Question

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

$\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=k$
Let foot of the perpendicular from $(-2,-1,0)$ to the plane $x+y+z-3=0$ is $A(\alpha ,\beta ,\gamma )$
$\therefore \frac{\alpha +2}{1}=\frac{\beta +1}{1}=\frac{\gamma }{1}=-\frac{(-2-1-3)}{3}=2$
So $A(\alpha ,\beta ,\gamma )=(0,1,2)$
Any point lie on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=k$ is of the form $(2k-2,-k-1,3k)$
So, $(2k-2)+(-k-1)+\left(3k\right)=3$
$\Rightarrow k=\frac{3}{2}$
So, the point of intersection of the plane and the line is $B(1,\frac{-5}{2},\frac{9}{2})$
$\therefore$ D.R's of projection line $AB=(2,-7,5)$
Thys locus of foot of perpendiculars is projection line i.e $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$.