The position vectors of A,B,C,D are` veca,b,2vec+3vecb and veca-2vecb` respectively show that `vec(DB)=3vecb - veca and vec(AC)=veca+3vecb`.

The position vector of foru points A,B,C,D are veca, vecb, 2veca+3vecb and veca-2vecb respectively. Expess the vectors vec(AC), vec(DB), vec(BC) and vec(CA) in terms of veca and vecb .

The position vectors of two points A and B are 3veca+2vecb and 2veca-vecb respectively. Find the vector vec(AB)

if veca , vecb ,vecc are three vectors such that veca +vecb + vecc = vec0 then

Find |veca-vecb| , if two vectors veca and vecb are such that |veca|=2,|vecb|=3 and veca.vecb=4 .

If veca , vecb are unit vectors such that |vec a+vecb|=-1 " then " |2veca -3vecb| =

If |veca+vecb|=|veca-vecb|, (veca,vecb!=vec0) show that the vectors veca and vecb are perpendicular to each other.

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

Show that the points having position vectors (veca-2vecb+3vecc),(-2veca+3vecb+2vecc),(-8veca+13vecb) re collinear whatever veca,vecb,vecc may be

Find |veca-vecb| , if two vectors veca and vecb are such that |veca| = 2,|vecb|=3 and veca.vecb = 4 .