Show that: `(veca+vecb).{(vecb+vecc)xx(vecc+veca)|=2{veca.(vecbxxvecc)}`

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

If veca,vecb,vecc are linearly independent vectors, then ((veca+2vecb)xx(2vecb+vecc).(5vecc+veca))/(veca.(vecbxxvecc)) is equal to

Show that: (veca-vecd)xx(vecb-vecc)+(vecb-vecd)xx(vecc-veca)+(vecc-vecd)xx(veca-vecb) is independent of vecd .

Prove that (vecbxxvecc)xx(veccxxveca)=[veca vecb vecc]vecc

Prove that (vecbxxvecc)xx(veccxxveca)=[veca vecb vecc]vecc

Prove that [ veca+ vecb , vecb+ vecc , vecc+ veca]=2[ veca , vecb , vecc] .

Prove that [veca+vecb vecb+vecc vecc+veca]=2[veca vecb vecc]

Let veca,vecb, vecc be any three vectors, Statement 1: [(veca+vecb, vecb+vecc,vecc+veca)]=2[(veca, vecb, vecc)] Statement 2: [(vecaxxvecb, vecbxxvecc, veccxxveca)]=[(veca, vecb, vecc)]^(2)

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

If vecb and vecc are two non-collinear such that veca ||(vecbxxvecc) . Then prove that (vecaxxvecb).(vecaxxvecc) is equal to |veca|^(2)(vecb.vecc)