JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between the lines \(\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}\) and \(\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}\) is
- A \(6 \sqrt{3}\)
- B \(4 \sqrt{3}\)
- C \(5 \sqrt{3}\)
- D \(8 \sqrt{3}\)
Answer & Solution
Correct Answer
(B) \(4 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5} \& \frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}\)…
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