Give a condition that the three vectors `veca,vecb` and `vecc` form the three sides of a triangle. What are other possibilities ?

Three vectors vecA, vecB and vecC add up to zero. Find which is false.

Prove that the necessary and sufficient condition for three vectors veca,vecb and vecc to be coplanar is that there exist scalars l,m,n (not all zero simultaneously) such that lveca+mvecb+nvecc=vec0 .

Given three vectors veca, vecb and vecc are non-zero and non-coplanar vectors. Then which of the following are coplanar.

If V is the volume of the parallelepiped having three coterminous edges as veca,vecb and vecc , then the volume of the parallelepiped having three coterminous edges as vecalpha = (veca.veca)veca+(veca.vecb)vecb+(veca.vecc)vecc , vecbeta=(vecb.veca)veca+(vecb.vecb)+(vecb.vecc)vecc and veclambda=(vecc.veca)veca+(vecc.vecb)vecb+(vecc.vecc)vecc is

The non-zero vectors veca,vecb and vecc are related by veca=8vecb and vecc=-7vecb . Then the angle between veca and vecc is :

Projection of vector veca on other vector vecb is equal to :

If veca=hati+2hatj+2hatk, |vecb|=5 and the angle between veca and vceb is pi/6 , then the area of the triangle formed by these two vectors as two sides is :

For any three vectors veca,vecb,vecc show that veca-vecb,vecb-vecc,vecc-veca are coplanar.