(i) If ` vec alpha +vec beta +vec gamma = 0 , prove that vec alpha× vec beta= vec beta x vec gamma= vec gamma x vec alpha`

If vec alpha+vec beta+vec gamma=0, prove the given equation

Let vec alpha=2i+3j-k and vec beta=i+j. If vec gamma is a unit vector,then the maximum value of [vec alpha xxvec beta,vec beta xxvec gamma,vec gamma xxvec alpha] is equal to

If |vec alpha + vec beta |=| vec alpha - vec beta|, then :

if vec alpha|(vec beta xxvec gamma), then (vec alpha xx beta)*(vec alpha xxvec gamma) equals to |vec alpha|^(2)(vec beta*vec gamma) b.|vec beta|^(2)(vec gamma*vec alpha) c.|vec gamma|^(2)(vec alpha*vec beta) d.|vec alpha||vec beta||vec gamma|

If vec alpha, vec beta, vecgamma be three unit vectors such that |vec alpha+vec beta+vec gamma|=1 and vec alpha is perpendicular to vec beta while vec gamma makes angles theta and phi with vec alpha and vec beta respectively, then the value of cos theta+cos phi is -

Statement 1: Let V_(1) be the volume of a parallelopiped ABCDEF having veca, vecb, vecc as three coterminous edges and V_(2) be the volume of the parallelopiped PQRSTU having three coterminous edges as vectors whose magnitudes are equal to the areas of three adjacent faces of the parallelopiped ABCDEF . Then V_(2)=2V_(1)^(2) Statement 2: For any three vectors vec(alpha), vec(beta), vec(gamma) [vec(alpha)xxvec(beta),vec(beta)xxvec(gamma),vec(gamma)xxvec(alpha)]=[(vec(alpha),vec(beta),vec(gamma))]^(2)

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 vector equation of the plane passing through vec a ,\ vec b , vec c\ i s\ vec r=alpha vec a+beta vec b+gamma vec c provided that a. alpha+beta+gamma=0 b. alpha+beta+gamma=1 c. alpha+beta=gamma d. alpha^2+beta^2+gamma^2=1