JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between the lines \(\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}\) and \(\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}\) is
- A \(\frac{18}{\sqrt{5}}\)
- B \(\frac{22}{3 \sqrt{5}}\)
- C \(\frac{46}{3 \sqrt{5}}\)
- D \(6 \sqrt{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{18}{\sqrt{5}}\)
Step-by-step Solution
Detailed explanation
\(\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}\) \(\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}\) \(A =(3,2,1) \quad B =(-3,6,5)\) \(\overline{n_{1}}=2 \hat{i}+3 \hat{j}-\hat{k}\) \(\overrightarrow{ n _{2}}=2 \hat{ i }+\hat{ j }-3 \hat{ k }\)…
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