JEE Advanced 2013 · Paper 2 · Mathematics Q59

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MathematicsList match+3 / -1Question 59 of 60

Consider the lines $L_1:\dfrac{x-1}2=\dfrac y{-1}=\dfrac{z+3}1$ , $L_2:\dfrac{x-4}1=\dfrac{y+3}1=\dfrac{z+3}2$ and the planes $P_1:7x+y+2z=3$ , $P_2:3x+5y-6z=4$ . Let $ax+by+cz=d$ be the equation of the plane passing through the point of intersection of lines $L_1$ and $L_2$ , and perpendicular to planes $P_1$ and $P_2$ .

Match List I with List II and select the correct answer using the code given below the lists:

List I

P
$a=$
Q
$b=$
R
$c=$
S
$d=$

List II

1
13
2
$-3$
3
1
4
$-2$

P → 3; Q → 2; R → 4; S → 1

P → 1; Q → 3; R → 4; S → 2

P → 3; Q → 2; R → 1; S → 4

P → 2; Q → 4; R → 1; S → 3

Answer and worked solution

Official answer: A

Intersection of the lines. $(1+2s,-s,-3+s)=(4+t,-3+t,-3+2t)$ gives $s=2$ , $t=1$ : the point $(5,-2,-1)$ .

Normal. $(7,1,2)\times(3,5,-6)=(-16,48,32)\propto(1,-3,-2)$ .

Plane. $x-3y-2z=5+6+2=13$ : $a=1$ → 3, $b=-3$ → 2, $c=-2$ → 4, $d=13$ → 1.

Code 3 2 4 1.

(A)

Topics:Equation of a plane Equation of a line in space Cross product

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JEE Advanced 2013 · Paper 2, Mathematics question 59