Question:
Let Q be the foot of perpendicular drawn from the point P(1, 2, 3) to the plane x + 2y + z = 14. If R is a point on the plane such that ∠PRQ = 60°, then the area of ΔPQR is equal to :
Let Q be the foot of perpendicular drawn from the point P(1, 2, 3) to the plane x + 2y + z = 14. If R is a point on the plane such that ∠PRQ = 60°, then the area of ΔPQR is equal to :
Updated On: Sep 24, 2024
\(\sqrt{\frac{3}{2}}\)
\(\sqrt3\)
\(2\sqrt3\)
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The Correct Option is B
Solution and Explanation
PQ=|\(\frac{1+4+3−14}{\sqrt6}\)|=√6
QR=\(\frac{PQ}{tan60^{\circ}}\)=\(\frac{\sqrt6}{\sqrt3}\)=\(\sqrt2\)
Area(ΔPQR)=\(\frac{1}{2}\)⋅PQ⋅QR=\(\sqrt3\)
So, the correct option is (B): \(\sqrt3\)
QR=\(\frac{PQ}{tan60^{\circ}}\)=\(\frac{\sqrt6}{\sqrt3}\)=\(\sqrt2\)
Area(ΔPQR)=\(\frac{1}{2}\)⋅PQ⋅QR=\(\sqrt3\)
So, the correct option is (B): \(\sqrt3\)
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