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

જો \(\mathrm{d}_1\) એ રેખાઓ \(x+1=2 y=-12 z, x=y+z=6 z-6\) વચ્ચેનું ન્યૂનતમ અંતર હોય તથા \(\mathrm{d}_2\) એ રેખાઓ \(\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}\) વચ્ચેનું ન્યુનતમ અંતર હોય, તો \(\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}\) નું મૂલ્ય ........... છે.

  1. A \(17\)
  2. B \(16\)
  3. C \(42\)
  4. D \(45\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(16\)

Step-by-step Solution

Detailed explanation

\(L_1: \frac{x+1}{1}=\frac{y}{1 / 2}=\frac{z}{-1 / 12}, L_2: \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}\) \(\mathrm{d}_1=\) shortest distance between \(\mathrm{L}_1 \& \mathrm{~L}_2\)…
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