Question

Find the direction cosines of the vector joining the points $P(1,2,-3)$ and $Q(-1,-2,1)$, which is directed from $P$ to $Q$.

• A. $(-\frac{1}{3},-\frac{2}{3},\frac{2}{3})$
• B. $(\frac{1}{3},\frac{2}{3},-\frac{2}{3})$
• C. $(-\frac{1}{3},2,\frac{2}{3})$
• D. $(-\frac{2}{3},2,1)$

Answer 1

The correct option is A $(-\frac{1}{3},-\frac{2}{3},\frac{2}{3})$

Given, $\overrightarrow{OP}=1\hat{i}+2\hat{j}-3\hat{k}$ and $\overrightarrow{OQ}=-1\hat{i}-2\hat{j}+1\hat{k}$,

where $O$ is the origin.

Now, $\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}$

$=(-1-1)\hat{I}+(-2-2)\hat{j}+(1-(-3\left)\right)\hat{k}$

$=-2\hat{i}-4\hat{j}+4\hat{k}$

$|\overrightarrow{PQ}|=\sqrt{(-2{)}^2+(-4{)}^2+(4{)}^2}=6$

Unit vector in the direction of $\overrightarrow{PQ}$ is,

$\widehat{PQ}=\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ|}}$

$=\frac{-2\hat{i}-4\hat{j}+4\hat{k}}{6}$

$=-\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}$

Hence direction cosines are $(\frac{-1}{3},\frac{-2}{3},\frac{2}{3})$.