JEE Advanced 2017 Paper 2 · Mathematics · Three Dimensional Geometry · medium

The line L: (x-1)/(2)=y/1=(z+1)/(-1) and the point P(0,1,2) are given. The perpendicular distance of the origin from the plane that contains L and…

Single correct option

The line

L:\ \dfrac{x-1}{2}=\dfrac{y}{1}=\dfrac{z+1}{-1}

and the point

P(0,1,2)

are given. The perpendicular distance of the origin from the plane that contains

L

and the point

P

A. $\dfrac{1}{10}$
B. $\dfrac{\sqrt2}{10}$
C. $\dfrac15$
D. $\dfrac{\sqrt2}{5}$

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Answer: (B) $\dfrac{\sqrt2}{10}$

The plane contains the point

A(1,0,-1)

L

with direction

\vec d=(2,1,-1)

, and also

P

. A normal is

\vec n=\vec d\times\overrightarrow{AP}

with

\overrightarrow{AP}=(-1,1,3)

. Computing,

\vec n=(1\cdot3-(-1)\cdot1,\ (-1)(-1)-2\cdot3,\ 2\cdot1-1\cdot(-1))=(4,-5,3)

. The plane is

4x-5y+3z=d_0

; using

A

4-0-3=1

, so the plane is

4x-5y+3z=1

, and

P

satisfies it (

-5+6=1

). Distance from the origin is

\dfrac{|1|}{\sqrt{16+25+9}}=\dfrac{1}{5\sqrt2}=\dfrac{\sqrt2}{10}

. Always confirm the point P also satisfies the final plane equation, which guards against sign mistakes in the cross product and in the constant term.

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