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

If veca and vecb are any two vectors , then prove that |vecaxxvecb|^(2)=|veca|^(2)|vecb|^(2)-(veca.vecb)^(2)=|{:(veca.veca,veca.vecb),(veca.vecb,vecb.vecb):}| or |vecaxxvecb|^(2)+(veca.vecb)^(2)=|veca|^(2)|vecb|^(2) (This is also known as Lagrange identily)

If veca and vecb are any two vectors , then prove that |vecaxxvecb|^(2)=|veca|^(2)|vecb|^(2)-(veca.vecb)^(2)=|{:(veca.veca,veca.vecb),(veca.vecb,vecb.vecb):}| or |vecaxxvecb|^(2)+(veca.vecb)^(2)=|veca|^(2)|vecb|^(2) (This is also known as Lagrange identily)

Prove that |vecaxxvecb|^2=|veca|^2|vecb|^2-(veca*vecb)^2=|:[veca*veca veca*vecb],[veca*vecb vecb*vecb]:| .

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

If veca" and "vecb are any two vectors, then (veca xx vecb)^(2)= |veca|^(2)|vecb|^(2)-(veca* vecb)^(2) .

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 .]

Prove that |veca xx vecb|^(2) = abs(veca)^(2)abs(vecb)^(2) - (veca * vecb)^(2)

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

Prove that (veca xx vecb)^2=|(veca.veca,veca.vecb),(veca.vecb,vecb.vecb)| .