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

Prove that (veca+vecb)*(veca+vecb)=|veca|^(2)+|vecb|^(2) , if and only if veca,vecb are perpendicular , given vecanevec0,vecbnevec0 . Choose the correct answer in Exercises 16 to 19.

veca. (veca xx vecb)=

(veca + vecb) xx (veca - vecb) is :

Let veca, vecb and vecc be non-zero vectors such that (veca xx vecb) xx vecc=1/3|vecb||vecc|veca. If theta is the acute angle between the vectors vecb and vecc, then sin theta equals.

If veca,vecb and vecc are non-coplaner,then value of veca{(vecbxxvecc)/(3vecb.(veccxxveca))}-vecb.{(veccxxveca)/(2vecc.(vecaxxvecb))} is

If veca and vecb are two unit vectors, then the vector (veca + vecb) xx (veca xx vecb) is parallel to the vector :

If (veca+vecb).(veca-vecb)=8 and |veca|=8(vecb) then find |vecb| .

Find |vecb| ,if (veca+vecb) . (veca-vecb)=8 and |veca|=8|vecb| .

If veca , vecb, vecc and vecd are unit vectors such that (veca xx vecb). (veccxx vecd) =1 and veca. Vecc= 1/2 then :