Mathematics · 3D Geometry

JEE Advanced 2020 — Paper 1 — Question 39

Let $L_{1}$ and $L_{2}$ be the following straight lines.

L_{1}: \frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{3} \text { and } L_{2}: \frac{x-1}{-3}=\frac{y}{-1}=\frac{z-1}{1}

Suppose the straight line

L: \frac{x-\alpha}{l}=\frac{y-1}{m}=\frac{z-\gamma}{-2}

lies in the plane containing $L_{1}$ and $L_{2}$ , and passes through the point of intersection of $L_{1}$ and $L_{2}$ . If the line $L$

bisects the acute angle between the lines $L_{1}$ and $L_{2}$ , then which of the following statements is/are TRUE?

Option A:
$\alpha-\gamma=3$
Correct
Option B:
$l+\mathrm{m}=2$
Correct
Option C:
$\alpha-\gamma=1$
Option D:
$l+\mathrm{m}=0$

Answer: A, B

Step-by-step solution

DCs of $L_{1}: \frac{1}{\sqrt{11}}, \frac{-1}{\sqrt{11}}, \frac{3}{\sqrt{11}}$

DCs of $L_{2}: \frac{-3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}, \frac{1}{\sqrt{11}}$

Since $\frac{1}{\sqrt{11}}\left(\frac{-3}{\sqrt{11}}\right)+\frac{-1}{\sqrt{11}}\left(\frac{-1}{\sqrt{11}}\right)+\frac{3}{\sqrt{11}} \frac{1}{\sqrt{11}}=\frac{1}{11}>0$

So DCs of acute angle bisectors are $\frac{-2}{\sqrt{11}}, \frac{-2}{\sqrt{11}}, \frac{4}{\sqrt{11}}$

{D.rs} can be written as $1,1,-2 \Rightarrow \ell=\mathrm{m}=1$

$L_{1}$ and $L_{2}$ intersect at $(1,0,1)$ which also lies on bisector $L$ . Point on $L$ having $y$ -coordinate equal to 1 is

$(2,1,-1) \Rightarrow \alpha=2, \gamma=-1$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 1
Subject: Mathematics
Chapter: 3D Geometry
Topic: Vector & Cartesian forms of lines and planes