Simplify : `(vecb+vecc)*{(vecc+veca)xx(veca+vecb)}`.

Simplify : {(vecb-vecc)xx(vecc-veca)}odot(veca-vecb)

Simplify : {(vecb+vecc)xx(vecc+veca))}odot(veca+vecb)

Simplify : [veca-vecb,vecb-vecc,vecc-veca] .

Prove that {(vecb+vecc)xx(vecc+veca)}.(veca+vecb)=2[veca,vecb,vecc]

If veca,vecb,vecc are any three vectors then prove that. vecaxx(vecb+vecc)+vecbxx(vecc+veca)+veccxx(veca+vecb)=0

Prove that veca*((vecb+vecc)xx(veca+vecb+vecc))=0 .

Show that the perpendicular distance of the point vecc from the line joining veca and vecb is : (|vecbxxvecc+veccxxveca+vecaxxvecb|)/(|vecb-veca|) .

If V is the volume of the parallelepiped having three coterminous edges as veca,vecb and vecc , then the volume of the parallelepiped having three coterminous edges as vecalpha = (veca.veca)veca+(veca.vecb)vecb+(veca.vecc)vecc , vecbeta=(vecb.veca)veca+(vecb.vecb)+(vecb.vecc)vecc and veclambda=(vecc.veca)veca+(vecc.vecb)vecb+(vecc.vecc)vecc is

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

If [vecaxxvecb vecbxxvecc veccxxveca]=lambda[veca vecb vecc]^2 , then lambda is equal to :