TITLE: Algebraic vector proof of Lagrange's identity
QUESTION [5 upvotes]: $$|v · w|
^2 + |v × w|
^2 = |v|^2 |w|^2 $$
Edit
Despite doing it multiple times it seems I have made a meal of the expansion see Jean-Claude's answer for a great explanation
Using $v = (v_1,v_2,v_3)$ and $w = (w_1, w_2, w_3)$ i have expanded the LHS and gotten
$$(v_2)^2(w_3)^2 + (v_3)^2(w_2)^2 + (v_3)^2(w_1)^2 + (v_1)^2(w_3)^2 + (v_1)^2(w_2)^2
 + (v_2)^2(w_1)^2 +(v_1)^2(w_1)^2 + (v_2)^2(w_2)^2 + (v_3)^2(w_3)^2 -\mathbf{2(v_2 w_3 w_2 v_3 + v_3 w_1 v_1 w_3 + v_1 w_2 v_2 w_1)}$$
Now this is RHS minus the bolded terms and I dont know how to get rid of the bolded terms.

REPLY [2 votes]: As both sides of the equality are homogeneous, you may assume that $v$ and $w$ are unit vectors. The two sides are also invariant under rotation of coordinate frame. So, you may further assume that $v=(1,0,0)^T$. The equality then reduces to $w_1^2+(w_2^2+w_3^2)=1$, which is true because $w$ is a unit vector.
Edit. Alternatively, write $w=u+z$, where $u\parallel v$ and $z\perp v$. Then
$$
|v\cdot w|^2+|v\times w|^2
=|v\cdot u|^2+|v\times z|^2
=|v|^2|u|^2+|v|^2|z|^2=|v|^2|w|^2.
$$