Question

The projection of the line $\frac{x+1}{1}=\frac{y}{2}=\frac{z-1}{3}$ on the plane $x-2y+z=6$ is the line of intersection of this plane with the plane

• A. $2x+y+2=0$
• B. $3x+y-z=2$
• C. $2x-3y+8z=3$
• D. none of these

Answer 1

Answer: (A)

Equation of the plane through $(-1,0,1)$ is
$a(x+1)+b(y-0)+c(z-1)=0...(i)$
which is parallel to the given line and perpendicular to the given plane
$-a+2b+3c=0...(ii)$
and $a-2b+c=0...(iii)$
From Eqs. (ii) and (iii), we get
c$=0,a=2b$
From Eq. $(i),2b(x+1)+by=0$
$\Rightarrow 2x+y+2=0$