Q. The point of intersection of the line passing through (0,0, 1) and intersecting the lines $x + 2y + \mathcal{z} = 1,\ - x + y - 2\mathcal{z} = 2$ and $x + y = 2,\ x + \mathcal{z} = 2$ with $xy$ plane is

A) $\left( \frac{5}{3}, - \frac{1}{3},0 \right)$
B) (1, 1, 0)
C) $\left( \frac{2}{3}, - \frac{1}{3},\ 0 \right)$
D) $\left( - \frac{5}{3},\frac{1}{3},0 \right)$
Explanation

(a)
Equation of line $x + 2y + \mathcal{z} - 1 + \lambda\left( - x + y - 2\mathcal{z} - 2 \right) = 0$ (i)
$x + y - 2 + \mu\left( x + \mathcal{z} - 2 \right) = 0$ (ii)
(0, 0, 1) lies on it $\Rightarrow \lambda = 0,\ \mu = - 2$
For point of intersection, $\mathcal{z} = 0$ and solve (i) and (ii)

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.