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:
Correct Option: 1
Solution:
Let point $P$ on the line is $(2 \lambda+1,-\lambda-1, \lambda)$ foot of perpendicular $Q$ is given by
$\frac{\mathrm{x}-2 \lambda-1}{1}=\frac{\mathrm{y}+\lambda+1}{1}=\frac{\mathrm{z}-\lambda}{1}=\frac{-(2 \lambda-3)}{3}$
$\because Q$ lies on $x+y+z=3 \& x-y+z=3$
$\Rightarrow x+z=3 \& y=0$
$\mathrm{y}=0 \Rightarrow \lambda+1=\frac{-2 \lambda+3}{3} \Rightarrow \lambda=0$
$\Rightarrow Q$ is $(2,0,1)$