Question

Let $P_1: 2x + y - z = 3$ and $P_2: x + 2y + z = 2$ be two planes. Then, which of the following statement(s) is (are) TRUE?

• A. The line of intersection of $P_1$ and $P_2$ has direction ratios $1, 2, -1$
• B. The line $\frac{3x - 4}{9} = \frac{1 - 3y}{9} = \frac{z}{3}$ is perpendicular to the line of intersection of $P_1$ and $P_2$
• C. The acute angle between $P_1$ and $P_2$ is $60^{\circ}$
• D. If $P_3$ is the plane passing through the point $(4, 2, -2)$ and perpendicular to the line of intersection of $P_1$ and $P_2$, then the distance of the point $(2, 1, 1)$ from the plane $P_3$ is $\frac{2 }{\sqrt{3}}$

Answer 1

Answer: (C), (D)

Equation of planes : $P_1 : 2x + y - z = 3 $
$P_2 : x + 2y + z = 2$
Let direction ratios of line of intersection of planes
$P_1$ and $P_2$ are < a, b, c >
$\therefore $ 2a + b - c = 0
and a + 2b + c = 0.
$\therefore $ Direction ratios = < 1, -1, 1 >
The given line is : $\frac{x -4/3}{3} = \frac{y - 1/3}{3} = \frac{z}{3}$
It is parallel to the line of intersection of $P_1$ and $P_2$.
$\therefore \, \, \cos \theta = | \frac{2 + 2 - 1}{\sqrt{6} \sqrt{6}} | = \frac{1}{2} \Rightarrow \theta = 60^{\circ}$
equation of plane $P_3$ is
1.(x - 4) - (y - 2) + (z + 2) = 0
$\therefore \, \, P_3 : x - y + z = 0 $
Distance of plane $P_3$ from point (2, 1, 1)
$ = | \frac{2 - 1 + 1 }{ \sqrt{1 + 1 + 1}} | = \frac{2}{\sqrt{3}} $ units