Mathematics · 3D Geometry

JEE Advanced 2018 — Paper 1 — Question 29

Let $P_{1}: 2 x+y-z=3$ and $P_{2}: x+2 y+z=2$ be two planes. Then, which of the following statement(s) is (are) TRUE ?

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

Answer: C, D

Step-by-step solution

(A) Direction ratios of line of intersection are given by

$(2 \hat{i}+\hat{j}-\hat{k}) \times(\hat{i}+2 \hat{j}+\hat{k})=3 \hat{i}-3 \hat{j}+3 \hat{k} \Rightarrow d r=(1,-1,1)$

(B) $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=3+3+3=9 \neq 0$

(C) Angle between $P_{1}, P_{2}=\cos ^{-1}\left|\frac{2+2-1}{\sqrt{6} \cdot \sqrt{6}}\right|=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}$

(D) $P_{3}: x-y+z=0$

Distance of $(2,1,1)$ from the plane $=\frac{2}{\sqrt{3}}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 1
Subject: Mathematics
Chapter: 3D Geometry
Topic: Intersection of lines, line & plane.