{ of a parallelogram }ABCD\text{ are represented by}$
$\displaystyle \text{the vectors } \widehat{i}+\widehat{j}-\widehat{k}\text{ and }\widehat{i}-\widehat{j}+\widehat{k},\text{ respectively. Find the area of the parallelogram.}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{AC}=\widehat{i}+\widehat{j}-\widehat{k},\qquad \overrightarrow{BD}=\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \text{Area of parallelogram }=\frac{1}{2}|\overrightarrow{AC}\times\overrightarrow{BD}|$
$\displaystyle \overrightarrow{AC}\times\overrightarrow{BD}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&1&-1\\1&-1&1\end{vmatrix}$
$\displaystyle =-2\widehat{j}-2\widehat{k}$
$\displaystyle |\overrightarrow{AC}\times\overrightarrow{BD}|=\sqrt{0^{2}+(-2)^{2}+(-2)^{2}}$
$\displaystyle =2\sqrt{2}$
$\displaystyle \therefore \text{Area of parallelogram }=\frac{1}{2}\times2\sqrt{2}$
$\displaystyle =\sqrt{2}\text{ square units}$
$\\$

$\displaystyle \textbf{Question 30. }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are three vectors such that }\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{c},$
$\displaystyle \overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{a}. \ \text{Prove that }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are mutually at right angle and } \\ |\overrightarrow{b}|=1,\ |\overrightarrow{c}|=|\overrightarrow{a}|.$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{c}$
$\displaystyle \therefore \overrightarrow{c}\perp\overrightarrow{a}\text{ and }\overrightarrow{c}\perp\overrightarrow{b}$
$\displaystyle \overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{a}$
$\displaystyle \therefore \overrightarrow{a}\perp\overrightarrow{b}\text{ and }\overrightarrow{a}\perp\overrightarrow{c}$
$\displaystyle \therefore \overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are mutually perpendicular.}$
$\displaystyle \text{Now, }|\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{c}|$
$\displaystyle \therefore |\overrightarrow{a}||\overrightarrow{b}|\sin90^\circ=|\overrightarrow{c}|$
$\displaystyle \therefore |\overrightarrow{a}||\overrightarrow{b}|=|\overrightarrow{c}| \qquad (1)$
$\displaystyle \text{Also, }|\overrightarrow{b}\times\overrightarrow{c}|=|\overrightarrow{a}|$
$\displaystyle \therefore |\overrightarrow{b}||\overrightarrow{c}|\sin90^\circ=|\overrightarrow{a}|$
$\displaystyle \therefore |\overrightarrow{b}||\overrightarrow{c}|=|\overrightarrow{a}| \qquad (2)$
$\displaystyle \text{Dividing (1) by (2),}$
$\displaystyle \frac{|\overrightarrow{a}||\overrightarrow{b}|}{|\overrightarrow{b}||\overrightarrow{c}|}=\frac{|\overrightarrow{c}|}{|\overrightarrow{a}|}$
$\displaystyle \therefore |\overrightarrow{a}|^{2}=|\overrightarrow{c}|^{2}$
$\displaystyle \therefore |\overrightarrow{a}|=|\overrightarrow{c}|$
$\displaystyle \text{Using this in (1),}$
$\displaystyle |\overrightarrow{a}||\overrightarrow{b}|=|\overrightarrow{a}|$
$\displaystyle \therefore |\overrightarrow{b}|=1$
$\displaystyle \therefore |\overrightarrow{b}|=1\text{ and }|\overrightarrow{c}|=|\overrightarrow{a}|$
$\\$

$\displaystyle \textbf{Question 31. }\text{If }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are perpendicular to each other, then prove that } \\ (\overrightarrow{a}\cdot\overrightarrow{b}\times\overrightarrow{c})^{2}=a^{2}b^{2}c^{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }\overrightarrow{b}\perp\overrightarrow{c},$
$\displaystyle |\overrightarrow{b}\times\overrightarrow{c}|=bc\sin90^\circ=bc$
$\displaystyle \text{Also, }\overrightarrow{b}\times\overrightarrow{c}\text{ is perpendicular to both }\overrightarrow{b}\text{ and }\overrightarrow{c}.$
$\displaystyle \text{Since }\overrightarrow{a}\text{ is also perpendicular to }\overrightarrow{b}\text{ and }\overrightarrow{c},$
$\displaystyle \overrightarrow{a}\text{ is parallel to }\overrightarrow{b}\times\overrightarrow{c}$
$\displaystyle \therefore \overrightarrow{a}\cdot(\overrightarrow{b}\times\overrightarrow{c})=a|\overrightarrow{b}\times\overrightarrow{c}|\cos0^\circ$
$\displaystyle =abc$
$\displaystyle \therefore (\overrightarrow{a}\cdot\overrightarrow{b}\times\overrightarrow{c})^{2}=a^{2}b^{2}c^{2}$
$\\$

$\displaystyle \textbf{Question 32. }\text{In any }\triangle ABC,\text{ prove by vector method, }\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle ABC,\text{ let }BC=a,\ CA=b,\ AB=c$
$\displaystyle \overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}$
$\displaystyle \therefore \overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}$
$\displaystyle \therefore |\overrightarrow{BC}|^{2}=|\overrightarrow{AC}-\overrightarrow{AB}|^{2}$
$\displaystyle a^{2}=b^{2}+c^{2}-2|\overrightarrow{AC}||\overrightarrow{AB}|\cos A$
$\displaystyle a^{2}=b^{2}+c^{2}-2bc\cos A$
$\displaystyle \therefore 2bc\cos A=b^{2}+c^{2}-a^{2}$
$\displaystyle \therefore \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}$
$\\$

$\displaystyle \textbf{Question 33. }\text{Find the position vectors of the points which divide}$
$\displaystyle \text{the line joining the two points }3\overrightarrow{a}-2\overrightarrow{b}\text{ and }2\overrightarrow{a}-5\overrightarrow{b}$
$\displaystyle \text{internally and externally in the ratio }3:2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P\text{ and }Q\text{ have position vectors }3\overrightarrow{a}-2\overrightarrow{b}\text{ and }2\overrightarrow{a}-5\overrightarrow{b}\text{ respectively.}$
$\displaystyle \text{If }R\text{ divides }PQ\text{ internally in the ratio }3:2,\text{ then}$
$\displaystyle \overrightarrow{OR}=\frac{3(2\overrightarrow{a}-5\overrightarrow{b})+2(3\overrightarrow{a}-2\overrightarrow{b})}{3+2}$
$\displaystyle =\frac{6\overrightarrow{a}-15\overrightarrow{b}+6\overrightarrow{a}-4\overrightarrow{b}}{5}$
$\displaystyle =\frac{12\overrightarrow{a}-19\overrightarrow{b}}{5}$
$\displaystyle \text{Hence, the position vector of the point dividing internally is}$
$\displaystyle \frac{12\overrightarrow{a}-19\overrightarrow{b}}{5}$
$\displaystyle \text{If }S\text{ divides }PQ\text{ externally in the ratio }3:2,\text{ then}$
$\displaystyle \overrightarrow{OS}=\frac{3(2\overrightarrow{a}-5\overrightarrow{b})-2(3\overrightarrow{a}-2\overrightarrow{b})}{3-2}$
$\displaystyle =6\overrightarrow{a}-15\overrightarrow{b}-6\overrightarrow{a}+4\overrightarrow{b}$
$\displaystyle =-11\overrightarrow{b}$
$\displaystyle \text{Hence, the position vector of the point dividing externally is}$
$\displaystyle -11\overrightarrow{b}$
$\\$

$\displaystyle \textbf{Question 34. }\text{Find the area of the triangle whose vertices are}$
$\displaystyle P(-1,2,-1),\ Q(3,-1,2)\text{ and }R(2,3,-1).$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{PQ}=(3+1)\widehat{i}+(-1-2)\widehat{j}+(2+1)\widehat{k}$
$\displaystyle =4\widehat{i}-3\widehat{j}+3\widehat{k}$
$\displaystyle \overrightarrow{PR}=(2+1)\widehat{i}+(3-2)\widehat{j}+(-1+1)\widehat{k}$
$\displaystyle =3\widehat{i}+\widehat{j}$
$\displaystyle \overrightarrow{PQ}\times\overrightarrow{PR}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\4&-3&3\\3&1&0\end{vmatrix}$
$\displaystyle =-3\widehat{i}+9\widehat{j}+13\widehat{k}$
$\displaystyle \therefore |\overrightarrow{PQ}\times\overrightarrow{PR}|=\sqrt{(-3)^2+9^2+13^2}=\sqrt{259}$
$\displaystyle \therefore \text{Area of triangle}=\frac{1}{2}\sqrt{259}$
$\\$

$\displaystyle \textbf{Question 35. }\text{If }\overrightarrow{a}=2\widehat{i}-3\widehat{j}+\widehat{k},\ \overrightarrow{b}=-\widehat{i}+\widehat{k}\text{ and }\overrightarrow{c}=2\widehat{j}-\widehat{k}\text{ are}$
$\displaystyle \text{three vectors, then find the area of the parallelogram having diagonals } \\ (\overrightarrow{a}+\overrightarrow{b})\text{ and }(\overrightarrow{b}+\overrightarrow{c}).$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}=(2\widehat{i}-3\widehat{j}+\widehat{k})+(-\widehat{i}+\widehat{k})$
$\displaystyle =\widehat{i}-3\widehat{j}+2\widehat{k}$
$\displaystyle \overrightarrow{b}+\overrightarrow{c}=(-\widehat{i}+\widehat{k})+(2\widehat{j}-\widehat{k})$
$\displaystyle =-\widehat{i}+2\widehat{j}$
$\displaystyle \text{Area of parallelogram}=\frac{1}{2}|(\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{b}+\overrightarrow{c})|$
$\displaystyle (\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{b}+\overrightarrow{c})=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-3&2\\-1&2&0\end{vmatrix}$
$\displaystyle =-4\widehat{i}-2\widehat{j}-\widehat{k}$
$\displaystyle \therefore |(\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{b}+\overrightarrow{c})|=\sqrt{16+4+1}=\sqrt{21}$
$\displaystyle \therefore \text{Area of parallelogram}=\frac{\sqrt{21}}{2}$
$\\$