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

\(\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}\) અને \(\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}\) વચ્ચેનું લઘુત્તમ અંતર \(\dots\dots\dots\)છે.

  1. A \(\frac{18}{\sqrt{5}}\)
  2. B \(\frac{22}{3 \sqrt{5}}\)
  3. C \(\frac{46}{3 \sqrt{5}}\)
  4. D \(6 \sqrt{3}\)
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Answer & Solution

Correct Answer

(A) \(\frac{18}{\sqrt{5}}\)

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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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