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 :
\(\sqrt{\frac{3}{2}}\)
\(\sqrt3\)
\(2\sqrt3\)
- 3
The Correct Option is B
Solution and Explanation
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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Concepts Used:
Distance of a Point from a Plane
The shortest perpendicular distance from the point to the given plane is the distance between point and plane. In simple terms, the shortest distance from a point to a plane is the length of the perpendicular parallel to the normal vector dropped from the particular point to the particular plane. Let's see the formula for the distance between point and plane.
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