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 $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}=\lambda ,($ 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 $\left(2,0,1\right).$