A line L passing through the point P(1,4,3) , is perpendicular to both the lines x-1/2=y+3/1=-z-2/4 and x+2/3=y-4/2=z+1/-2 . If the position vector of point Q on L is (a 1 , a 2 , a 3 ) such that (P Q) 2 =357 , then ( a + a 2 + a 3 ) can be:

Mathematics · 3D Geometry

JEE Advanced 2019 — Paper 2 — Question 21

A line $L$ passing through the point $P(1,4,3)$ , is perpendicular to both the lines $\dfrac{x-1}{2}=\dfrac{y+3}{1}=-\dfrac{z-2}{4}$ and $\dfrac{x+2}{3}=\dfrac{y-4}{2}=\dfrac{z+1}{-2}$ .

If the position vector of point $Q$ on $L$ is $\left(a_{1}, a_{2}, a_{3}\right)$ such that $(P Q)^{2}=357$ , then ( $\mathrm{a}+\mathrm{a}_{2}+\mathrm{a}_{3}$ ) can be:

Option A:
16
Option B:
15
Correct
Option C:
2
Option D:
1
Correct

Answer: B, D

Step-by-step solution

Equation of the line passing through $\mathrm{P}(1,4,3)$ is: $\dfrac{\mathrm{x}-1}{\mathrm{a}}=\dfrac{\mathrm{y}-4}{\mathrm{~b}}=\dfrac{\mathrm{z}-3}{\mathrm{c}}$

Since equation (i) is perpendicular to $\dfrac{x-1}{2}=\dfrac{y+3}{1}=\dfrac{z-2}{4}$ and $\dfrac{x+2}{3}=\dfrac{y-4}{2}=\dfrac{z+1}{-2}$

Hence $2 a+b+4 c=0$ and $3 a+2 b-2 c=0$

\begin{gathered} \therefore \dfrac{a}{-2-8}=\dfrac{b}{12+4}=\dfrac{c}{4-3} \Rightarrow \dfrac{a}{-10}=\dfrac{b}{16}=\dfrac{c}{1} \end{gathered}

Hence the equation of the lines is $\dfrac{x-1}{-10}=\dfrac{y-4}{16}=\dfrac{z-3}{1}$

Now any point Q on (2) can be taken as ( $1-10 \lambda, 16 \lambda+4, \lambda+3$ )

$\therefore$ Distance of Q from $\mathrm{P}(1,4,3)=(10 \lambda)^{2}+(16 \lambda)^{2}+\lambda^{2}=357$

$\Rightarrow(100+256+1) \lambda^{2}=357$

$\Rightarrow \quad \lambda=1$ or -1

$\therefore \mathrm{Q}$ is $(-9,20,4)$ or $(11,-12,2)$

Hence $a_{1}+a_{2}+a_{3}=15$ or 1

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 2
Subject: Mathematics
Chapter: 3D Geometry
Topic: Straight Lines in 3D Geometry