Question

The co-ordinates of the point $P$ on the line $\overrightarrow{r}=(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})+\lambda (-\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k})$ which is nearest to the origin is

• A. $(\frac{2}{3},\frac{4}{3},\frac{2}{3})$
• B. $(-\frac{2}{3},-\frac{4}{3},\frac{2}{3})$
• C. $(\frac{2}{3},\frac{4}{3},-\frac{2}{3})$
• D. $(\frac{2}{3},\frac{2}{3},\frac{4}{3})$

Answer 1

The correct option is A $(\frac{2}{3},\frac{4}{3},\frac{2}{3})$

$\overrightarrow{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda (-\hat{i}+\hat{j}-\hat{k})$

the shortest distance between a line and a point is the perpendicular distance.Hence,the point nearest to origin on line will be its foot of perpendicular.

$\therefore O(0,0,0)$ ,let $P$ be foot of $\perp$.

$\therefore P$ lies on line.

$\therefore \overrightarrow{P}=(\hat{i}+\hat{j}+\hat{k})+\lambda (-\hat{i}+\hat{j}-\hat{k})\overrightarrow{O}=0\hat{i}+0\hat{j}+0\hat{k}$

$\overrightarrow{OP}\perp$ line $\Rightarrow \overrightarrow{OP}=(1-\lambda )\hat{i}+(1+\lambda )\hat{j}+(1-\lambda )\hat{k}$

$\therefore \overrightarrow{OP}.(-\hat{i}+\hat{j}-\hat{k})=0$

$\Rightarrow \left(\right(1-\lambda )\hat{i}+(1+\lambda )\hat{j}+(1-\lambda \left)\hat{k}\right).(-\hat{i}+\hat{j}-\hat{k})=0\Rightarrow \lambda =\frac{1}{3}\therefore \overrightarrow{P}=\frac{2}{3}\hat{i}+\frac{4}{3}\hat{j}+\frac{2}{3}\hat{k}$

$\therefore$ Required point is $(\frac{2}{3},\frac{4}{3},\frac{2}{3})$