Show that `(veca - vecb) `x` (veca + vecb) = 2( veca `x` vecb)`

Show that (veca-vecb) xx(veca+vecb)=2(veca xx vecb)

Show (veca-vecb)x(veca+vecb)=2(vecaxxvecb)

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

If a and b are perpendicular vectors show that (veca+vecb)^2 = (veca-vecb)^2 . [(veca+vecb)^2 means (veca+vecb).(veca+vecb) , so does (veca-vecb)^2 .]

|veca pm vecb|^2 = |veca|^2 + |vecb|^2 pm 2|veca||vecb|cos theta and (veca + vecb).(veca - vecb) = |veca|^2 - |vecb|^2

If veca, vecb and vecc are vectors such that veca. vecb = veca.vecc, veca xx vecb = veca xx vecc, a ne 0. then show that vecb = vecc.

Prove that: |(veca+vecb)xx(veca-vecb)|=2ab if veca_|_vecb

Prove that: |(veca+vecb)xx(veca-vecb)|=2ab if veca_|_vecb

If (veca + vecb).(veca-vecb) = 0 show that |veca| = |vecb| .