Q. Equation of a line in the plane $\pi \equiv 2x - y + \mathcal{z} - 4 = 0$ which is perpendicular to the line $l$ whose equation is $\frac{x - 2}{1} = \frac{y - 2}{- 1} = \frac{\mathcal{z} - 3}{- 2}$ and which passes through the point of intersection of $l$ and $\pi$ is
A) $\frac{x - 2}{1} = \frac{y - 1}{5} = \frac{\mathcal{z} - 1}{- 1}$
B) $\frac{x - 1}{3} = \frac{y - 3}{5} = \frac{\mathcal{z} - 5}{- 1}$
C) $\frac{x + 2}{2} = \frac{y + 1}{- 1} = \frac{\mathcal{z} + 1}{1}$
D) $\frac{x - 2}{2} = \frac{y - 1}{- 1} = \frac{\mathcal{z}–1}{1}$
Explanation
(b)
Let direction ratios of the line be $(a,\ b,\ c)$, then
$2a - b + c = 0$ and $a - b - 2c = 0$ i.e, $\frac{a}{3} = \frac{b}{5} = \frac{c}{- 1}$
Therefore, direction ratios of the line are $(3,\ 5,\ - 1)$
Any point on the given line is $(2 + \lambda,\ 2 - \lambda,\ 3 - 2\lambda)$, it lies on the given plane $\pi$ if
$2(2 + \lambda) - (2 - \lambda) + (3 - 2\lambda) = 4$
$\Rightarrow 4 + 2\lambda - 2 + \lambda + 3 - 2\lambda = 4\ \Rightarrow \lambda = - 1$
Therefore, the point of intersection of the line and the plane is (1,3, 5)
Therefore, equation of the required line is
$\frac{x - 1}{3} = \frac{y - 3}{5} = \frac{\mathcal{z} - 5}{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.