Question

Let $P_1:2x+y-z=3$ and $P_2:x+2y+z=2$ be two planes. Then, which of the following statements(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)

The correct options are
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 $\frac{2}{\sqrt{3}}$

The direction ratios of line of intersection of two planes is the cross product of their normal vectors.
$$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 1& 2& 1\end{array}\right|$$$=3\hat{i}-3\hat{j}+3\hat{k}=3(\hat{i}-\hat{j}+\hat{k})$
Direction ratio is $(1,-1,1)$.

For option 2:
$\text{Consider}\frac{3x-4}{9}=\frac{1-3y}{9}=\frac{z}{3}$ and $(1,-1,1)$
$(9,9,3)\cdot (1,-1,1)=3\ne 0$
$\therefore$ Option 2 is wrong.

For option 3:
Angle between $P_1$ and $P_2$ is given by $\cos \theta =\left|\frac{2\times 1+1\times 2+(-1)\times 1}{\sqrt{6}\sqrt{6}}\right|$
$\Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta ={60}^{\circ }$

For option 4:
Since $P_3$ is perpendicular to the line of intersection of $P_1$ and $P_2$.
Normal vector of $P_3=(1,-1,1)$
Equation of required plane is $1(x-4)-1(y-2)+1(z+2)=0$
$or,x-y+z=0$
Distance of the point $(2,1,1)\text{from}{P}_3$
$=\left|\frac{2-1+1}{\sqrt{3}}\right|=\frac{2}{\sqrt{3}}$