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

રેખાઓ \(\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}\) અન \(\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}\) વચ્યેનું ન્યૂનતમ અંતર \(...........\) છે.

  1. A \(3 \sqrt{6}\)
  2. B \(6 \sqrt{3}\)
  3. C \(6 \sqrt{2}\)
  4. D \(2 \sqrt{6}\)
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Answer & Solution

Correct Answer

(A) \(3 \sqrt{6}\)

Step-by-step Solution

Detailed explanation

\(S _{ d }=\left|\frac{(\overrightarrow{ a }-\overrightarrow{ b }) \times\left(\overrightarrow{ n }_1 \times \overrightarrow{ n }_2\right)}{\left|\overrightarrow{ n }_1 \times \overrightarrow{ n }_2\right|}\right|\) \(\overline{ a }=(4,-2,-3)\) \(\overline{ b }=(1,3,4)\)…
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