JEE Advanced · Mathematics · 31. 3D Geometry
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
- A
\(2 / \sqrt{75}\)
- B
\(7 / \sqrt{75}\)
- C
\(13 / \sqrt{75}\)
- D
\(23 / \sqrt{75}\)
Answer & Solution
Correct Answer
(C)
\(13 / \sqrt{75}\)
Step-by-step Solution
Detailed explanation
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
\[
=\left|\frac{1+7-5+10}{\sqrt{1+49+25}}\right|=\frac{13}{\sqrt{75}}
\]
\[
\begin{array}{rlrl}
\Rightarrow \quad(x+1)+7(y+2)-5(z+1) & =0 \\
& x+7 y-5 z+10 & =0
\end{array}
\]
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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