JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between the lines \(\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}\) and \(\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}\) is
- A \(\frac{7}{2} \sqrt{30}\)
- B \(3\sqrt{30}\)
- C \(3\)
- D \({2} \sqrt{30}\)
Answer & Solution
Correct Answer
(B) \(3\sqrt{30}\)
Step-by-step Solution
Detailed explanation
Shortest distance \(=\frac{\left|\begin{array}{ccc}{6} & {15} & {-3} \\ {3} & {-1} & {1} \\ {-3} & {2} & {4}\end{array}\right|}{\sqrt{11 \times 29-49}}=\frac{270}{\sqrt{270}}\) \(=\sqrt{270}=3 \sqrt{30}\)
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