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

રેખાઓ \(\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}\) અને \(\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}\) વચ્ચેનું લઘુતમ અંતર જે રેખા છે તે સમતલ \(P: a x-y-\) \(z=0\), \((a>0)\) સાથે \(\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)\) નો ખૂણો બનાવે છે. જો બિંદુ \((1,1,-5)\) નું સમતલ \(P\) માં પ્રતિબિંબ \((\alpha, \beta, \gamma)\) હોય તો \(\alpha+\beta-\gamma\) ની કિમંત  \(........\) થાય.

  1. A \(4\)
  2. B \(5\)
  3. C \(2\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(3\)

Step-by-step Solution

Detailed explanation

\(DR's\) of line of shortest distance \(\left|\begin{array}{lll}\hat{ i } & \hat{ j } & \hat{ k } \\0 & 1 & 1 \\2 & 2 & 1\end{array}\right|=-\hat{ i }+2 \hat{ j }-2 \hat{ k }\)angle between line and plane is \(\cos ^{-1} \sqrt{\frac{2}{27}}=\alpha\)…
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