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JEE Advanced 2008Paper 2

Exam: JEE Advanced
Year: 2008
Paper: Paper 2
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Advanced 2008 · Paper 2 · Mathematics · Three Dimensional Geometry

Paragraph: Consider the lines: L_1: (x+1)/(3)=(y+2)/(1)=(z+1)/(2),...

Paragraph: Consider the lines:

L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}, L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}

Question: The distance of the point

(1,1,1)

from the plane passing through the point

(-1,-2,-1)

and whose normal is perpendicular to both the lines

L_1

and

L_2

, is

Options

A $2 / \sqrt{75}$
B $7 / \sqrt{75}$
C $13 / \sqrt{75}$
D $23 / \sqrt{75}$

Correct answer

C. $13 / \sqrt{75}$

Step-by-step solution

The equation of the plane passing through the point

(-1,-2,-1)

and whose normal is perpendicular to both the given lines

L_1 and L_2

may be written as

\begin{array}{rlrl} \Rightarrow \quad(x+1)+7(y+2)-5(z+1) & =0 \\ & x+7 y-5 z+10 & =0 \end{array}

The distance of the point(1,1,1) from the plane =

\left|\frac{1+7-5+10}{\sqrt{1+49+25}}\right|=\frac{13}{\sqrt{75}}

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