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

ધારો કે બિંદુ \((-1, \alpha, \beta)\) એ રેખાઓ \(\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}\) અને \(\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}\) વચ્ચેના ન્યૂનતમ અંતર વાળી રેખા પર આવેલ છે. તો \((\alpha-\beta)^2=\) .........

  1. A \(65\)
  2. B \(45\)
  3. C \(32\)
  4. D \(25\)
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Correct Answer

(D) \(25\)

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\( \mathrm{P}(-3 \lambda-2,4 \lambda+2,2 \lambda+5) \) \( \mathrm{Q}(-\mu-2,2 \mu-6,1) \) \( \mathrm{DRS} \text { of } \mathrm{PQ}=(3 \lambda-\mu, 2 \mu-4 \lambda-8,-2 \lambda-4)\) \(D R S\) of…
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