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

જો બે રેખાઓ \(l_{1}: \frac{ x -2}{3}=\frac{ y +1}{-2}, z =2\) અને \(l_{2}: \frac{x-1}{1}=\frac{2 y+3}{\alpha}=\frac{z+5}{2}\) પરસ્પર લંબ હોય,તો રેખાઓ \(l_{2}\) અને \(l_{3}: \frac{1- x }{3}=\frac{2 y -1}{-4}=\frac{ z }{4}\) વચ્ચેનો ખૂણો \(\dots\dots\dots\)છે.

  1. A \(\cos ^{-1}\left(\frac{29}{4}\right)\)
  2. B \(\sec ^{-1}\left(\frac{29}{4}\right)\)
  3. C \(\cos ^{-1}\left(\frac{2}{29}\right)\)
  4. D \(\cos ^{-1}\left(\frac{2}{\sqrt{29}}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sec ^{-1}\left(\frac{29}{4}\right)\)

Step-by-step Solution

Detailed explanation

\(l_{1}: \frac{ x -2}{3}=\frac{ y +1}{-2}=\frac{ z -2}{0}\) \(l_{2}: \frac{x-1}{1}=\frac{y+3 / 2}{\alpha / 2}=\frac{z+5}{2}\) \(l_{3}: \frac{x-1}{-3}=\frac{y-1 / 2}{-2}=\frac{z-0}{4}\)…
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