Class 12: Vectors – Miscellaneous Problems

$\displaystyle \textbf{Question 1: } \text{Observe the vector diagram shown below and the expression that} \\ \text{follows.}$ $\displaystyle (1)\ \overrightarrow{SO} \hspace{1cm} (2)\ \overrightarrow{OS} \hspace{1cm} (3)\ -\overrightarrow{SO} \hspace{1cm} (4)\ -\overrightarrow{OS}$
$\displaystyle (a)\ \text{only (1)} \hspace{1cm} (b)\ \text{only (2)} \hspace{1cm} (c)\ \text{only (1) and (4)} \hspace{1cm} (d)\ \text{only (2) and (3)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d)}$
$\displaystyle \text{(1) } \overrightarrow{SO}\text{ does not follow because initial point is }O\text{ and terminal point is }S$
$\displaystyle \text{(2) } \overrightarrow{OS}\text{ follows because }O\text{ is initial point and }S\text{ is terminal point}$
$\displaystyle \text{(3) }-\overrightarrow{SO}\text{ follows because }-\overrightarrow{SO}\text{ or }\overrightarrow{OS}\text{ is negative vector of }\overrightarrow{SO}$
$\displaystyle \text{(4) }-\overrightarrow{OS}\text{ does not follow because }-\overrightarrow{OS}\text{ or }\overrightarrow{SO}\text{ is negative vector of }\overrightarrow{OS}$
$\\$

$\displaystyle \textbf{Question 2: } \text{If } \overrightarrow{u} = \widehat{i},\ \overrightarrow{v} = \widehat{j} \text{ and } \overrightarrow{w} = \widehat{k} \text{ are unit vectors, what is the angle} \\ \text{between } (\overrightarrow{w} \times \overrightarrow{v}) \text{ and } \overrightarrow{u}?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \overrightarrow{u}=\widehat{i},\ \overrightarrow{v}=\widehat{j}\text{ and }\overrightarrow{w}=\widehat{k}$
$\displaystyle \therefore \overrightarrow{w}\times\overrightarrow{v}=(\widehat{k}\times\widehat{j})=-\widehat{i}$
$\displaystyle \text{Let }\theta\text{ be the angle between }(\overrightarrow{w}\times\overrightarrow{v})\text{ and }\overrightarrow{u}$
$\displaystyle \therefore \cos\theta=\frac{(-\widehat{i})\cdot\widehat{i}}{\sqrt{(-1)^{2}}\sqrt{(1)^{2}}}=-1$
$\displaystyle \Rightarrow \cos\theta=\cos\pi \Rightarrow \theta=\pi$
$\\$

$\displaystyle \textbf{Question 3: } \text{Find a vector of magnitude of } 10 \text{ units and parallel to the vector } \\ 2\widehat{i} + 3\widehat{j} - \widehat{k}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } \overrightarrow{a}=\alpha\widehat{i}+\beta\widehat{j}+\gamma\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+3\widehat{j}-\widehat{k}$
$\displaystyle \text{According to the question, } \overrightarrow{a}=\lambda\overrightarrow{b}\quad \cdots(i)$
$\displaystyle \Rightarrow |\overrightarrow{a}|=|\lambda\overrightarrow{b}|\Rightarrow |\overrightarrow{a}|=|\lambda||\overrightarrow{b}|$
$\displaystyle \Rightarrow 10=\lambda\sqrt{(2)^{2}+(3)^{2}+(-1)^{2}}$
$\displaystyle \Rightarrow \lambda=\frac{10}{\sqrt{14}}$
$\displaystyle \Rightarrow \overrightarrow{a}=\frac{10}{\sqrt{14}}(2\widehat{i}+3\widehat{j}-\widehat{k})$
$\displaystyle \Rightarrow \overrightarrow{a}=\frac{20}{\sqrt{14}}\widehat{i}+\frac{30}{\sqrt{14}}\widehat{j}-\frac{10}{\sqrt{14}}\widehat{k}$
$\\$

$\displaystyle \textbf{Question 4: } \text{Find the position vector of a point } R \text{ which divides the line joining} \\ \text{the two-points } P \text{ and } Q \text{ with position vectors } 2\widehat{i} + \widehat{j} \text{ and } \widehat{i} - 2\widehat{j}, \text{ respectively in the} \\ \text{ratio of } 2:1 \text{ externally.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The position vector of point }R\text{ dividing the line segment joining }P\text{ and }Q\text{ in the ratio }m:n\text{ is given by}$
$\displaystyle \frac{m\overrightarrow{OQ}-n\overrightarrow{OP}}{m-n}\quad \text{[externally]}$
$\displaystyle \text{Position vectors of }P\text{ and }Q\text{ are given as}$
$\displaystyle \overrightarrow{OP}=2\widehat{i}+\widehat{j}\text{ and }\overrightarrow{OQ}=\widehat{i}-2\widehat{j}$
$\displaystyle \text{The position vector of point }R\text{ dividing }PQ\text{ externally in the ratio }2:1\text{ is}$
$\displaystyle \overrightarrow{OR}=\frac{2(\widehat{i}-2\widehat{j})-1(2\widehat{i}+\widehat{j})}{2-1}$
$\displaystyle \Rightarrow \overrightarrow{OR}=2\widehat{i}-4\widehat{j}-2\widehat{i}-\widehat{j}$
$\displaystyle \Rightarrow \overrightarrow{OR}=-5\widehat{j}$
$\\$

$\displaystyle \textbf{Question 5: } \text{If the vectors } a\widehat{i} + 3\widehat{j} - 2\widehat{k} \text{ and } 3\widehat{i} - 4\widehat{j} + b\widehat{k} \text{ are collinear, then } (a,b) \text{ is} \\ \text{equal to }$
$\displaystyle (a)\ \left(\frac{9}{4}, \frac{8}{3}\right) \hspace{1cm} (b)\ \left(-\frac{9}{4}, \frac{8}{3}\right)$
$\displaystyle (c)\ \left(\frac{9}{4}, -\frac{8}{3}\right) \hspace{1cm} (d)\ \left(-\frac{9}{4}, -\frac{8}{3}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) If }a\widehat{i}+3\widehat{j}-2\widehat{k}\text{ and }3\widehat{i}-4\widehat{j}+b\widehat{k}\text{ are collinear,}$
$\displaystyle \text{then } \frac{a}{3}=\frac{3}{-4}=\frac{-2}{b}$
$\displaystyle \text{Now, } \frac{a}{3}=\frac{3}{-4}$
$\displaystyle \Rightarrow -4a=9 \Rightarrow a=-\frac{9}{4}$
$\displaystyle \text{and } \frac{3}{-4}=\frac{-2}{b}$
$\displaystyle \Rightarrow 3b=8 \Rightarrow b=\frac{8}{3}$
$\displaystyle \therefore (a,b)=\left(-\frac{9}{4},\frac{8}{3}\right)$
$\\$

$\displaystyle \textbf{Question 6: } \text{Find the unit vector parallel to the vector: } 3\widehat{i} + 6\widehat{j} - 2\widehat{k}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \overrightarrow{r}=3\widehat{i}+6\widehat{j}-2\widehat{k}$
$\displaystyle |\overrightarrow{r}|=\sqrt{3^{2}+6^{2}+(-2)^{2}}$
$\displaystyle =\sqrt{9+36+4}=\sqrt{49}=7$
$\displaystyle \therefore \widehat{r}=\frac{\overrightarrow{r}}{|\overrightarrow{r}|}=\frac{3\widehat{i}+6\widehat{j}-2\widehat{k}}{7}$
$\\$

$\displaystyle \textbf{Question 7: } \text{If } \overrightarrow{a}=\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k} \text{ and } \overrightarrow{b}=-\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k}$
$\displaystyle \text{and } \overrightarrow{c}=3\overrightarrow{i}+\overrightarrow{j}, \text{ find } t \text{ such that } \overrightarrow{a}+t\overrightarrow{b} \text{ is perpendicular to } \overrightarrow{c}.$
$\displaystyle \text{(a) } 0 \quad \text{(b) } 5 \quad \text{(c) } 4 \quad \text{(d) } 2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given, }\overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k},\ \overrightarrow{b}=-\widehat{i}+2\widehat{j}+\widehat{k}$
$\displaystyle \text{and }\overrightarrow{c}=3\widehat{i}+\widehat{j}$
$\displaystyle \text{Here, }\overrightarrow{a}+t\overrightarrow{b}\perp\overrightarrow{c}\Rightarrow(\overrightarrow{a}+t\overrightarrow{b})\cdot\overrightarrow{c}=0$
$\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{c}+t(\overrightarrow{b}\cdot\overrightarrow{c})=0$
$\displaystyle \Rightarrow t=-\frac{\overrightarrow{a}\cdot\overrightarrow{c}}{\overrightarrow{b}\cdot\overrightarrow{c}}$
$\displaystyle =-\frac{(\widehat{i}+2\widehat{j}+3\widehat{k})\cdot(3\widehat{i}+\widehat{j})}{(-\widehat{i}+2\widehat{j}+\widehat{k})\cdot(3\widehat{i}+\widehat{j})}$
$\displaystyle =-\frac{3+2}{-3+2}=5$
$\\$

$\displaystyle \textbf{Question 8: } \text{What will be the value of } m, \text{ if the vector } 2\overrightarrow{i}+m\overrightarrow{j}+\overrightarrow{k}$
$\displaystyle \text{is perpendicular to } 2\overrightarrow{i}-\overrightarrow{j}+3\overrightarrow{k}? \quad$
$\displaystyle \text{(a) } 7 \quad \text{(b) } 0 \quad \text{(c) } 1 \quad \text{(d) } -1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Let }\overrightarrow{a}=2\widehat{i}+m\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}-\widehat{j}+3\widehat{k}$
$\displaystyle \text{For two vectors to be perpendicular, }\overrightarrow{a}\cdot\overrightarrow{b}=0$
$\displaystyle \therefore (2\widehat{i}+m\widehat{j}+\widehat{k})\cdot(2\widehat{i}-\widehat{j}+3\widehat{k})=0$
$\displaystyle \Rightarrow 4-m+3=0$
$\displaystyle \Rightarrow m=7$
$\\$

$\displaystyle \textbf{Question 9: } \text{Find the angle between the vectors } \overrightarrow{a}=6\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k},$
$\displaystyle \overrightarrow{b}=2\overrightarrow{i}-9\overrightarrow{j}+6\overrightarrow{k}. \quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{a}=6\widehat{i}+2\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}-9\widehat{j}+6\widehat{k}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(6\widehat{i}+2\widehat{j}+3\widehat{k})\cdot(2\widehat{i}-9\widehat{j}+6\widehat{k})$
$\displaystyle =12-18+18=12$
$\displaystyle |\overrightarrow{a}|=\sqrt{6^{2}+2^{2}+3^{2}}=\sqrt{36+4+9}=\sqrt{49}=7$
$\displaystyle |\overrightarrow{b}|=\sqrt{2^{2}+(-9)^{2}+6^{2}}=\sqrt{4+81+36}=\sqrt{121}=11$
$\displaystyle \therefore \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{12}{7\times11}=\frac{12}{77}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{12}{77}\right)$
$\\$

$\displaystyle \textbf{Question 10. } \text{If } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are perpendicular vectors, } |\overrightarrow{a}+\overrightarrow{b}|=13$
$\displaystyle \text{and } |\overrightarrow{a}|=5, \text{ then find the value of } |\overrightarrow{b}|. \quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }|\overrightarrow{a}+\overrightarrow{b}|=13$
$\displaystyle \Rightarrow |\overrightarrow{a}+\overrightarrow{b}|^{2}=169$
$\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})=169$
$\displaystyle \Rightarrow |\overrightarrow{a}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^{2}=169$
$\displaystyle \Rightarrow |\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}=169\quad[\because \overrightarrow{a}\perp\overrightarrow{b}\Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}=0]$
$\displaystyle \Rightarrow |\overrightarrow{b}|^{2}=169-|\overrightarrow{a}|^{2}=169-25=144$
$\displaystyle \Rightarrow |\overrightarrow{b}|=12$
$\\$

$\displaystyle \textbf{Question 11: } \text{Given below are two vectors in their component form} \\ \text{such that } |\overrightarrow{p}\times\overrightarrow{q}|=1. \overrightarrow{p}=x\overrightarrow{i}-\frac{y}{4}\overrightarrow{j} \text{ and } \overrightarrow{q}=y\overrightarrow{i}+\frac{x}{9}\overrightarrow{j}.$
$\displaystyle \text{Find the relation between the variables } x \text{ and } y. \quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{p}=x\widehat{i}-\frac{y}{4}\widehat{j},\ \overrightarrow{q}=y\widehat{i}+\frac{x}{9}\widehat{j}\text{ and }|\overrightarrow{p}\times\overrightarrow{q}|=1$
$\displaystyle \overrightarrow{p}\times\overrightarrow{q}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\x&-\frac{y}{4}&0\\y&\frac{x}{9}&0\end{vmatrix}$
$\displaystyle =\widehat{k}\left(x\cdot\frac{x}{9}-\left(-\frac{y}{4}\cdot y\right)\right)$
$\displaystyle =\widehat{k}\left(\frac{x^{2}}{9}+\frac{y^{2}}{4}\right)$
$\displaystyle \Rightarrow |\overrightarrow{p}\times\overrightarrow{q}|=\frac{x^{2}}{9}+\frac{y^{2}}{4}$
$\displaystyle \Rightarrow 1=\frac{x^{2}}{9}+\frac{y^{2}}{4}$
$\displaystyle \Rightarrow 4x^{2}+9y^{2}=36$
$\\$

$\displaystyle \textbf{Question 12: } \text{The position vectors of the points } A \text{ and } B \text{ are } (2,-3,2) \text{ and } (2,3,1)$
$\displaystyle \text{respectively, then find the area of } \triangle OAB. \quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\overrightarrow{OA}=2\widehat{i}-3\widehat{j}+2\widehat{k}\text{ and }\overrightarrow{OB}=2\widehat{i}+3\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{OA}\times\overrightarrow{OB}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&-3&2\\2&3&1\end{vmatrix}$
$\displaystyle =\widehat{i}((-3)(1)-2\cdot3)-\widehat{j}(2\cdot1-2\cdot2)+\widehat{k}(2\cdot3-(-3)\cdot2)$
$\displaystyle =-9\widehat{i}+2\widehat{j}+12\widehat{k}$
$\displaystyle \Rightarrow |\overrightarrow{OA}\times\overrightarrow{OB}|=\sqrt{(-9)^{2}+2^{2}+12^{2}}=\sqrt{81+4+144}=\sqrt{229}$
$\displaystyle \text{Area of }\triangle OAB=\frac{1}{2}|\overrightarrow{OA}\times\overrightarrow{OB}|=\frac{\sqrt{229}}{2}$
$\\$

$\displaystyle \textbf{Question 13: } \text{For any two non-zero vectors } \overrightarrow{a} \text{ and } \overrightarrow{b}, \text{ if } \\ |\overrightarrow{a}\cdot\overrightarrow{b}|=|\overrightarrow{a}\times\overrightarrow{b}|, \text{then find the angle between them.} \quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }|\overrightarrow{a}\cdot\overrightarrow{b}|=|\overrightarrow{a}\times\overrightarrow{b}|$
$\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\cos\theta=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta$
$\displaystyle \Rightarrow \tan\theta=1$
$\displaystyle \Rightarrow \theta=\frac{\pi}{4}$
$\displaystyle \therefore \text{Angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }\frac{\pi}{4}$
$\\$

$\displaystyle \textbf{Question 14: } \text{Find } \lambda \text{ if the scalar projection of } \overrightarrow{a}=\lambda\overrightarrow{i}+\overrightarrow{j}+4\overrightarrow{k} \text{ on}$
$\displaystyle \overrightarrow{b}=2\overrightarrow{i}+6\overrightarrow{j}+3\overrightarrow{k} \text{ is } 4 \text{ units.} \quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that projection of }\overrightarrow{a}\text{ on }\overrightarrow{b}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}$
$\displaystyle \Rightarrow 4=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}\quad ...(i)$
$\displaystyle \text{Now, }\overrightarrow{a}\cdot\overrightarrow{b}=2\lambda+6+12=2\lambda+18$
$\displaystyle \text{Also, }|\overrightarrow{b}|=\sqrt{2^{2}+6^{2}+3^{2}}=\sqrt{4+36+9}=\sqrt{49}=7$
$\displaystyle \text{On putting in Eq. (i), we obtain}$
$\displaystyle 4=\frac{2\lambda+18}{7}$
$\displaystyle \Rightarrow 2\lambda+18=28$
$\displaystyle \Rightarrow 2\lambda=10$
$\displaystyle \Rightarrow \lambda=5$
$\\$

$\displaystyle \textbf{Question 15: } \text{Find the area of the parallelogram whose adjacent sides are given by }$
$\displaystyle \text{the vectors } \overrightarrow{a}=3\overrightarrow{i}+\overrightarrow{j}+4\overrightarrow{k} \text{ and} \overrightarrow{b}=\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}. \quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have adjacent sides of parallelogram, given by}$
$\displaystyle \overrightarrow{a}=3\widehat{i}+\widehat{j}+4\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&1&4\\1&-1&1\end{vmatrix}$
$\displaystyle =\widehat{i}(1\cdot1-4(-1))-\widehat{j}(3\cdot1-4\cdot1)+\widehat{k}(3(-1)-1\cdot1)$
$\displaystyle =5\widehat{i}+\widehat{j}-4\widehat{k}$
$\displaystyle \text{Area of parallelogram}=|\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{25+1+16}=\sqrt{42}$
$\\$

$\displaystyle \textbf{Question 16: } \text{Show that } (\overrightarrow{a}\times\overrightarrow{b})^{2}=\frac{|\overrightarrow{a}\cdot\overrightarrow{a}\;\;\overrightarrow{a}\cdot\overrightarrow{b}|}{|\overrightarrow{a}\cdot\overrightarrow{b}\;\;\overrightarrow{b}\cdot\overrightarrow{b}|}. \quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{To prove: }(\overrightarrow{a}\times\overrightarrow{b})^{2}=\begin{vmatrix}\overrightarrow{a}\cdot\overrightarrow{a}&\overrightarrow{a}\cdot\overrightarrow{b}\\\overrightarrow{a}\cdot\overrightarrow{b}&\overrightarrow{b}\cdot\overrightarrow{b}\end{vmatrix}$
$\displaystyle \text{LHS }=|\overrightarrow{a}\times\overrightarrow{b}|^{2}=(\overrightarrow{a}\times\overrightarrow{b})\cdot(\overrightarrow{a}\times\overrightarrow{b})$
$\displaystyle =|\overrightarrow{a}|^{2}|\overrightarrow{b}|^{2}\sin^{2}\theta$
$\displaystyle =|\overrightarrow{a}|^{2}|\overrightarrow{b}|^{2}(1-\cos^{2}\theta)$
$\displaystyle =|\overrightarrow{a}|^{2}|\overrightarrow{b}|^{2}-|\overrightarrow{a}|^{2}|\overrightarrow{b}|^{2}\cos^{2}\theta$
$\displaystyle =|\overrightarrow{a}|^{2}|\overrightarrow{b}|^{2}-(\overrightarrow{a}\cdot\overrightarrow{b})^{2}$
$\displaystyle =\begin{vmatrix}\overrightarrow{a}\cdot\overrightarrow{a}&\overrightarrow{a}\cdot\overrightarrow{b}\\\overrightarrow{a}\cdot\overrightarrow{b}&\overrightarrow{b}\cdot\overrightarrow{b}\end{vmatrix}=\text{RHS}$
$\displaystyle \therefore \text{Proved}$
$\\$

$\displaystyle \textbf{Question 17: } \text{Find the area of the triangle whose adjacent sides are}$
$\displaystyle \overrightarrow{i}+4\overrightarrow{j}-\overrightarrow{k} \text{ and } \overrightarrow{i}+\overrightarrow{j}+2\overrightarrow{k}. \quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=\widehat{i}+4\widehat{j}-\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}+\widehat{j}+2\widehat{k}$
$\displaystyle \text{Area of triangle}=\frac{1}{2}|\overrightarrow{a}\times\overrightarrow{b}|$
$\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&4&-1\\1&1&2\end{vmatrix}$
$\displaystyle =\widehat{i}(4\cdot2-(-1)\cdot1)-\widehat{j}(1\cdot2-(-1)\cdot1)+\widehat{k}(1\cdot1-4\cdot1)$
$\displaystyle =9\widehat{i}-3\widehat{j}-3\widehat{k}$
$\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{9^{2}+(-3)^{2}+(-3)^{2}}=\sqrt{81+9+9}=\sqrt{99}=3\sqrt{11}$
$\displaystyle \therefore \text{Area}=\frac{1}{2}\times3\sqrt{11}=\frac{3\sqrt{11}}{2}$
$\\$

$\displaystyle \textbf{Question 18: } \text{Find the position vector of a point which divides the join of points with} \\ \text{position vectors } \overrightarrow{a} - 2\overrightarrow{b} \text{ and } 2\overrightarrow{a} + \overrightarrow{b} \text{ externally in the ratio 2:1.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let given position vectors are }$
$\displaystyle \overrightarrow{OA}=\overrightarrow{a}-2\overrightarrow{b}\text{ and }\overrightarrow{OB}=2\overrightarrow{a}+\overrightarrow{b}$
$\displaystyle \text{Let }\overrightarrow{OC}\text{ be position vector of }C\text{ dividing }OA\text{ and }OB\text{ externally in ratio }2:1$
$\displaystyle \therefore \overrightarrow{OC}=\frac{2\overrightarrow{OB}-1\overrightarrow{OA}}{2-1}$
$\displaystyle =\frac{2(2\overrightarrow{a}+\overrightarrow{b})-(\overrightarrow{a}-2\overrightarrow{b})}{1}$
$\displaystyle =4\overrightarrow{a}+2\overrightarrow{b}-\overrightarrow{a}+2\overrightarrow{b}$
$\displaystyle =3\overrightarrow{a}+4\overrightarrow{b}$
$\\$

$\displaystyle \textbf{Question 19: } \text{If } \overrightarrow{a} = 4\widehat{i} - \widehat{j} + \widehat{k} \text{ and } \overrightarrow{b} = 2\widehat{i} - 2\widehat{j} + \widehat{k}, \text{ then find a unit vector parallel to} \\ \text{the vector } \overrightarrow{a} + \overrightarrow{b}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, vectors are }$
$\displaystyle \overrightarrow{a}=4\widehat{i}-\widehat{j}+\widehat{k},\ \overrightarrow{b}=2\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{a}+\overrightarrow{b}=(4\widehat{i}-\widehat{j}+\widehat{k})+(2\widehat{i}-2\widehat{j}+\widehat{k})$
$\displaystyle =6\widehat{i}-3\widehat{j}+2\widehat{k}$
$\displaystyle \text{and }|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{6^{2}+(-3)^{2}+2^{2}}$
$\displaystyle =\sqrt{36+9+4}=\sqrt{49}=7$
$\displaystyle \therefore \text{Unit vector parallel to }\overrightarrow{a}+\overrightarrow{b}\text{ is}$
$\displaystyle \frac{\overrightarrow{a}+\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}|}=\frac{6\widehat{i}-3\widehat{j}+2\widehat{k}}{7}$
$\\$