Let $\mathbf v$ and $\mathbf w$ be vectors in $3-$space. Then $$(\mathbf v\bullet\mathbf w)^2+|\mathbf v\times\mathbf w|^2=|\mathbf v|^2 |\mathbf w|^2, \quad \text{Lagrange's identity}.$$
Neglecting the proof, I am curious to know the practical meaning of this identity in the exercises and what it is used for.
In a particular case...involved in 3d surface relationships ... in the derivation of geodesic curvature $=kg$ formula in differential geometry. This vector is orthogonal to the curve.
It may be found in text-books of differential geometry, maybe in Weatherburn's.
Practically, vector addition of components cos ( dot product) and sin ( cross product ) of angles between the individual vectors should result in their (squared) scalar product of individual vector norms.
$$ \vec a \times \vec b= a\;b \sin \theta $$ $$ \vec a \cdot \vec b= a\;b \cos \theta $$
Eliminate $\theta$
$$|\vec a \cdot \vec b |^2 +| \vec a \times \vec b |^2=a^2 b^2 $$
Among many examples of such Pythagorean resolution is the curvature vector of a small circle on the sphere can be generalized to any surface in $ \mathbb R^3 $
$$k^2=k_g^2+k_n^2$$
Situations not fully known to me is total energy $ E$ conserved in dynamics of rotation and translation:
$$ E^2 = Torque^2+ Work ^2 = |F\times r|^2 +|F \cdot r|^2$$
Wiki for Lagrange