Question

A perpendicular is drawn from the point $P(5,4,2)$ to the line $\overrightarrow{r}=-\hat{i}+3\hat{j}+\hat{k}+\lambda (2\hat{i}+3\hat{j}-\hat{k})$. Then the correct option(s) among the following is/are:

• A. The coordinates of the foot of perpendicular is $(1,6,0)$
• B. The length of the perpendicular $\sqrt{24}\text{units}$
• C. The length of the perpendicular $\sqrt{26}\text{units}$
• D. The image of point $P$ is $(-3,8,-2)$

Answer 1

Answer: (A), (B), (D)

The image of point $P$ is $(-3,8,-2)$

Given line equation is $\overrightarrow{r}=-\hat{i}+3\hat{j}+\hat{k}+\lambda (2\hat{i}+3\hat{j}-\hat{k})$
Any point on the line can be taken as $A=(2\lambda -1,3\lambda +3,-\lambda +1)$
Assuming ${}^{\prime }{A}^{\prime }$ as the foot of perpendicular from $P(5,4,2)$, we have D.R's of $AP=(2\lambda -6,3\lambda -1,-\lambda -1).$
$\therefore AP.(2\hat{i}+3\hat{j}-\hat{k})=0$
$\Rightarrow (2\lambda -6)\times 2+(3\lambda -1)\times 3+(-\lambda -1)\times (-1)=0$
$\Rightarrow 4\lambda -12+9\lambda -3+\lambda +1=0$
$\Rightarrow 14\lambda -14=0$
$\Rightarrow \lambda =1$
The coordinates of the foot of perpendicular are $A(1,6,0)$
The length of the perpendicular =$|AP|=$
$\sqrt{(5-1{)}^2+(4-6{)}^2+(2-0{)}^2}=\sqrt{16+4+4}=\sqrt{24}$
The foot of perpendicular would be the midpoint of $P$ and the image of $P$ in the line
$\therefore (5\hat{i}+4\hat{j}+2\hat{k})+(x\hat{i}+y\hat{j}+z\hat{k})=2\times (\hat{i}+6\hat{j})$
$\Rightarrow x=2-5=-3,y=12-4=8,z=0-2=-2$
The image of point $P$ is $(-3,8,-2)$