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JEE Mains · Maths · STD 12 - 11. three dimension geometry

Shortest distance between the lines \(\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}\) and \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}\) is

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

Answer & Solution

Correct Answer

(B) \(4 \sqrt{3}\)

Step-by-step Solution

Detailed explanation

\(\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5} \quad \vec{a}=\hat{i}-8 \hat{j}+4 \hat{k}\) \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3} \quad \vec{b}=\hat{i}+2 \hat{j}+6 \hat{k}\) \(\vec{p}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \overrightarrow{ q }=2 \hat{i}+\hat{j}-3 \hat{k}\)…
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