Q. The plane, which passes through the point (3, 2, 0) and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{\mathcal{z}–4}{4}$ is:
A) $x - y + \mathcal{z} = 1$
B) $x + y + \mathcal{z} = 5$
C) $x + 2y - \mathcal{z} = 1$
D) $2x - y + \mathcal{z} = 5$
Explanation
(a)
The required plane is $\left| \begin{matrix}
x - 3 & y - 6 & \mathcal{z} - 4 \\
3 - 3 & 2 - 6 & 0 - 4 \\
1 & 5 & 4
\end{matrix} \right| = 0$
$\Rightarrow \ \left| \begin{matrix}
x - 3 & y - z - 2 & \mathcal{z} - 4 \\
0 & 0 & - 4 \\
1 & 1 & 4
\end{matrix} \right| = 0$ (Operating $C_{2} \rightarrow C_{2} - C_{3}$)
$\Rightarrow 4\left( x - 3 - y + \mathcal{z} + 2 \right) = 0$
$\Rightarrow x - y + \mathcal{z} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.