A point O is the centre of a circle circunscribed about a triangle ABC. Then , `vec OA sin 2A + bvec OB sin 2B + vec OC sin 2C ` is equal to

Options

A. `(vec(OA) + vec(OB) + vec(OC)) sin 2A`
B. ` 3 vec(OG)`, where G is the centroid of triangle ABC
C. `vec0`
D. none of these

Answer

C

Explanation

The position vector of the point O with respect to itself is
`(vec(OA) sin 2A + vec(OB) sin2B + vec(OC) sin 2C)/(sin 2A + sin 2B + sin 2C)`
`rArr (vec(OA) sin 2A + vec(OB) sin 2B + vec(OC) sin 2C)/( sin2A + sin 2B + sin 2C) = vec0`
or `" "vec(OA) sin 2A + vec(OB) sin 2B + vec(OC) sin 2C = vec0`