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 shortest distance between \(L_1\) and \(L_2\) is
- A
0
- B
\(17 / \sqrt{3}\)
- C
\(41 / 5 \sqrt{3}\)
- D
\(17 / 5 \sqrt{3}\)
Answer & Solution
Correct Answer
(D)
\(17 / 5 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
The shortest distance between \(L_1\) and \(L_2\) is
\[
\begin{array}{r}
\left|\frac{\{(2-(-1)) \hat{\mathbf{i}}+(2-2) \hat{\mathbf{j}}+(3-(-1)) \hat{\mathbf{k}}\} \cdot(-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{5 \sqrt{3}}\right| \\
=\left|\frac{(3 \hat{\mathbf{i}}+4 \hat{\mathbf{k}}) \cdot(-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{5 \sqrt{3}}\right|=\frac{17}{5 \sqrt{3}}
\end{array}
\]
\[
\begin{array}{r}
\left|\frac{\{(2-(-1)) \hat{\mathbf{i}}+(2-2) \hat{\mathbf{j}}+(3-(-1)) \hat{\mathbf{k}}\} \cdot(-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{5 \sqrt{3}}\right| \\
=\left|\frac{(3 \hat{\mathbf{i}}+4 \hat{\mathbf{k}}) \cdot(-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{5 \sqrt{3}}\right|=\frac{17}{5 \sqrt{3}}
\end{array}
\]
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