Class 12: Algebra of Vectors – CBSE Board Problems

$\displaystyle \textbf{Question 1:} \text{Find the position vector of a point }R\text{ which divides the line} \\ \text{segment joining } P\text{ and }Q\text{ whose position vectors are }2\overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{a}-3\overrightarrow{b}, \\ \text{externally in the ratio }1:2. \text{ Also, show that }P\text{ is the mid-point of the line} \\ \text{segment }RQ. \hspace{2.2cm} \text{[CBSE 2010]}$

$\displaystyle \text{Answer:}$
$\displaystyle \text{It is given that }R\text{ divides }PQ\text{ externally in the ratio }1:2.$
$\displaystyle \therefore\ \text{The position vector of }R =\frac{1\times(\overrightarrow{a}-3\overrightarrow{b})-2(2\overrightarrow{a}+\overrightarrow{b})}{1-2} =3\overrightarrow{a}+5\overrightarrow{b}$
$\displaystyle \text{Now,}$
$\displaystyle \frac{\text{Position vector of }R+\text{Position vector of }Q}{2} =\frac{3\overrightarrow{a}+5\overrightarrow{b}+\overrightarrow{a}-3\overrightarrow{b}}{2}$
$\displaystyle =2\overrightarrow{a}+\overrightarrow{b} =\text{Position vector of }P$
$\displaystyle \text{Hence, }P\text{ is the mid-point of }PQ.$

$\displaystyle \textbf{Question 2:} \text{The two vectors }\widehat{i}+\widehat{i}\text{ and }3\widehat{i}-\widehat{j}+4\widehat{k}\text{ represent the sides }AB\text{ and }AC \\ \text{respectively of triangle }ABC.\ \text{Find the length of the median through }A. \\ \text{[CBSE 2015, 2016]}$

$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }D\text{ be the mid-point of side }BC\text{ of triangle }ABC.\ \text{Then,}$
$\displaystyle \overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AD}$
$\displaystyle \Rightarrow (\widehat{j}+\widehat{i})+(3\widehat{i}-\widehat{j}+4\widehat{k}) =2\overrightarrow{AD}$
$\displaystyle \Rightarrow \overrightarrow{AD}=2\widehat{i}+0\widehat{j}+2\widehat{k}$
$\displaystyle \Rightarrow \left|\overrightarrow{AD}\right| =\sqrt{4+0+4}=2\sqrt{2}$

$\displaystyle \textbf{Question 3. }\text{The magnitude of the vector }6\widehat{i}-2\widehat{j}+3\widehat{k}\text{ is} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }1 \qquad \text{(b) }5$
$\displaystyle \text{(c) }7 \qquad \text{(d) }12$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) Let }\overrightarrow{a}=6\widehat{i}-2\widehat{j}+3\widehat{k}$
$\displaystyle \text{Magnitude of }\overrightarrow{a}=|\overrightarrow{a}|=\sqrt{(6)^2+(-2)^2+(3)^2}$
$\displaystyle =\sqrt{36+4+9}=\sqrt{49}=7$
$\\$

$\displaystyle \textbf{Question 4. }\text{Two vectors }\overrightarrow{a}=a_{1}\widehat{i}+a_{2}\widehat{j}+a_{3}\widehat{k}\text{ and}$
$\displaystyle \overrightarrow{b}=b_{1}\widehat{i}+b_{2}\widehat{j}+b_{3}\widehat{k}\text{ are collinear, if} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}=0$
$\displaystyle \text{(b) }\frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=\frac{a_{3}}{b_{3}}$
$\displaystyle \text{(c) }a_{1}=b_{1},a_{2}=b_{2},a_{3}=b_{3}$
$\displaystyle \text{(d) }a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) We know that the two vectors }\overrightarrow{a}=a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k}$
$\displaystyle \text{and }\overrightarrow{b}=b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}\text{ are collinear iff }$
$\displaystyle \frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}$
$\\$

$\displaystyle \textbf{Question 5. }\text{If }\overrightarrow{a}+\overrightarrow{b}=\widehat{i}\text{ and }\overrightarrow{a}=2\widehat{i}-2\widehat{j}+2\widehat{k},\text{ then }|\overrightarrow{b}|\text{ equals} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }\sqrt{14} \qquad \text{(b) }3$
$\displaystyle \text{(c) }\sqrt{12} \qquad \text{(d) }\sqrt{17}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given, }\overrightarrow{a}+\overrightarrow{b}=\widehat{i}\text{ and }\overrightarrow{a}=2\widehat{i}-2\widehat{j}+2\widehat{k}$
$\displaystyle \text{Let }\overrightarrow{b}=b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{a}+\overrightarrow{b}=\widehat{i}$
$\displaystyle \Rightarrow (2\widehat{i}-2\widehat{j}+2\widehat{k})+(b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k})=\widehat{i}+0\widehat{j}+0\widehat{k}$
$\displaystyle \text{On comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k},\text{ we get}$
$\displaystyle 2+b_1=1\Rightarrow b_1=-1$
$\displaystyle -2+b_2=0\Rightarrow b_2=2$
$\displaystyle 2+b_3=0\Rightarrow b_3=-2$
$\displaystyle \therefore \overrightarrow{b}=b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}=-\widehat{i}+2\widehat{j}-2\widehat{k}$
$\displaystyle |\overrightarrow{b}|=\sqrt{(-1)^2+(2)^2+(-2)^2}=\sqrt{1+4+4}=\sqrt9=3$
$\\$

$\displaystyle \textbf{Question 6. }\text{A unit vector along the vector }4\widehat{i}-3\widehat{k}\text{ is} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }\frac{1}{7}(4\widehat{i}-3\widehat{k}) \qquad \text{(b) }\frac{1}{5}(4\widehat{i}-3\widehat{k})$
$\displaystyle \text{(c) }\frac{1}{\sqrt{7}}(4\widehat{i}-3\widehat{k}) \qquad \text{(d) }\frac{1}{\sqrt{5}}(4\widehat{i}-3\widehat{k})$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Let }\overrightarrow{a}=4\widehat{i}-3\widehat{k}$
$\displaystyle \text{Now, unit vector along }\overrightarrow{a}\text{ is}$
$\displaystyle \widehat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{4\widehat{i}-3\widehat{k}}{\sqrt{(4)^2+(0)^2+(-3)^2}}$
$\displaystyle =\frac{4\widehat{i}-3\widehat{k}}{\sqrt{16+9}}=\frac{4\widehat{i}-3\widehat{k}}{\sqrt{25}}=\frac{1}{5}(4\widehat{i}-3\widehat{k})$
$\\$

$\displaystyle \textbf{Question 7. }\text{In }\triangle ABC,\ \overrightarrow{AB}=\widehat{i}+\widehat{j}+2\widehat{k}\text{ and }\overrightarrow{AC}=3\widehat{i}-\widehat{j}+4\widehat{k}.\text{ If }D\text{ is}$
$\displaystyle \text{mid-point of }BC,\text{ then vector }\overrightarrow{AD}\text{ is equal to} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }4\widehat{i}+6\widehat{k} \qquad \text{(b) }2\widehat{i}-2\widehat{j}+2\widehat{k}$
$\displaystyle \text{(c) }\widehat{i}-\widehat{j}+\widehat{k} \qquad \text{(d) }2\widehat{i}+3\widehat{k}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) By using the triangle law of vector addition, we have}$
$\displaystyle \overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}$
$\displaystyle =(3\widehat{i}-\widehat{j}+4\widehat{k})-(\widehat{i}+\widehat{j}+2\widehat{k})=2\widehat{i}-2\widehat{j}+2\widehat{k}$
$\displaystyle \therefore \overrightarrow{BD}=\frac{1}{2}\overrightarrow{BC}\qquad [\text{given}]$
$\displaystyle =\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \text{In }\triangle ABD,\text{ by using triangle law of vector addition, we have}$
$\displaystyle \overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=(\widehat{i}+\widehat{j}+2\widehat{k})+(\widehat{i}-\widehat{j}+\widehat{k})=2\widehat{i}+3\widehat{k}$
$\\$

$\displaystyle \textbf{Question 8. }\text{If a vector makes an angle of }\frac{\pi}{4}\text{ with the positive}$
$\displaystyle \text{directions of both }X\text{-axis and }Y\text{-axis, then the angle}$
$\displaystyle \text{which it makes with positive }Z\text{-axis is} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }\frac{\pi}{4} \qquad \text{(b) }\frac{3\pi}{4}$
$\displaystyle \text{(c) }\frac{\pi}{2} \qquad \text{(d) }0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) Let the required angle be }\theta.$
$\displaystyle \therefore l=\cos\frac{\pi}{4},\ m=\cos\frac{\pi}{4}\text{ and }n=\cos\theta$
$\displaystyle \text{We know that }l^2+m^2+n^2=1$
$\displaystyle \Rightarrow \cos^2\frac{\pi}{4}+\cos^2\frac{\pi}{4}+\cos^2\theta=1$
$\displaystyle \Rightarrow \frac{1}{2}+\frac{1}{2}+\cos^2\theta=1\Rightarrow \cos^2\theta=0\Rightarrow \theta=\frac{\pi}{2}$
$\\$

$\displaystyle \textbf{Question 9. }\text{Unit vector along }\overrightarrow{PQ},\text{ where coordinates of }P\text{ and }Q$
$\displaystyle \text{respectively are }(2,1,-1)\text{ and }(4,4,-7)\text{ is} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }2\widehat{i}+3\widehat{j}-6\widehat{k} \qquad \text{(b) }-2\widehat{i}-3\widehat{j}+6\widehat{k}$
$\displaystyle \text{(c) }\frac{2\widehat{i}}{7}-\frac{3\widehat{j}}{7}+\frac{6\widehat{k}}{7} \qquad \text{(d) }\frac{2\widehat{i}}{7}+\frac{3\widehat{j}}{7}-\frac{6\widehat{k}}{7}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) Given, points are }P(2,1,-1)\text{ and }Q(4,4,-7).$
$\displaystyle \text{Here, }x_1=2,\ y_1=1,\ z_1=-1\text{ and }x_2=4,\ y_2=4,\ z_2=-7$
$\displaystyle \text{So, vector }\overrightarrow{PQ}=(x_2-x_1)\widehat{i}+(y_2-y_1)\widehat{j}+(z_2-z_1)\widehat{k}$
$\displaystyle =(4-2)\widehat{i}+(4-1)\widehat{j}+(-7+1)\widehat{k}$
$\displaystyle =2\widehat{i}+3\widehat{j}-6\widehat{k}$
$\displaystyle |\overrightarrow{PQ}|=\sqrt{2^2+3^2+(-6)^2}=\sqrt{4+9+36}=\sqrt{49}=7$
$\displaystyle \text{Hence, the unit vector in the direction of }\overrightarrow{PQ}\text{ is}$
$\displaystyle \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\frac{2\widehat{i}+3\widehat{j}-6\widehat{k}}{7}=\frac{2}{7}\widehat{i}+\frac{3}{7}\widehat{j}-\frac{6}{7}\widehat{k}$
$\\$

$\displaystyle \textbf{Question 10. }\text{Position vector of the mid-point of line segment }AB\text{ is}$
$\displaystyle 3\widehat{i}+2\widehat{j}-3\widehat{k}.\text{ If position vector of the point }A\text{ is}$
$\displaystyle 2\widehat{i}+3\widehat{j}-4\widehat{k},\text{ then position vector of the point }B\text{ is} \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }\frac{5}{2}\widehat{i}+\frac{5}{2}\widehat{j}-\frac{7}{2}\widehat{k} \qquad \text{(b) }4\widehat{i}+\widehat{j}-2\widehat{k}$
$\displaystyle \text{(c) }5\widehat{i}+5\widehat{j}-7\widehat{k} \qquad \text{(d) }\frac{\widehat{i}}{2}-\frac{\widehat{j}}{2}+\frac{\widehat{k}}{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given, position vector of the mid-point of line segment }=3\widehat{i}+2\widehat{j}-3\widehat{k}$
$\displaystyle \text{and position vector of point }A=2\widehat{i}+3\widehat{j}-4\widehat{k}$
$\displaystyle \text{Let }M\text{ be the mid-point of line segment }AB$
$\displaystyle \text{and position vector of }B\text{ is }(x\widehat{i}+y\widehat{j}+z\widehat{k})$
$\displaystyle \text{By mid-point of line segment, we have}$
$\displaystyle \frac{x+2}{2}=3\Rightarrow x=4$
$\displaystyle \frac{y+3}{2}=2\Rightarrow y=1$
$\displaystyle \frac{z-4}{2}=-3\Rightarrow z=-2$
$\displaystyle \text{Hence, the position vector of point }B=4\widehat{i}+\widehat{j}-2\widehat{k}$
$\\$

$\displaystyle \textbf{Question 11. }\text{Find all the vectors of magnitude }3\sqrt{3}\text{ which are}$
$\displaystyle \text{collinear to vector }\widehat{i}+\widehat{j}+\widehat{k}. \hspace{2.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{r}=\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle |\overrightarrow{r}|=\sqrt{1^2+1^2+1^2}=\sqrt3$
$\displaystyle \text{Now, unit vector of }\overrightarrow{r}=\widehat{r}=\frac{1}{\sqrt3}\widehat{i}+\frac{1}{\sqrt3}\widehat{j}+\frac{1}{\sqrt3}\widehat{k}$
$\displaystyle \therefore \text{All the vectors of magnitude }3\sqrt3\text{ which are}$
$\displaystyle \text{collinear to }\overrightarrow{r},\text{ are}$
$\displaystyle =3\sqrt3\left(\frac{1}{\sqrt3}\widehat{i}+\frac{1}{\sqrt3}\widehat{j}+\frac{1}{\sqrt3}\widehat{k}\right)$
$\displaystyle =3\widehat{i}+3\widehat{j}+3\widehat{k}$
$\\$

$\displaystyle \textbf{Question 12. }\text{Position vectors of the points }A,B\text{ and }C\text{ as shown in}$
$\displaystyle \text{the figure below are }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c},\text{ respectively.} \hspace{2.2cm}\text{[CBSE 2023]}$ $\displaystyle \text{If }\overrightarrow{AC}=\frac{5}{4}\overrightarrow{AB},\text{ then express }\overrightarrow{c}\text{ in terms of }\overrightarrow{a}\text{ and }\overrightarrow{b}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given condition is,}$
$\displaystyle \overrightarrow{AC}=\frac{5}{4}\overrightarrow{AB}\Rightarrow \overrightarrow{c}-\overrightarrow{a}=\frac{5}{4}(\overrightarrow{b}-\overrightarrow{a})$
$\displaystyle \Rightarrow 4(\overrightarrow{c}-\overrightarrow{a})=5(\overrightarrow{b}-\overrightarrow{a})$
$\displaystyle \Rightarrow 4\overrightarrow{c}-4\overrightarrow{a}=5\overrightarrow{b}-5\overrightarrow{a}$
$\displaystyle \Rightarrow 4\overrightarrow{c}=5\overrightarrow{b}-\overrightarrow{a}$
$\displaystyle \Rightarrow \overrightarrow{c}=\frac{1}{4}(5\overrightarrow{b}-\overrightarrow{a})$
$\\$

$\displaystyle \textbf{Question 13. }\text{Find the position vector of a point which divides the}$
$\displaystyle \text{join of points with position vectors }\overrightarrow{a}-2\overrightarrow{b}\text{ and }2\overrightarrow{a}+\overrightarrow{b}$
$\displaystyle \text{externally in the ratio }2:1. \hspace{2.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let given position vectors are }\overrightarrow{OA}=\overrightarrow{a}-2\overrightarrow{b}\text{ and}$
$\displaystyle \overrightarrow{OB}=2\overrightarrow{a}+\overrightarrow{b}.$
$\displaystyle \text{Let }\overrightarrow{OC}\text{ be the position vector of a point }C\text{ which}$
$\displaystyle \text{divides the join of points, with position vectors }\overrightarrow{OA}$
$\displaystyle \text{and }\overrightarrow{OB},\text{ externally in the ratio }2:1.$
$\displaystyle \therefore \overrightarrow{OC}=\frac{2\overrightarrow{OB}-1\overrightarrow{OA}}{2-1}\qquad [\text{by external section formula}]$
$\displaystyle =\frac{2(2\overrightarrow{a}+\overrightarrow{b})-1(\overrightarrow{a}-2\overrightarrow{b})}{1}$
$\displaystyle =4\overrightarrow{a}+2\overrightarrow{b}-\overrightarrow{a}+2\overrightarrow{b}=3\overrightarrow{a}+4\overrightarrow{b}$
$\\$

$\displaystyle \textbf{Question 14. }\text{If }\overrightarrow{a}=4\widehat{i}-\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}-2\widehat{j}+\widehat{k},\text{ then find a unit}$
$\displaystyle \text{vector parallel to the vector }\overrightarrow{a}+\overrightarrow{b}. \hspace{2.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, vectors are}$
$\displaystyle \overrightarrow{a}=4\widehat{i}-\widehat{j}+\widehat{k},\quad \overrightarrow{b}=2\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{a}+\overrightarrow{b}=(4\widehat{i}-\widehat{j}+\widehat{k})+(2\widehat{i}-2\widehat{j}+\widehat{k})$
$\displaystyle =6\widehat{i}-3\widehat{j}+2\widehat{k}$
$\displaystyle \text{and }|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{(6)^2+(-3)^2+(2)^2}$
$\displaystyle =\sqrt{36+9+4}=\sqrt{49}=7\text{ units}$
$\displaystyle \therefore \text{The unit vector parallel to the vector }\overrightarrow{a}+\overrightarrow{b}\text{ is}$
$\displaystyle \frac{\overrightarrow{a}+\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}|}=\frac{6\widehat{i}-3\widehat{j}+2\widehat{k}}{7}$
$\\$

$\displaystyle \textbf{Question 15. }\text{The two vectors }\widehat{j}+\widehat{k}\text{ and }3\widehat{i}-\widehat{j}+4\widehat{k}\text{ represent the}$
$\displaystyle \text{two sides }\overrightarrow{AB}\text{ and }\overrightarrow{AC}\text{ respectively of }\triangle ABC.\text{ Find the}$
$\displaystyle \text{length of the median through }A. \hspace{2.2cm}\text{[CBSE 2016; CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{AB}=\widehat{j}+\widehat{k}\text{ and }\overrightarrow{AC}=3\widehat{i}-\widehat{j}+4\widehat{k}$
$\displaystyle \text{Clearly, median vector, }\overrightarrow{AD}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}$
$\displaystyle =\frac{(\widehat{j}+\widehat{k})+(3\widehat{i}-\widehat{j}+4\widehat{k})}{2}=\frac{3\widehat{i}+5\widehat{k}}{2}$
$\displaystyle \text{Now, length of median}=|\overrightarrow{AD}|$
$\displaystyle =\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{5}{2}\right)^2}=\frac{1}{2}\sqrt{9+25}=\frac{\sqrt{34}}{2}$
$\displaystyle =\frac{\sqrt{17}\times\sqrt2}{2}=\sqrt{\frac{17}{2}}\text{ units}$
$\\$

$\displaystyle \textbf{Question 16. }\text{Write the direction ratios of the vector }3\overrightarrow{a}+2\overrightarrow{b},\text{ where}$
$\displaystyle \overrightarrow{a}=\widehat{i}+\widehat{j}-2\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}-4\widehat{j}+5\widehat{k}. \hspace{2.2cm}\text{[CBSE 2015C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Clearly, }3\overrightarrow{a}+2\overrightarrow{b}=3(\widehat{i}+\widehat{j}-2\widehat{k})+2(2\widehat{i}-4\widehat{j}+5\widehat{k})$
$\displaystyle =(3\widehat{i}+3\widehat{j}-6\widehat{k})+(4\widehat{i}-8\widehat{j}+10\widehat{k})$
$\displaystyle =7\widehat{i}-5\widehat{j}+4\widehat{k}$
$\displaystyle \text{Hence, the direction ratios of vector }3\overrightarrow{a}+2\overrightarrow{b}\text{ are }7,-5\text{ and }4.$
$\\$

$\displaystyle \textbf{Question 17. }\text{Find the unit vector in the direction of the sum of the}$
$\displaystyle \text{vectors }2\widehat{i}+3\widehat{j}-\widehat{k}\text{ and }4\widehat{i}-3\widehat{j}+2\widehat{k}. \hspace{2.2cm}\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=2\widehat{i}+3\widehat{j}-\widehat{k}\text{ and }\overrightarrow{b}=4\widehat{i}-3\widehat{j}+2\widehat{k}$
$\displaystyle \text{Now, sum of two vectors,}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}=(2\widehat{i}+3\widehat{j}-\widehat{k})+(4\widehat{i}-3\widehat{j}+2\widehat{k})=6\widehat{i}+\widehat{k}$
$\displaystyle \therefore \text{Required unit vector }=\frac{\overrightarrow{a}+\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}|}$
$\displaystyle =\frac{6\widehat{i}+\widehat{k}}{\sqrt{6^2+1^2}}=\frac{6\widehat{i}+\widehat{k}}{\sqrt{36+1}}=\frac{6\widehat{i}+\widehat{k}}{\sqrt{37}}$
$\displaystyle =\frac{6}{\sqrt{37}}\widehat{i}+\frac{1}{\sqrt{37}}\widehat{k}$
$\\$

$\displaystyle \textbf{Question 18. }\text{Find a vector in the direction of vector }2\widehat{i}-3\widehat{j}+6\widehat{k}$
$\displaystyle \text{which has magnitude }21\text{ units.} \hspace{2.2cm}\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=2\widehat{i}-3\widehat{j}+6\widehat{k}$
$\displaystyle \text{Then, }|\overrightarrow{a}|=\sqrt{(2)^2+(-3)^2+(6)^2}$
$\displaystyle =\sqrt{4+9+36}=\sqrt{49}=7\text{ units}$
$\displaystyle \text{The unit vector in the direction of the given vector }\overrightarrow{a}\text{ is}$
$\displaystyle \widehat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{1}{7}(2\widehat{i}-3\widehat{j}+6\widehat{k})=\frac{2}{7}\widehat{i}-\frac{3}{7}\widehat{j}+\frac{6}{7}\widehat{k}$
$\displaystyle \text{Now, the vector of magnitude is equal to }21\text{ units and}$
$\displaystyle \text{in the direction of }\overrightarrow{a}\text{ is given by}$
$\displaystyle 21\widehat{a}=21\left(\frac{2}{7}\widehat{i}-\frac{3}{7}\widehat{j}+\frac{6}{7}\widehat{k}\right)=6\widehat{i}-9\widehat{j}+18\widehat{k}$
$\\$

$\displaystyle \textbf{Question 19. }\text{Find a vector }\overrightarrow{a}\text{ of magnitude }5\sqrt{2},\text{ making an angle of}$
$\displaystyle \frac{\pi}{4}\text{ with }X\text{-axis, }\frac{\pi}{2}\text{ with }Y\text{-axis and an acute angle }\theta$
$\displaystyle \text{with }Z\text{-axis.} \hspace{2.2cm}\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, we have }l=\cos\frac{\pi}{4},\ m=\cos\frac{\pi}{2}\text{ and }n=\cos\theta$
$\displaystyle \Rightarrow l=\frac{1}{\sqrt2},\ m=0\text{ and }n=\cos\theta$
$\displaystyle \text{We know that, }l^2+m^2+n^2=1$
$\displaystyle \Rightarrow \left(\frac{1}{\sqrt2}\right)^2+(0)^2+n^2=1\Rightarrow \frac{1}{2}+n^2=1$
$\displaystyle \Rightarrow n^2=1-\frac{1}{2}=\frac{1}{2}\Rightarrow n=\pm\frac{1}{\sqrt2}\Rightarrow n=\frac{1}{\sqrt2}$
$\displaystyle [\because \theta \text{ is an acute angle with Z-axis}]$
$\displaystyle \therefore \cos\theta=\frac{1}{\sqrt2}\Rightarrow \theta=\frac{\pi}{4}$
$\displaystyle \text{Thus, the DC's of a line are }\frac{1}{\sqrt2},0,\frac{1}{\sqrt2}$
$\displaystyle \therefore \text{Vector }\overrightarrow{a}=|\overrightarrow{a}|(l\widehat{i}+m\widehat{j}+n\widehat{k})$
$\displaystyle =5\sqrt2\left(\frac{1}{\sqrt2}\widehat{i}+0\widehat{j}+\frac{1}{\sqrt2}\widehat{k}\right)\qquad [\because |\overrightarrow{a}|=5\sqrt2,\ \text{given}]$
$\displaystyle =5\widehat{i}+5\widehat{k}$
$\\$

$\displaystyle \textbf{Question 20. }\text{Write a unit vector in the direction of the sum of the vectors }\overrightarrow{a}=2\widehat{i}+2\widehat{j}-5\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+\widehat{j}-7\widehat{k}.\hspace{2.2cm}\text{[CBSE 2014 C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}=(2+2)\widehat{i}+(2+1)\widehat{j}+(-5-7)\widehat{k}$
$\displaystyle =4\widehat{i}+3\widehat{j}-12\widehat{k}$
$\displaystyle \left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{4^{2}+3^{2}+(-12)^{2}}$
$\displaystyle =\sqrt{16+9+144}=\sqrt{169}=13$
$\displaystyle \text{Hence, the required unit vector is}$
$\displaystyle \frac{\overrightarrow{a}+\overrightarrow{b}}{\left|\overrightarrow{a}+\overrightarrow{b}\right|}=\frac{1}{13}(4\widehat{i}+3\widehat{j}-12\widehat{k})$
$\displaystyle =\frac{4}{13}\widehat{i}+\frac{3}{13}\widehat{j}-\frac{12}{13}\widehat{k}$

$\displaystyle \textbf{Question 21. }\text{Find the value of }p\text{ for which the vectors }3\widehat{i}+2\widehat{j}+9\widehat{k}$
$\displaystyle \text{and }\widehat{i}-2p\widehat{j}+3\widehat{k}\text{ are parallel.} \hspace{2.2cm}\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }3\widehat{i}+2\widehat{j}+9\widehat{k}\text{ and }\widehat{i}-2p\widehat{j}+3\widehat{k}\text{ are two parallel}$
$\displaystyle \text{vectors, so their direction ratios will be proportional.}$
$\displaystyle \therefore \frac{3}{1}=\frac{2}{-2p}=\frac{9}{3}\Rightarrow \frac{2}{-2p}=\frac{3}{1}$
$\displaystyle \Rightarrow -6p=2\Rightarrow p=\frac{2}{-6}=-\frac{1}{3}$
$\\$

$\displaystyle \textbf{Question 22. }\text{Write the value of cosine of the angle which the vector}$
$\displaystyle \overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}\text{ makes with }Y\text{-axis.} \hspace{2.2cm}\text{[CBSE 2014C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{Now, unit vector in the direction of }\overrightarrow{a}\text{ is}$
$\displaystyle \widehat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{\widehat{i}+\widehat{j}+\widehat{k}}{\sqrt{(1)^2+(1)^2+(1)^2}}$
$\displaystyle =\frac{\widehat{i}+\widehat{j}+\widehat{k}}{\sqrt3}=\frac{1}{\sqrt3}\widehat{i}+\frac{1}{\sqrt3}\widehat{j}+\frac{1}{\sqrt3}\widehat{k}$
$\displaystyle \therefore \text{Cosine of the angle which given vector makes with Y-axis is }\frac{1}{\sqrt3}$
$\\$

$\displaystyle \textbf{Question 23. }\text{Find the angle between }X\text{-axis and the vector }\widehat{i}+\widehat{j}+\widehat{k}. \hspace{0.2cm}\text{[CBSE 2014C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{Now, unit vector in the direction of }\overrightarrow{a}\text{ is}$
$\displaystyle \widehat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{\widehat{i}+\widehat{j}+\widehat{k}}{\sqrt{1^2+1^2+1^2}}$
$\displaystyle \Rightarrow \widehat{a}=\frac{\widehat{i}+\widehat{j}+\widehat{k}}{\sqrt3}\Rightarrow \widehat{a}=\frac{1}{\sqrt3}\widehat{i}+\frac{1}{\sqrt3}\widehat{j}+\frac{1}{\sqrt3}\widehat{k}$
$\displaystyle \text{So, angle between X-axis and the vector }\widehat{i}+\widehat{j}+\widehat{k}\text{ is }\cos\alpha=\frac{1}{\sqrt3}$
$\displaystyle \Rightarrow \alpha=\cos^{-1}\left(\frac{1}{\sqrt3}\right)$
$\\$

$\displaystyle \textbf{Question 24. }\text{Write a vector in the direction of the vector }\widehat{i}-2\widehat{j}+2\widehat{k}\text{ that has magnitude }9\text{ units.}\hspace{2.2cm}\text{[CBSE 2014 C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=\widehat{i}-2\widehat{j}+2\widehat{k}$
$\displaystyle |\overrightarrow{a}|=\sqrt{1^{2}+(-2)^{2}+2^{2}}=\sqrt{9}=3$
$\displaystyle \text{Unit vector in the direction of }\overrightarrow{a}\text{ is}$
$\displaystyle \frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{1}{3}\left(\widehat{i}-2\widehat{j}+2\widehat{k}\right)$
$\displaystyle \text{Hence, a vector of magnitude }9\text{ in the same direction is}$
$\displaystyle 9\cdot\frac{1}{3}\left(\widehat{i}-2\widehat{j}+2\widehat{k}\right)$
$\displaystyle =3\widehat{i}-6\widehat{j}+6\widehat{k}$
$\\$

$\displaystyle \textbf{Question 25. }\text{Write a unit vector in the direction of vector }\overrightarrow{PQ},\text{ where}$
$\displaystyle P\text{ and }Q\text{ are the points }(1,3,0)\text{ and }(4,5,6),\text{ respectively.} \hspace{2.2cm}\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, points are }P(1,3,0)\text{ and }Q(4,5,6).$
$\displaystyle \text{Here, }x_1=1,\ y_1=3,\ z_1=0$
$\displaystyle \text{and }x_2=4,\ y_2=5,\ z_2=6$
$\displaystyle \text{So, vector }\overrightarrow{PQ}=(x_2-x_1)\widehat{i}+(y_2-y_1)\widehat{j}+(z_2-z_1)\widehat{k}$
$\displaystyle =(4-1)\widehat{i}+(5-3)\widehat{j}+(6-0)\widehat{k}$
$\displaystyle =3\widehat{i}+2\widehat{j}+6\widehat{k}$
$\displaystyle \therefore \text{Magnitude of given vector}$
$\displaystyle |\overrightarrow{PQ}|=\sqrt{3^2+2^2+6^2}=\sqrt{9+4+36}=\sqrt{49}=7\text{ units}$
$\displaystyle \text{Hence, the unit vector in the direction of }\overrightarrow{PQ}\text{ is}$
$\displaystyle \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\frac{3\widehat{i}+2\widehat{j}+6\widehat{k}}{7}=\frac{3}{7}\widehat{i}+\frac{2}{7}\widehat{j}+\frac{6}{7}\widehat{k}$
$\\$

$\displaystyle \textbf{Question 26. }\text{If a unit vector }\overrightarrow{a}\text{ makes angle }\frac{\pi}{3}\text{ with }\widehat{i},\ \frac{\pi}{4}$
$\displaystyle \text{with }\widehat{j}\text{ and an acute angle }\theta\text{ with }\widehat{k},\text{ then find the value of }\theta.$
$\displaystyle \hspace{2.2cm}\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, we have}$
$\displaystyle l=\cos\frac{\pi}{3},\quad m=\cos\frac{\pi}{4}\text{ and }n=\cos\theta$
$\displaystyle \Rightarrow l=\frac{1}{2},\quad m=\frac{1}{\sqrt2}\text{ and }n=\cos\theta$
$\displaystyle \text{We know that }l^2+m^2+n^2=1$
$\displaystyle \Rightarrow \left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt2}\right)^2+n^2=1$
$\displaystyle \Rightarrow n^2=1-\frac{1}{4}-\frac{1}{2}\Rightarrow n^2=\frac{4-1-2}{4}=\frac{1}{4}$
$\displaystyle \Rightarrow n=\pm\frac{1}{2}$
$\displaystyle \Rightarrow \cos\theta=\pm\frac{1}{2}$
$\displaystyle \text{But }\theta\text{ is an acute angle, therefore }\cos\theta=\frac{1}{2}$
$\displaystyle \Rightarrow \theta=\frac{\pi}{3}$
$\\$

$\displaystyle \textbf{Question 27. }\text{Write a unit vector in the direction of the sum of vectors }\overrightarrow{a}=2\widehat{i}-\widehat{j}+2\widehat{k}\text{ and }\overrightarrow{b}=-\widehat{i}+\widehat{j}+3\widehat{k}.\hspace{2.2cm}\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}=(2-1)\widehat{i}+(-1+1)\widehat{j}+(2+3)\widehat{k}$
$\displaystyle =\widehat{i}+5\widehat{k}$
$\displaystyle \left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{1^{2}+0^{2}+5^{2}}=\sqrt{26}$
$\displaystyle \text{Therefore, the required unit vector is}$
$\displaystyle \frac{\overrightarrow{a}+\overrightarrow{b}}{\left|\overrightarrow{a}+\overrightarrow{b}\right|}=\frac{\widehat{i}+5\widehat{k}}{\sqrt{26}}$
$\\$

$\displaystyle \textbf{Question 28. }\text{If }\overrightarrow{a}=x\widehat{i}+2\widehat{j}-z\widehat{k}\text{ and }\overrightarrow{b}=3\widehat{i}-y\widehat{j}+\widehat{k}\text{ are two}$
$\displaystyle \text{equal vectors, then write the value of }x+y+z. \hspace{2.2cm}\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=\overrightarrow{b}\Rightarrow x\widehat{i}+2\widehat{j}-z\widehat{k}=3\widehat{i}-y\widehat{j}+\widehat{k}$
$\displaystyle \text{On comparing the coefficient of components, we get}$
$\displaystyle x=3,\ y=-2,\ z=-1$
$\displaystyle \text{Now, }x+y+z=3-2-1=0$
$\\$

$\displaystyle \textbf{Question 29. }\text{P and Q are two points with position vectors }3\overrightarrow{a}-2\overrightarrow{b}\text{ and } \\ \overrightarrow{a}+\overrightarrow{b}\text{ respectively. Write the position} \text{vector of a point }R\text{ which divides the} \\ \text{line segment } PQ\text{ in the ratio }2:1\text{ externally.}\hspace{2.2cm}\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{OP}=3\overrightarrow{a}-2\overrightarrow{b},\qquad \overrightarrow{OQ}=\overrightarrow{a}+\overrightarrow{b}$
$\displaystyle \text{Since }R\text{ divides }PQ\text{ externally in the ratio }2:1,$
$\displaystyle \overrightarrow{OR}=\frac{2\overrightarrow{OQ}-1\overrightarrow{OP}}{2-1}$
$\displaystyle =2(\overrightarrow{a}+\overrightarrow{b})-(3\overrightarrow{a}-2\overrightarrow{b})$
$\displaystyle =2\overrightarrow{a}+2\overrightarrow{b}-3\overrightarrow{a}+2\overrightarrow{b}$
$\displaystyle =-\overrightarrow{a}+4\overrightarrow{b}$
$\displaystyle \therefore \text{The required position vector is }-\overrightarrow{a}+4\overrightarrow{b}.$
$\\$

$\displaystyle \textbf{Question 30. }\text{L and M are two points with position vectors }2\overrightarrow{a}-\overrightarrow{b}\text{ and } \\ \overrightarrow{a}+2\overrightarrow{b}\text{ respectively. Write the position vector of} \text{a point }N\text{ which divides the} \\ \text{line segment }LM\text{ in the ratio }2:1\text{ externally.}\hspace{2.2cm}\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{OL}=2\overrightarrow{a}-\overrightarrow{b},\qquad \overrightarrow{OM}=\overrightarrow{a}+2\overrightarrow{b}$
$\displaystyle \text{Since }N\text{ divides }LM\text{ externally in the ratio }2:1,$
$\displaystyle \overrightarrow{ON}=\frac{2\overrightarrow{OM}-1\overrightarrow{OL}}{2-1}$
$\displaystyle =2(\overrightarrow{a}+2\overrightarrow{b})-(2\overrightarrow{a}-\overrightarrow{b})$
$\displaystyle =2\overrightarrow{a}+4\overrightarrow{b}-2\overrightarrow{a}+\overrightarrow{b}$
$\displaystyle =5\overrightarrow{b}$
$\displaystyle \therefore \text{The required position vector is }5\overrightarrow{b}.$
$\\$

$\displaystyle \textbf{Question 31. }\text{ }A\text{ and }B\text{ are two points with position vectors }2\overrightarrow{a}-3\overrightarrow{b}$
$\displaystyle \text{ and }6\overrightarrow{b}-\overrightarrow{a},\text{ respectively. Write the position vector of}$
$\displaystyle \text{a point }P\text{ which divides the segment } \text{ internally in the ratio }1:2. \hspace{0.2cm}\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }A\text{ and }B\text{ are two points with position vectors}$
$\displaystyle 2\overrightarrow{a}-3\overrightarrow{b}\text{ and }6\overrightarrow{b}-\overrightarrow{a},\text{ respectively. Also, point }P$
$\displaystyle \text{divides the line segment }AB\text{ in the ratio }1:2\text{ internally.}$
$\displaystyle \therefore \text{Position vector of a point }P$
$\displaystyle =\frac{1\times(6\overrightarrow{b}-\overrightarrow{a})+2\times(2\overrightarrow{a}-3\overrightarrow{b})}{1+2}$
$\displaystyle \qquad [\text{by internal section formula}]$
$\displaystyle =\frac{6\overrightarrow{b}-\overrightarrow{a}+4\overrightarrow{a}-6\overrightarrow{b}}{3}$
$\displaystyle =\frac{3\overrightarrow{a}}{3}=\overrightarrow{a}$
$\\$

$\displaystyle \textbf{Question 32. }\text{Find the sum of the vectors }$
$\displaystyle \overrightarrow{a}=\widehat{i}-2\widehat{j}+\widehat{k},\ \overrightarrow{b}=-2\widehat{i}+4\widehat{j}+5\widehat{k}\text{ and } \overrightarrow{c}=\widehat{i}-6\widehat{j}-7\widehat{k}. \hspace{0.2cm}\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, vectors are }\overrightarrow{a}=\widehat{i}-2\widehat{j}+\widehat{k},$
$\displaystyle \overrightarrow{b}=-2\widehat{i}+4\widehat{j}+5\widehat{k}\text{ and }\overrightarrow{c}=\widehat{i}-6\widehat{j}-7\widehat{k}.$
$\displaystyle \text{Sum of the vectors }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ is}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=(\widehat{i}-2\widehat{j}+\widehat{k})+(-2\widehat{i}+4\widehat{j}+5\widehat{k})$
$\displaystyle +(\widehat{i}-6\widehat{j}-7\widehat{k})$
$\displaystyle =-4\widehat{j}-\widehat{k}$
$\\$

$\displaystyle \textbf{Question 33. }\text{Find the sum of the following vectors }\overrightarrow{a}=\widehat{i}-3\widehat{k}, \\ \ \overrightarrow{b}=2\widehat{j}-\widehat{k},\ \overrightarrow{c}=2\widehat{i}-3\widehat{j}+2\widehat{k}.\hspace{2.2cm}\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=(\widehat{i}-3\widehat{k})+(2\widehat{j}-\widehat{k})+(2\widehat{i}-3\widehat{j}+2\widehat{k})$
$\displaystyle =(1+2)\widehat{i}+(2-3)\widehat{j}+(-3-1+2)\widehat{k}$
$\displaystyle =3\widehat{i}-\widehat{j}-2\widehat{k}$
$\displaystyle \therefore \text{The required sum is }3\widehat{i}-\widehat{j}-2\widehat{k}.$
$\\$

$\displaystyle \textbf{Question 34. }\text{Find the sum of the following vectors }\overrightarrow{a}=\widehat{i}-2\widehat{j}, \\ \ \overrightarrow{b}=2\widehat{i}-3\widehat{j},\ \overrightarrow{c}=2\widehat{i}+3\widehat{k}.\hspace{2.2cm}\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=(\widehat{i}-2\widehat{j})+(2\widehat{i}-3\widehat{j})+(2\widehat{i}+3\widehat{k})$
$\displaystyle =(1+2+2)\widehat{i}+(-2-3)\widehat{j}+3\widehat{k}$
$\displaystyle =5\widehat{i}-5\widehat{j}+3\widehat{k}$
$\displaystyle \therefore \text{The required sum is }5\widehat{i}-5\widehat{j}+3\widehat{k}.$
$\\$

$\displaystyle \textbf{Question 35. }\text{Find the scalar components of }\overrightarrow{AB}\text{ with initial point}$
$\displaystyle A(2,1)\text{ and terminal point }B(-5,7). \hspace{2.2cm}\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, initial point is }A(2,1)\text{ and terminal point is}$
$\displaystyle B(-5,7),\text{ then scalar components of }\overrightarrow{AB}\text{ are}$
$\displaystyle x_2-x_1=-5-2=-7\text{ and }y_2-y_1=7-1=6.$
$\\$

$\displaystyle \textbf{Question 36. }\text{For what value of }\alpha,\text{ the vectors }2\widehat{i}-3\widehat{j}+4\widehat{k}\text{ and}$
$\displaystyle \alpha\widehat{i}+6\widehat{j}-8\widehat{k}\text{ are collinear?} \hspace{2.2cm}\text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let given vectors are }\overrightarrow{a}=2\widehat{i}-3\widehat{j}+4\widehat{k}$
$\displaystyle \text{and}$
$\displaystyle \overrightarrow{b}=a\widehat{i}+6\widehat{j}-8\widehat{k}$
$\displaystyle \text{We know that vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are said to be collinear,}$
$\displaystyle \text{if }\overrightarrow{a}=k\overrightarrow{b},\text{ where }k\text{ is a scalar.}$
$\displaystyle \therefore 2\widehat{i}-3\widehat{j}+4\widehat{k}=k(a\widehat{i}+6\widehat{j}-8\widehat{k})$
$\displaystyle \text{On comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k},\text{ we get}$
$\displaystyle 2=ka\text{ and }-3=6k$
$\displaystyle \Rightarrow k=-\frac{1}{2}$
$\displaystyle \Rightarrow 2=-\frac{1}{2}a$
$\displaystyle \Rightarrow a=-4$
$\\$

$\displaystyle \textbf{Question 37. }\text{Write the direction cosines of vector }-2\widehat{i}+\widehat{j}-5\widehat{k}. \hspace{2.2cm}\text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=-2\widehat{i}+\widehat{j}-5\widehat{k}$
$\displaystyle \therefore \text{DC's of }\overrightarrow{a}\text{ are}$
$\displaystyle \frac{-2}{\sqrt{(-2)^2+(1)^2+(-5)^2}},\ \frac{1}{\sqrt{(-2)^2+(1)^2+(-5)^2}}$
$\displaystyle \text{ and }\frac{-5}{\sqrt{(-2)^2+(1)^2+(-5)^2}}$
$\displaystyle \text{i.e., }\frac{-2}{\sqrt{30}},\ \frac{1}{\sqrt{30}},\ \frac{-5}{\sqrt{30}}$
$\\$