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 coordinates of $Q$ are

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

Answer 1

Answer: (A)

The given line is $\frac{x - 1}{2}=\frac{y + 1}{- 1}=\frac{z}{1}=\lambda ,\left(\right.$ let
$\Rightarrow \frac{x - 1}{2}=\lambda ,\frac{y + 1}{- 1}=\lambda ,\frac{z}{1}=\lambda $
$\Rightarrow x=2\lambda +1,y=-\lambda -1,z=\lambda .$
Hence, any point on the line is $\left(2 \lambda + 1 , - \lambda - 1 , \, \lambda \right).$
Thus, the point $P$ on the line is $\left(2 \lambda + 1 , - \lambda - 1 , \, \lambda \right).$
We know that the foot of perpendicular from a point $\left(x_{1} , y_{1} , z_{1}\right)$ on the plane $ax+by+cz+d=0$ is $\frac{x - x_{1}}{a}=\frac{y - y_{1}}{b}=\frac{z - z_{1}}{c}=-\frac{\left(a x_{1} + b y_{1} + c z_{1}\right)}{a^{2} + b^{2} + c^{2}}.$
Thus, the foot of perpendicular $Q$ from $P$ on the plane $x+y+z-3=0$ is given by
$\frac{x - 2 \lambda - 1}{1}=\frac{y + \lambda + 1}{1}=\frac{z - \lambda }{1}=-\frac{\left(2 \lambda - 3\right)}{3}$
$\because Q$ lies on $x+y+z=3 \, \& \, x-y+z=3\,$
On adding the two equations, we get $y=0\,$
$\Rightarrow \frac{\lambda + 1}{1}=\frac{- 2 \lambda + 3}{3}$
$\Rightarrow \lambda =0$
Hence, we have $\frac{x - 1}{1}=\frac{z}{1}=-\frac{\left(- 3\right)}{3}$
$\Rightarrow x=2,z=1\,$
So, the coordinate of $Q$ are $\left(2 , 0 , 1\right).$