If ` vecalpha+ vecbeta+ vecgamma=a vecdeltaa n d vecbeta+ vecgamma+ vecdelta=b vecalpha, vecalphaa n d vecdelta` are non-colliner, then ` vecalpha+ vecbeta+ vecgamma+ vecdelta` equals a. `a vecalpha` b. `b vecdelta` c. `0` d. `(a+b) vecgamma`

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

let vecalpha=i+2j-k , vecbeta=2i-j+3k , vecgamma=2i+j+6k be three vectors . fi vecalpha and vecbeta are both perpendicular to the vecdelta and vecdelta.vecgamma=10, then what is the magnitude of vecdelta

vecalpha and vecbeta are two unit vectos and vecr is a vector such that vecr,vecalpha=0 and sqrt(2)(vecrxxvecbeta)=3(vecrxxvecalpha)-vecbeta , then (1)/(|vecr|^(2)) equal

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|

If a(vecalphaxxvecbeta)=b(vecbetaxxvecgamma)+c(vecgammaxxvecalpha)=vec0 and at least one of a,b and c is non zero then vectors vecalpha, vecbeta, vecgamma are (A) parallel (B) coplanar (C) mutually perpendicular (D) none of these

Statement 1: vec a , vec b ,a n d vec c are three mutually perpendicular unit vectors and vec d is a vector such that vec a , vec b , vec ca n d vec d are non-coplanar. If [ vec d vec b vec c]=[ vec d vec a vec b]=[ vec d vec c vec a]=1,t h e n vec d= vec a+ vec b+ vec c Statement 2: [ vec d vec b vec c]=[ vec d vec a vec b]=[ vec d vec c vec a] =>vec d is equally inclined to veca,vecb,vecc.

Findthe value of vecalphaxx(vecbetaxxvecgamma) , where, vecalpha=2veci-10vecj+2veck, vecbeta=3veci+vecj+2veck, vecgamma =2veci+vecj+3veck

If veca,vecb,vecc are mutually perpendicular vector and veca=alpha(vecaxxvecb)+beta(vecbxxvecc)+gamma(veccxxveca) and [veca vecb vecc]=1 then vecalpha+vecbeta+vecgamma= (A) |veca|^2 (B) -|veca|^2 (C) 0 (D) none of these

If vecalpha is a constant vectro and vecgamma is the position vector of a variable point (x,y,z), show that (vecgamma-vecalpha) vecalpha =0 is the equation of a plane through fixed point vec(alpha)

Given three vectors vec a=6 hat i-3 hat j , vec b=2 hat i-6 hat ja n d vec c=-2 hat i+21 hat j such that vecalpha= vec a+ vec b+ vec c Then the resolution of the vector vecalpha into components with respect to vec aa n d vec b is given by a. 3 vec a-2 vec b b. 3 vec b-2 vec a c. 2 vec a-3 vec b d. vec a-2 vec b