Solved `lamdavecr+(veca.vecr)vecb=vecc,lamda!=0`

For non zero vectors veca,vecb,vecc|(vecaxxvecb).vecc|=|veca||vecb||vec| holds if and only if (A) veca.vecb=0,vecb.vecc=0 (B) vecb.vecc=0,vecc.veca=0 (C) vecc.veca=0,veca.vecb=0 (D) veca.vecb=vecb.vecc=vecc.veca=0

Solve veca.vecr=x, vecb.vecr=y, vecc.vecr=z , where veca,vecb,vecc are given non coplanar vectors.

If [(2veca+4vecb,vecc,vecd)]=lamda[(veca,vecc,vecd)]+mu[(vecb,vecc,vecd)] , then lamda+mu=

If vecaxxvecb=veccxxvecd and vecaxxvecc=vecbxxvecd then (A) (veca-vecd)=lamda(vecb-vecc) (B) veca+vecd=lamda(vecb+vecc) (C) (veca-vecb)=lamda(vecc+vecd) (D) none of these

If vecP = (vecbxxvecc)/([vecavecbvecc]).vecq=(veccxxveca)/([veca vecb vecc])and vecr = (vecaxxvecb)/([veca vecbvecc]), " where " veca,vecb and vecc are three non- coplanar vectors then the value of the expression (veca + vecb + vecc ). (vecq+ vecq+vecr) is

The two lines vecr=veca+veclamda(vecbxxvecc) and vecr=vecb+mu(veccxxveca) intersect at a point where veclamda and mu are scalars then (A) veca,vecb,vecc are non coplanar (B) |veca|=|vecb|=|vecc| (C) veca.vecc=vecb.vecc (D) lamda(vecb xxvecc)+mu(vecc xxveca)=vecc

If veca, vecb, vecc are non coplanar vectors and lamda is a real number, then [(lamda(veca+vecb), lamda^(2)vecb, lamdavecc)]=[(veca, vecb+vecc,vecb)] for

Statement 1: If the vectors veca and vecc are non collinear, then the lines vecr=6veca-vecc+lamda(2vecc-veca) and vecr=veca-vecc+mu(veca+3vecc) are coplanar. Statement 2: There exists lamda and mu such that the two values of vecr in statement -1 become same