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

જેના માટે રેખાઓ \(\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}\) તથા \(\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2 \alpha}\) વચ્ચેનું ન્યૂનતમ અંતર \(\sqrt{2}\) હોય, તેવા \(\alpha\) ના તમામ મૂલ્યોનો સરવાળો ___ છે.

  1. A 8
  2. B -6
  3. C 6
  4. D -8
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Answer & Solution

Correct Answer

(B) -6

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Detailed explanation

\(\sqrt{2}=\frac{\left|\begin{array}{ccc}-1 & 1 & 3 \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2 \alpha\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & j & k \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2 \alpha\end{array}\right|}\)…
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