Question

Let there are two planes ${{P}_{1}}:2x+y-z=3$ and ${{P}_{2}}:x+2y+z=2$. Then which of the following is true
This question has multiple answers
A. The line of intersection of ${{P}_{1}}$ and ${{P}_{2}}$ has directional ratios 1, 2, -1
B. The line $\dfrac{3x-4}{9}=\dfrac{1-3y}{9}=\dfrac{z}{3}$ 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 a plane passing through the point $\left( 4,2,-2 \right)$ and perpendicular to the line of intersection of ${{P}_{1}}$ and ${{P}_{2}}$. Then the distance of the point $\left( 2,1,1 \right)$ from the plane is $\dfrac{2}{\sqrt{3}}$

Answer 1

To solve this problem, we should know the properties of the plane and the perpendicular condition. The directional ratios of the normal of a plane with the equation $ax+by+cz=d$ are given by $a,b,c$ respectively. Using this, we can find the directional ratios of the normal of the two planes. We know that the line of intersection of the two planes will be perpendicular to the normal of both the planes. Two lines with directional ratios $a,b,c$ and $d,e,f$ are perpendicular when $ad+be+cf=0$. Using this condition, we can find the directional ratios of the line of intersection of the two lines. Option A can be directly verified.

Complete step-by-step answer:
The directional ratios of the line $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ are a, b, c respectively. Using this property and the perpendicular property, we can verify option-B. Angle between the planes is same as angle between their normal. Angle between two lines whose directional ratios are $a,b,c$ and $d,e,f$ is given by $\theta ={{\cos }^{-1}}\left( \dfrac{ad+be+cf}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\times \sqrt{{{d}^{2}}+{{e}^{2}}+{{f}^{2}}}} \right)$. Using this relation, we can find the angle between the planes and verify option-C. The directional ratios of the line of intersection of the two planes ${{P}_{1}}$ and ${{P}_{2}}$ are the directional ratios of the normal of the plane ${{P}_{3}}$. Equation of a plane whose directional ratios are $a,b,c$ and passing through the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ is given by $a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0$. The distance of the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ from the plane $ax+by+cz+d=0$ is given by $\text{distance}=\dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$. Using this formula, we can verify option-D

The given planes are ${{P}_{1}}:2x+y-z=3$ and ${{P}_{2}}:x+2y+z=2$.
. The directional ratios of the normal of a plane with the equation $ax+by+cz=d$ are given by $a,b,c$ respectively. Using this,
The directional ratios of ${{P}_{1}}$ are $2,1,-1$.
The directional ratios of ${{P}_{2}}$ are $1,2,1$.
We know that the line of intersection of the two planes will be perpendicular to the normal of both the planes.
Two lines with directional ratios $a,b,c$ and $d,e,f$ are perpendicular when $ad+be+cf=0$.
Using this condition, we can find the directional ratios of the line of intersection of the two lines. Let the directional ratios of the line of intersection are $a,b,c$, we get
With the normal directional ratios of ${{P}_{1}}$
$2a+b-c=0$
With the normal directional ratios of ${{P}_{1}}$
$a+2b+c=0$
Let us consider a = 1, we get
$\begin{aligned} & b-c=-2 \\\ & 2b+c=-1 \\\ \end{aligned}$
Adding them, we get
$\begin{aligned} & 3b=-3 \\\ & b=-1 \\\ \end{aligned}$
Substituting b = -1 in one of them, we get
$\begin{aligned} & -1-c=-2 \\\ & -c=-1 \\\ & c=1 \\\ \end{aligned}$
So, the directional ratios of the line of intersection are $1,-1,1$.
Option-A is wrong.
Let us consider option-B
Given line is $\dfrac{3x-4}{9}=\dfrac{1-3y}{9}=\dfrac{z}{3}$, which can be rewritten as $\dfrac{x-\dfrac{4}{3}}{3}=\dfrac{y-\dfrac{1}{3}}{-3}=\dfrac{z}{3}$
The directional ratios of the line $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ are a, b, c respectively
The directional ratios of the given line are $3,-3,3$, when divided by 3 the directional ratios become $1,-1,1$. The two lines have same directional ratios which mean that they are parallel.
Option-B is wrong.
Let us consider option-C
Angle between two lines whose directional ratios are $a,b,c$ and $d,e,f$ is given by $\theta ={{\cos }^{-1}}\left( \dfrac{ad+be+cf}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\times \sqrt{{{d}^{2}}+{{e}^{2}}+{{f}^{2}}}} \right)$.
Angle between the given two planes is
$\theta ={{\cos }^{-1}}\left( \dfrac{2\times 1+1\times 2-1\times 1}{\sqrt{{{2}^{2}}+{{1}^{2}}+{{\left( -1 \right)}^{2}}}\times \sqrt{{{1}^{2}}+{{2}^{2}}+{{1}^{2}}}} \right)={{\cos }^{-1}}\left( \dfrac{3}{\sqrt{6}\times \sqrt{6}} \right)={{\cos }^{-1}}\left( \dfrac{3}{6} \right)={{\cos }^{-1}}\left( \dfrac{1}{2} \right)={{60}^{\circ }}$
Option-C is correct.

Let us consider option-D
The directional ratios of the line of intersection of the two planes ${{P}_{1}}$ and ${{P}_{2}}$ are the directional ratios of the normal of the plane ${{P}_{3}}$. The directional ratios of the normal of ${{P}_{3}}$ are $1,-1,1$. The plane passes through the point $\left( 4,2,-2 \right)$.
Equation of a plane whose directional ratios are $a,b,c$ and passing through the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ is given by $a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0$.
We get the equation of the line as
$\begin{aligned} & 1\left( x-4 \right)-1\left( y-2 \right)+1\left( z+2 \right)=0 \\\ & x-4-y+2+z+2=0 \\\ & x-y+z=0 \\\ \end{aligned}$
The distance of the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ from the plane $ax+by+cz+d=0$ is given by $\text{distance}=\dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$.
The distance of the plane ${{P}_{3}}$ from $\left( 2,1,1 \right)$ is
$\text{distance}=\dfrac{2-1+1}{\sqrt{{{1}^{2}}+{{1}^{2}}+{{1}^{2}}}}=\dfrac{2}{\sqrt{3}}$
So, option-D is correct.

So, the correct answers are “Option C and D”.

Note: Students might commit a mistake in calculating the directional ratios of the line given in option-B. The directional ratios are obtained as the denominators in the equalities when the coefficients of x, y, z are 1. When there are other values, we should adjust the equation such that we multiply and divide the equation by the same number and change the coefficients to 1.