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

The correct option is D

$\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

To find the foot of perpendiculars and find its locus.
formula used
Foot of perpendicular from $(x_1,y_1,z_1)toax+by+cz+d=0be(x_2,y_2,z_2),then\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{z_2-z_1}{c}=\frac{-(a{x}_1+b{y}_1+c{z}_1+d)}{a^2+b^2+c^2}Anypointon\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=\lambda \Rightarrow x=2\lambda -2,y=-\lambda -1,z=3\lambda$
Let foot of perpendicular from $(2\lambda -2,-\lambda -1,3\lambda )tox+y+z=3be(x_2,y_2,z_2).\therefore \frac{x_2(2\lambda -2)}{1}=\frac{y_2-(-\lambda -1)}{1}=\frac{z_2-\left(3\lambda \right)}{1}=-\frac{(2\lambda -2-\lambda -1+3\lambda -3)}{1+1+1}\Rightarrow x_2-2\lambda +2=y_2+\lambda +1=z_2-3\lambda =2-\frac{4\lambda }{3}\therefore x_2=\frac{2\lambda }{3},y_2=1-\frac{7\lambda }{3},z_2=2+\frac{5\lambda }{3}\Rightarrow \lambda =\frac{x_2-0}{\frac{2}{3}}=\frac{y_2-1}{\frac{-7}{3}}=\frac{z_2-2}{\frac{5}{3}}$
Hence, foot of perpendicular lie on
$\frac{x}{\frac{2}{3}}=\frac{y-1}{\frac{-7}{3}}=\frac{z-2}{\frac{5}{3}}\Rightarrow \frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$