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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

  1. A  \(6 \sqrt{3}\)
  2. B  \(4 \sqrt{3}\)
  3. C \(5 \sqrt{3}\)
  4. D  \(8 \sqrt{3}\)
Verified Solution

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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