Question

A perpendicular is drawn from a point on the line $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}$ to the plane $x+y+z=3$ such that the foot of the perpendicular $Q$ also lies on the plane $x-y+z=3$. Then the co-ordinates of $Q$ are :

• A. $(2,0,1)$
• B. $(1,0,2)$
• C. $(-1,0,4)$
• D. $(4,0,-1)$

Answer 1

The correct option is A $(2,0,1)$

$\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}=\lambda$
Then the any point $P$ on the line is
$(2\lambda +1,-\lambda -1,\lambda )$

Let point $Q=(x_2,y_2,z_2)$ and $P=(x_1,y_1,z_1)$

Then, foot of perpendicular $Q$ drawn from point $P$ to the plane $ax+by+cz+d=0$ is given by
$\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{z_2-z_1}{c}$
$=-\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}$

Foot of perpendicular $Q$ is given by
$\frac{x-2\lambda -1}{1}=\frac{y+\lambda +1}{1}=\frac{z-\lambda }{1}=-\frac{(2\lambda -3)}{3}$

$Q$ lies on $x+y+z=3$ and $x-y+z=3$
$\Rightarrow x+z=3$ and $y=0$
$\therefore y=0\Rightarrow \lambda +1=\frac{-2\lambda +3}{3}\Rightarrow \lambda =0$

$Q=(2,0,1)$