JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between the lines \(\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}\) and \(\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}\) is equal to \(............\)
- A \(13\)
- B \(12\)
- C \(14\)
- D \(16\)
Answer & Solution
Correct Answer
(C) \(14\)
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Shortest distance between the lines…
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