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