given that `veca. vecb = veca.vecc, veca xx vecb= veca xx vecc and veca ` is not a zero vector. Show that `vecb=vecc`.

[ veca + vecb vecb + vecc vecc + veca ]=[ veca vecb vecc ] , then

if veca + vecb + vecc=0 , then show that veca xx vecb = vecb xx vecc = vecc xx veca .

If veca , vecb and vecc are three vectors such that vecaxx vecb =vecc, vecb xx vecc= veca, vecc xx veca =vecb then prove that |veca|= |vecb|=|vecc|

If veca + 2 vecb + 3 vecc = vec0 " then " veca xx vecb + vecb xx vecc + vecc xx veca=

If veca, vecb,vecc are three non-coplanar vectors such that veca xx vecb=vecc,vecb xx vecc=veca,vecc xx veca=vecb , then the value of |veca|+|vecb|+|vecc| is

If veca, vecb, vecc and vecd ar distinct vectors such that veca xx vecc = vecb xx vecd and veca xx vecb = vecc xx vecd . Prove that (veca-vecd).(vecc-vecb)ne 0, i.e., veca.vecb + vecd.vecc nevecd.vecb + veca.vecc.

If veca, vecb, vecc are unit vectors such that veca. vecb =0 = veca.vecc and the angle between vecb and vecc is pi/3 , then find the value of |veca xx vecb -veca xx vecc|

for any three vectors, veca, vecb and vecc , (veca-vecb) . (vecb -vecc) xx (vecc -veca) = 2 veca.vecb xx vecc .

If veca, vecb and vecc are three non-coplanar non-zero vectors, then prove that (veca.veca) vecb xx vecc + (veca.vecb) vecc xx veca + (veca.vecc)veca xx vecb = [vecb vecc veca] veca

Prove that veca. [(vecb + vecc) xx (veca + 3 vecb + 4 vecc) ]= [{:(veca,vecb,vecc):}]