Q. The projection of the line $\frac{x + 1}{- 1} = \frac{y}{2} = \frac{\mathcal{z} - 1}{3}$ on the plane $x - 2y + \mathcal{z} = 6$ is the line of intersection of this plane with the plane
A) $2x + y + 2 = 0$
B) $3x + y - \mathcal{z} = 2$
C) $2x - 3y + 8\mathcal{z} = 3$
D) None of these
Explanation
(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., $2b(x + 1) + by = 0$
$\Rightarrow 2x + y + 2 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.