Question

Perpendicular are drawn from points on the line $\frac{x+2}{2} = \frac{y+1}{-1} = \frac{z}{3} $ to the plane $x+ y + z = 3$. The feet of perpendiculars lie on the line

• A. $\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$
• B. $\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$
• C. $\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$
• D. $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

Answer 1

Answer: (D)

PLAN To find the foot of perpendiculars and find its locus.
Formula used
Foot of perpendicular from $(x_1,y_1,z_1)$ to
$ax + by + cz +d = 0$ be $(x_2 , y_2 ,z_2)$ , then
$\frac{x_{2}-x_{1}}{a}=\frac{y_{2}-y_{1}}{b}=\frac{z_{2}-z_{1}}{c}$
=$\frac{\left(ax_{1}+by_{1}+cz_{1}+d\right)}{a^{2 }+b^{2}+c^{2}}$
Any point on $\frac{x+2}{2} = \frac{y + 1}{-1} = \frac{z}{3} = \lambda$
$\Rightarrow x = 2 \lambda - 2 , y = - \lambda -1 , z = 3 \lambda$
Let foot of perpendicular from $(2 \lambda - 2 , - \lambda - 1 , 3 \lambda)$
to $x + y+ z = 3 $ be $(x_2,y_2 ,z_2)$.
$\therefore \frac{x_{2}-\left(2\lambda-2\right)}{1}=\frac{y_{2}-\left(-\lambda-1\right)}{1}=\frac{z_{2}-\left(3\lambda\right)}{1}$
= $- \frac{\left(2\lambda-2-1+2\lambda-1+3\lambda-3\right)}{1+1+1}$
$\Rightarrow x_2 - 2\lambda +2 = y_2 + \lambda +1 = z_2 - 3\lambda = 2 - \frac{4\lambda}{3}$
$\therefore x_2 = \frac{2\lambda}{3} , y_2 = 1 - \frac{7\lambda}{3} , z_2 = 2+ \frac{5\lambda}{3}$
$\Rightarrow \lambda = \frac{x_2 - 0}{2/3} = \frac{y_2 - 1}{-7/3} = \frac{z_2 - 2}{5/3} $
Hence, foot of perpendicular lie on
$\frac{x }{2/3} = \frac{y -1}{-7 /3} = \frac{z-2}{5/3}\Rightarrow \frac{x}{2}=\frac{y-1}{-7} = \frac{z-2}{5}$