Question

Projection of the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ on the plane $3x-6y+3z=108$ is

• A. $\frac{x}{1}=\frac{y+24}{2}=\frac{z+12}{3}$
• B. $\frac{x-3}{3}=\frac{y+6}{-6}=\frac{z-3}{3}$
• C. $\frac{x-6}{1}=\frac{y+12}{2}=\frac{z-6}{3}$
• D. $\frac{x-6}{3}=\frac{y+12}{-6}=\frac{z-6}{3}$

Answer 1

Answer: (A), (C)

$\frac{x-6}{1}=\frac{y+12}{2}=\frac{z-6}{3}$

Clearly given line is $\perp$ to normal of the plane
$\therefore$ Projection line d.r.'s $=(1,2,3)$
Now, let foot of perpendicular from $(0,0,0)$ to the plane $3x-6y+3z=108$ is $(a,b,c)$
$\therefore \frac{a}{3}=\frac{b}{-6}=\frac{c}{3}=\frac{-(-108)}{54}=2$
$\Rightarrow (a,b,c)=(6,-12,6)$
$\therefore$ Projection line equation is
$\frac{x-6}{1}=\frac{y+12}{2}=\frac{z-6}{3}$ which is also same as
$\frac{x}{1}=\frac{y+24}{2}=\frac{z+12}{3}$
(Since, $(0,-24,-12)$ is a point on projection line )