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

ધારોકે બિંદુ \(A (4,3,1)\) થી સમતલ \(P : x-y+2 z+3=0\) પરનો લંબપપાદ \(N\) છે.જો \(B (5\), \(\alpha, \beta), \alpha, \beta \in Z\) એ સમતલ \(P\) પરનું એવું બિંદું હોય કે જેથી ત્રિકોણ \(ABN\) નું ક્ષેત્રફળ \(3 \sqrt{2}\) થાય,તો \(\alpha^2+\beta^2+\alpha \beta=.........\)

  1. A \(6\)
  2. B \(5\)
  3. C \(7\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(7\)

Step-by-step Solution

Detailed explanation

\(AN =\sqrt{6}\) \(5-\alpha+2 \beta+3=0\) \(\alpha=8+2 \beta\) \(N\) is given by \(\frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}=\frac{-(4-3+2+3)}{1+1+4}\) \(x =3, y =4, z =-1\) \(N \text { is }(3,4,-1)\) \(BN =\sqrt{4+(\alpha-4)^2+(\beta+1)^2}\)…
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