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 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)

Any point $B$ on line is $(2\lambda -2,-\lambda -1,3\lambda )$
Point $B$ lies on the plane for some $\lambda$.
$\Rightarrow (2\lambda -2)+(-\lambda -1)+3\lambda =3$ or $\lambda =3/2$
$\Rightarrow B\equiv (1,-5/2,9/2)$
The foot of the perpendicular from point $(-2,-1,0)$ on the plane is the point $A(0,1,2)$.
$\Rightarrow$ Direction ratio of $AB=\left(1,\frac{-7}{2},\frac{5}{2}\right)\equiv (2,-7,5)$ Hence, feet of perpendicular lies on the line $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}.$