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JEE Main 2015 Paper

Section: Mathematics

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Question : 42 of 90

Marks: +1, -0

The equation of the plane containing the line

2x-5y+z=3;\; x+y+4z=5

, and parallel to the plane,

x+3y+6z=1

$x+3y+6z=-7$
$x+3y+6z=7$
$2x+6y+12z=-13$
$2x+6y+12z=13$

Solution:

Let equation of plane is

(2x-5y+z-3)+\lambda(x+y+4z-5)=0

As plane is parallel to

x+3y+6z-1=0

\frac{2+\lambda}{1}=\frac{\lambda-5}{3}=\frac{1+4\lambda}{6}

\Rightarrow 6+3l=\lambda-5

11-2\lambda

\lambda=-\frac{11}{2}

Also

6\lambda-30=3+12\lambda

-6\lambda=33

\lambda=-\frac{11}{2}

so the equation of required plane is

(4x-10y+2z-6)-11(x+y+4x-5)=0

\Rightarrow -7x-21y-42z+49=0

\Rightarrow x+3y=6z-7=0

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