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
$ax+by+cz+d=0be(x_2,y_2,z_2)$ then
$\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{z_2-z_1}{c}$
$=\frac{-(a{x}_1+b{y}_1+c{z}_1+d)}{a^2+b^2+c^2}$
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 $(x_2,y_2,z_2)$.
$\therefore \frac{x_2-(2\lambda -2)}{1}=\frac{y_2-(-\lambda -1)}{1}=\frac{z_2-(3\lambda )}{1}$
$=\frac{(2\lambda -2-\lambda -1+3\lambda -3)}{1+1+1}$
$\Rightarrow x_2-2\lambda +2=y_2+\lambda +1=z_2-3\lambda =2-\frac{4\lambda }{3}$
$\therefore$ $x_2=\frac{2\lambda }{3},y_2=1-\frac{7\lambda }{3},+2=z_2=2+\frac{5\lambda }{3}$
$\Rightarrow$ $\lambda =\frac{x_2-0}{2/3},=1-\frac{y_2-1}{-7/3},=+\frac{z_2-2}{5/3}$
Hence, foot of perpendicular lie on
$=\frac{x}{2/3}=\frac{y-1}{-7/3}=\frac{z-2}{5/3}\Rightarrow \frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$