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

ધારો કે રેખાઓ \(L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0\) અને \(L_1: x+1=y-1=4-z\) વચ્ચેનું લધુતમ અંતર \(2 \sqrt{6}\) છે.જો \((\alpha, \beta, \gamma)\) એ \(L\) પર હોય, તો નીચેનાં પૈકી કયું શક્ય નથી ?

  1. A \(\alpha+2 \gamma=24\)
  2. B \(2 \alpha+\gamma=7\)
  3. C \(2 \alpha-\gamma=9\)
  4. D \(\alpha-2 \gamma=19\)
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Answer & Solution

Correct Answer

(A) \(\alpha+2 \gamma=24\)

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\(\overline{ b _1} \times \overline{ b _2}=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ -2 & 0 & 1 \\ 1 & 1 & -1\end{array}\right|=-\hat{ i }-\hat{ j }-2 \hat{ k }\)…
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