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 -

Let vec(a), vec(b), vec(c) be three vectors in the xyz space such that

1.	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 -
Answer» 1 2 3 6 Solution :`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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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 -

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