Consider a base `veca, vecb, vecc and a` vector `-2veca + 3vecb - vecc` Compute the co-ordinates of this vector relatively to the base `vecp, vecq, vecr` where `vec p=2vec a-3evc b, vec q=vec a-2vec b+2vec b+vec c,vec r=-3vec a+vec b+2vec c.`

Consider a base vec a,vec b,vec c and a vector -2vec a+3vec b-vec c compute the co-ordinates of this vector relatively to the base vec p,vec q,vec r where vec p=2vec a-3evcb,vec q=vec a-2vec b+2vec b+vec r,vec r=-3vec a+vec b+2vec c

Show that for any three vectors veca , vec b and vec c [ vec a + vec b, vecb + vec c , vec c + vec a ] =2[vec a , vec b , vecc] .

If veca,vecb,vecc are coplanar vectors then find value of [veca-vecb+vec2c vecb-vec c+2veca veca+2vecb-vec c]

If veca,vecb,vecc are coplanar vectors then find value of [veca-vecb+vec2c vecb-vec c+2veca veca+2vecb-vec c]

veca,vecb and vecc are three non-coplanar vectors and r is any arbitrary vector. Prove that [[vecb, vecc,vec r]]veca + [[vecc, veca, vecr]]vecb +[[veca,vec b,vec r]]vecc = [[veca,vec b, vecc]]vecr

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

If vec a, vec b, vec c are unit vectors such that vec a+ vec b+ vec c = vec 0 , then find the value of vec a* vecb + vec b* vec c +vec c* veca .

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

Let veca , vecb, vec c be three non coplanar vectors , and let vecp , vecq " and " vec r be the vectors defined by the relation vecp = (vecb xx vec c )/([veca vecb vec c ]), vec q = (vec c xx vec a)/([veca vecb vec c ]) " and " vec r = (vec a xx vec b)/([veca vecb vec c ]) Then the value of the expension (vec a + vec b) .vec p + (vecb + vec c) .q + (vec c + vec a) . vec r is equal to

If vec a, vec b, vec c are linearly independent vectors and Delta = |(vec a, vec b, vec c), (vec a. vec a, vec a.vec b, veca . vec c), (vec a. vec c, vec b . vecc, vec c. vec c)| then ,