If the position vectors of the vertices of a triangle are `vec(O), vec(a) and vec(b) and Delta` is its area, then prove that `4 Delta^(2)= |vec(a)|^(2) |vec(b)|^(2)- (vec(a).vec(b))^(2)`.

If A, B, C are the vertices of a triangle then vec(A)B + vec(B)C + vec(C)A =

If |vec(a)| = |vec(b)|=2 , then (vec(a) xx vec(b))^(2) + (vec(a).vec(b))^(2)=

If D is the midpoint of the side BC of Delta ABC then vec(A)B + vec(A)C =

Observe the following statements: Assertion (A) : If vertices of a triangle are A(vec(a)), B(vec(b)), C(vec(c )) , then length of altitude through A is (|vec(a) xx vec(b) + vec(b) xx vec(c)+ vec(c ) xx vec(a)|)/(|vec(b)-vec(c )|) Reason (R ): Area of triangle is Delta = (1)/(2) (Base) (Height) = (1)/(2) |vec(AB) xx vec(AC)|

If vec(a) + 2vec(b) + 4vec(c ) = vec(0) then show that vec(a) xx vec(b) =4 (vec(b) xx vec(c ))= 2(vec(c ) xx vec(a))

If vec(a) and vec(b) are unit vectors, then the vectors (vec(a) + vec(b)) xx (vec(a) xx (vec(b)) is parallel to the vector.

If |vec(a)|=2, |vec(b)|=4 , then (|vec(a) xx vec(b)|^(2))/(1-cos^(2)(vec(a), vec(b)))=

vec(a) and vec(b) are non-zero vectors such that vec(a) xx vec(b)= vec(b) xx vec(a) " then" vec(a), vec(b) are

If vec(a), vec(b) are unit vectors satisfying (5vec(a) + 3vec(b)) xx (3vec(a) -7vec(b))=vec(0) , then (vec(a), vec(b)) is