JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(Q\) be the foot of perpendicular drawn from the point \(P (1,2,3)\) to the plane \(x +2 y + z =14\). If \(R\) is a point on the plane such that \(\angle PRQ =60^{\circ}\), then the area of \(\triangle PQR\) is equal to.
- A \(\frac{\sqrt{3}}{2}\)
- B \(\sqrt{3}\)
- C \(2 \sqrt{3}\)
- D \(3\)
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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