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)=`

In a quadrilateral ABCD, vec(AB) + vec(DC) =

If ABCD is a quadrilateral, then vec(BA) + vec(BC)+vec(CD) + vec(DA)=

In a parallelogram ABCD. Prove that vec(AC)+ vec (BD) = 2 vec(BC)

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

If vec(AB) = vecb and vec(AC) =vecc then the length of the perpendicular from A to the line BC is

veca+vecb+vecc=vec0, |veca|=3, |vecb|=5,|vecc|=9 ,find the angle between veca and vecc .

Ab, AC and AD are three adjacent edges of a parallelpiped. The diagonal of the praallelepiped passing through A and direqcted away from it is vector veca . The vector of the faces containing vertices A, B , C and A, B, D are vecb and vecc , respectively , i.e. vec(AB) xx vec(AC)=vecb and vec(AD) xx vec(AB) = vecc the projection of each edge AB and AC on diagonal vector veca is |veca|/3 vector vec(AB) is

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

If ABCDE is a pentagon, then vec(AB) + vec(AE) + vec(BC) + vec(DC) + vec(ED) + vec(AC) is equal to