JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between the lines \(\frac{x+7}{-6}=\frac{y-6}{7}=z\) and \(\frac{7-x}{2}=y-2=z-6\) is
- A \(2 \sqrt{29}\)
- B \(1\)
- C \(\sqrt{\frac{37}{29}}\)
- D \(\frac{\sqrt{29}}{2}\)
Answer & Solution
Correct Answer
(A) \(2 \sqrt{29}\)
Step-by-step Solution
Detailed explanation
\(L_{1}: \frac{x+7}{-6}=\frac{y-6}{7}=\frac{z-0}{1}\) Any point on it \(\vec{a}_{1}(-7,6,0)\) and \(L_{1}\) is parallel to \(\vec{b}_{1}(-6,7,1)\) \(L_{2}: \frac{x-7}{-2}=\frac{y-2}{1}=\frac{z-6}{1}\) Any point on it, \(\vec{a}_{2}(7,2,6)\) and \(L_{2}\) is parallel to…
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