The resolved part of the vector `veca` along the vector `vecb is veclamda` and that perpendicular to `vecb is vecmu`. Then (A) `veclamda=((veca.vecb).veca)/veca^2` (B) `veclamda=((veca.vecb).vecb)/vecb^2` (C) `vecmu=((vecb.vecb0veca-(veca.vecb)vecb)/vecb^2` (D) `vecmu=(vecbxx(vecaxxvecb))/vecb^2`

If vecA+vecB=vecR and 2vecA+vecB is perpendicular to vecB then

If vecc=vecaxxvecb and vecb=veccxxveca then (A) veca.vecb=vecc^2 (B) vecc.veca.=vecb^2 (C) veca_|_vecb (D) veca||vecbxxvecc

If vectors veca and vecb are two adjacent sides of parallelograsm then the vector representing the altitude of the parallelogram which is perpendicular to veca is (A) vecb+(vecbxxveca)/(|veca|^2) (B) (veca.vecb)/(vecb|^2) (C) vecb-(vecb.veca)/(|veca|)^2) (D) (vecaxx(vecbxxveca))/(vecb|^20

If vecax(vecaxxvecb)=vecbxx(vecbxxvecc) and veca.vecb!=0 , and [(veca,vecb,vecc)]=

The vectors veca and vecb are not perpendicular and vecac and vecd are two vectors satisfying : vecbxxvecc=vecbxxvecd and veca.vecd=0. Then the vecd is equal to (A) vecc+(veca.vecc)/(veca.vecb)vecb (B) vecb+(vecb.vecc)/(veca.vecb)vecc (C) vecc-(veca.vecc)/(veca.vecb)vecb (D) vecb-(vecb.vecc)/(veca.vecb)vecc

Show that (veca xx vecb)^(2) = |veca| ^(2) |vecb|^(2) - (veca.vecb)^(2) = |(veca.veca)/(veca. vecb)(veca.vecb)/(vecb.vecb)|

If veca is perpendicular to vecb then the vector vecaxx[vecaxx{vecaxx(vecaxxvecb)}] is equla (A) |veca|^2vecb (B) |veca|vecb (C) |veca|^3vecb (D) |veca|^4vecb

l veca . m vecb = lm(veca . vecb) and veca . vecb = 0 then veca and vecb are perpendicular if veca and vecb are not null vector

The vector (veca-vecb)xx(veca+vecb) is equal to (A) 1/2 (vecaxxvecb) (B) vecaxxvecb (C) 2(veca+vecb) (D) 2(vecaxxvecb)