The position vectors of A, B,C and D are `vec a , vec b` , `vec 2a+ vec 3b` and `vec a - vec 2b` respectively show that `vec DB=3 vec b -vec a` and `vec AC =vec a + vec 3b`

The position vectors of A,B,C and D are vec a,vec b,vec 2a+vec 3b and vec a-vec 2b respectively show that vec DB=3vec b-vec a and vec AC=vec a+vec 3b

[vec a + vec b, vec b + vec c, vec c + vec a] = 2 [vec a, vec b, vec c]

The position vector of A,B are vec a and vec b respectively.The position vector of C is (5(vec a)/(3))-vec b then

If vec a,vec b and vec c be any three vectors then show that vec a+(vec b+vec c)=(vec a+vec b)+vec c

If |veca| = 1, |vec b| = 2, |vec c| = 3 and vec a +vec b +vec c = 0 , then the value of vec a * vec b +vec b. vec c +vec c* vec a equals

If the position vectors of three points are vec a-2vec b+3vec c,2vec a+3vec b-4vec c,-7vec b+10vec c then the three points are

[vec a+vec b,vec b+vec c,vec c+vec a]=2[vec a,vec b,vec c]

Statement 1: vec a , vec b ,a n d vec c are three mutually perpendicular unit vectors and vec d is a vector such that vec a , vec b , vec ca n d vec d are non-coplanar. If [ vec d vec b vec c]=[ vec d vec a vec b]=[ vec d vec c vec a]=1,t h e n vec d= vec a+ vec b+ vec c Statement 2: [ vec d vec b vec c]=[ vec d vec a vec b]=[ vec d vec c vec a] =>vec d is equally inclined to veca,vecb,vecc.

vec a * {(vec b + vec c) xx (vec a + 2vec b + 3vec c)} = [vec with bvec c]

If? B, c are three vectors such that | vec a | = 5, | vec b | = 12 and | vec c | = 13 and vec a + vec b + vec c = 0 then vec a * vec b + vec b * vec c + vec c * vec a