JEE Advanced 2016 · Paper 2 · Mathematics Q42

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MathematicsSingle correct+3 / -1Question 42 of 54

Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3$ . Then the equation of the plane passing through $P$ and containing the straight line $\dfrac x1=\dfrac y2=\dfrac z1$ is

x+y-3z=0

3x+z=0

x-4y+7z=0

2x-y=0

Answer and worked solution

Official answer: C

Image of (3, 1, 7) in $x-y+z=3$ .
$P=(3,1,7)-2\cdot\frac{3-1+7-3}{3}(1,-1,1)=(-1,5,3).$

The plane. It contains the origin, the direction $(1,2,1)$ and the point $(-1,5,3)$ . Its normal is
$(1,2,1)\times(-1,5,3)=(1,-4,7).$
So the plane is $x-4y+7z=0$ .

(C)

Topics:Equation of a plane

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JEE Advanced 2016 · Paper 2, Mathematics question 42