Given that `(vecbeta-vecalpha).(vecbeta+vecalpha)=8` and `vecalpha.vecbeta=2` Also `|vec alpha|=1` then angle between `(vecbeta-vecalpha)` and `(vecbeta+vecalpha)` is

Given that (vec beta-vec alpha) * (vec beta + vec alpha) = 8 and vec alpha * vec beta = 2 Also | vec alpha | = 1 then angle between (vec beta-vec alpha) and (vec beta + vec alpha) is

The angle between 3(vec a xxvec b) and (1)/(2)(vec b-(a.b)vec a) is -

A parallelogram is constructed on the vectors veca=3vecalpha-vecbeta, vecb=vecalpha+3vecbeta . If |vecalpha|=|vecbeta|=2 and angle between vecalpha and vecbeta is pi/3 then the length of a diagonal of the parallelogram is (A) 4sqrt(5) (B) 4sqrt(3) (C) 4sqrt(7) (D) none of these

Given that (vec x-widehat a)(vec x+widehat a)=8 and vec x*widehat a=2, then the angle between (vec x-widehat a) and (vec x+widehat a) is:

If alpha+beta+gamma= a vecdelta and vecbeta+vecgamma+vecdelta = b vecalpha and alpha, vecbeta, vecgamma are non coplanar and vecalpha is not parallel to vecdelta then vecalpha+vecbeta+vecgamma+vecdelta equals (A) avecalpha (B) bvecdelta (C) 0 (D) (a+b)vecgamma

If vec a=(-1,1,2);vec b=(2,1,-1);vec c=(-2,1,3) then the angle between 2vec a-vec c and vec a+vec b is

If vecalpha||(vecbxxvecgamma), then (vecalphaxxvecbeta).(vecalphaxxvecgamma)= (A) |vecalpha|^2(vecbeta.vecgamma) (B) |vecbeta|^2(vecgamma.vecalpha) (C) |vecgamma|^2(vecalpha.vecbeta) (D) |vecalpha||vecbeta||vecgamma|