Question

Find the coordinates of the foot of perpendicular and the length of the perpendicular drawn from the point $P(5,4,2)$ to the line $\overrightarrow{r}=-\hat{i}+3\hat{j}+\hat{k}+\mu (2\hat{i}+3\hat{j}-\hat{k})$. Also find the image of $P$ is this line.

Answer 1

Any point on the line can be written in parametric form as $(2\lambda -1,3\lambda +3,-\lambda +1)$

Assuming this as the foot of perpendicular from $(5,4,2)$, we can equate the dot product of this vector and the line direction to zero.

$\therefore \left(\right(2\lambda -1-5)\hat{i}+(3\lambda +3-4)\hat{j}+(-\lambda +1-2\left)\hat{k}\right).(2\hat{i}+3\hat{j}-\hat{k})=0$

$\Rightarrow (2\lambda -6)\times 2+(3\lambda -1)\times 3+(-\lambda -1)\times (-1)=0$

$\Rightarrow 4\lambda -12+9\lambda -3+\lambda +1=0$

$\Rightarrow 14\lambda -14=0$

$\Rightarrow \lambda =1$

The coordinates of the point are thus $(1,6,0)$

The length of the perpendicular can be found out by $\sqrt{(5-1{)}^2+(4-6{)}^2+(2-0{)}^2}=\sqrt{16+4+4}=\sqrt{24}$

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 thus $(-3,8,-2)$