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

અહી \(Q\) એ બિંદુ \(P (1,2,3)\) નો સમતલ \(x +2 y + z =14\) પરનો લંબપાદ છે. જો \(R\) એ સમતલ પર છે કે જેથી \(\angle PRQ =60^{\circ}\) હોય તો  \(\triangle PQR\) નું ક્ષેત્રફળ મેળવો.

  1. A \(\frac{\sqrt{3}}{2}\)
  2. B \(\sqrt{3}\)
  3. C \(2 \sqrt{3}\)
  4. D \(3\)
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Answer & Solution

Correct Answer

(B) \(\sqrt{3}\)

Step-by-step Solution

Detailed explanation

\(x+2 y+z=14\) Length of perpendicular \(PQ =\left|\frac{1+4+3-14}{\sqrt{6}}\right|=\sqrt{6}\) \(QR =( PQ ) \cot 60^{\circ}=\sqrt{2}\) \(\therefore\) Area of \(\triangle PQR =\frac{1}{2}( PQ )( QR )=\sqrt{3}\)
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