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Let vec(a), vec(b), vec(c) be three vect...

Let `vec(a), vec(b), vec(c)` be three vectors in the xyz space such that `vec(a)xxvec(b)=vec(b)xxvec(c)=vec(c)xx vec(a) ne 0` If A, B, C are points with position vector `vec(a), vec(b), vec(c)` respectively, then the number of possible position of the centroid of triangle ABC is -

A

1

B

2

C

3

D

6

Text Solution

Verified by Experts

The correct Answer is:
A
`vec(a)xxvec(b)+vec(c)xxvec(b)=0` similarly `vec(b)+vec(c)=lambda_(2) vec(a)`
`vec(a)+vec(c)=lambda_(1)vec(b)" "vec(b)+vec(a)=lambda_(3)vec(c)`
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