In quadrilateral `ABCD, vec(AB)=veca, vec(BC)=vecb, vec(AD)=vecb-veca` If `M` is the mid point of `BC` and `N` is a point on `DM` such that `DN=4/5 DM`, then `vec(AN)=`

ABCD is a quadrilateral with vec(AB) = veca, vec(AD) = vecb and vec(AC)=2veca+3vecb. If its area is alpha times the area of the parallelogram with AB and AD as adjacent sides, then alpha=

If vec A . vecB = vecA*vecB , find |vecA - vecB|

If ABCDEF is a regular hexagon with vec(AB) = veca and vec(BC)= vecb, then vec(CE) equals

If D is the mid-point of the side BC of a triangle ABC, prove that vec AB+vec AC=2vec AD .

Prove that : veca*(vecb+vec c)xx(veca+2vecb+3vec c)=[veca vecb vec c]

In the figure, M is the mid-point of [AB] and N is the mid-point of [CD] and O is the mid-point of [MN]. Prove that : vec(BC)+vec(AD)=2vec(MN) .

If [veca+vecb, vecb+vec c, vec c+veca]=8 then [veca, vecb, vec c] is

If veca .vecb = veca.vec cand veca xx vecb = veca xx vec c,veca ne0 , then

Prove that veca*(vecb+vec c)xx (veca+3vecb+2vec c)=-(veca vecb vecc )