Geometry of $|A||B| = |A * B| + |A \times B|$

Question

Geometry of $(|A||B|)^2 = |A * B|^2 + |A \times B|^2$ where $*$ represents the dot product

This identity is easy to verify algebraically once we recall that $|A * B| = |A||B|||\cos(\theta)|$ and $|A \times B| = |A||B||\sin(\theta)|$. Can somebody help me understand this identity in a more geometric way? Why should we expect these values to form a pythagorean triple? Thank you

Answer 1

When $B$ is a unit, $A\cdot B$ is (the magnitude of) the projection of $A$ onto $B$, and $|A\times B|$ is (the magnitude of) the rejection of $A$ from $B$.

The projection of a vector onto another is probably familiar, but the rejection is likely less so. The rejection of $A$ from $B$ is formed by

1. Forming the plane $P$ spanned by $A, B$.
2. Finding the orthogonal complement $B'$ within $P$ of the span of $B$; in other words $B' = (\mathbb RB)^\perp\cap P$.
3. Projecting $A$ onto $B'$.

You'll see easily that the projection and rejection form legs of a right triangle, and their hypotenuse, i.e. their sum, is $A$.

So the real mystery is how the cross product is related to rejection. What we can notice is that the rejection is $A$ minus the projection, $$ A - \frac{A\cdot B}{|B|^2}B, $$ but $B\cdot B/|B|^2 = 1$ so this can be written $$ B\times\frac{(A\times B)}{|B|^2}. $$ Notice the similarity with projection. When $B$ is a unit we can write $$ A = B(A\cdot B) + B\times(A\times B). $$ Both terms are clearly orthogonal to each other, so $$ |A|^2 = |B(A\cdot B)|^2 + |B\times(A\times B)|^2. $$ The first term is of course $(A\cdot B)^2$. For the second term, $B$ is orthogonal to $A\times B$ so $$ |B\times(A\times B)|^2 = |B|^2|A\times B|^2 = |A\times B|^2. $$ All together $$ |A|^2 = (A\cdot B)^2 + |A\times B|^2, $$ as desired.

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This is less mysterious once you realize the cross product is a stand-in for the plane $P$ via an orthogonal vector; algebraically it is the Hodge star of the exterior product: $$ A\times B = \star(A\wedge B). $$ The Hodge star takes $k$-vectors (in this case a $2$-vector or bivector) representing $k$-dimensional subspaces $X$ to $(n-k)$-vectors (in this case $1$-vectors or just vectors) representing $X^\perp$, where $n$ is the overall dimension of the space. Notice how this construction of the cross product is specific to $n=3$.

As Imsteffan notes in the comments there is a very close connection here with geometric algebra, which is a Clifford algebra interpreted (canonically) as a geometric product on the exterior algebra (the algebra of multivectors, $k$-vectors for all $k$, with the exterior product). Roughly, the Clifford algebra for $A \mapsto |A|^2$ is the associative algebra where the only rules are $A^2 = |A|^2$ and anything that can be derived from that. We see that vectors have inverses: $$ A^{-1} = A/A^2 = A/|A|^2. $$ If I also state as fact that the geometric product of two vectors is $$ AB = A\cdot B + A\wedge B $$ which you can think of as a "bag" containing a scalar $A\cdot B$ and a bivector $A\wedge B$, then we see $$ A = AB^{-1}B = (A\cdot B^{-1})B + (A\wedge B^{-1})B. $$ The first term is exactly the projection onto $B$: $$ (A\cdot B^{-1})B = \frac{A\cdot B}{|B|^2}B. $$ This means the second term must also be a vector and be the rejection from $B$. As we know, $A\wedge B$ is related to $A\times B$, and in fact we now see $$ (A\wedge B^{-1})B = B\times(A\times B^{-1}). $$

Alternatively, there is a very natural way to extend the inner product to an inner product on all multivectors, and which makes $k$-vectors and $l$-vectors orthogonal when $k\ne l$. Then we directly have $$ |AB|^2 = |A\cdot B|^2 + |A\wedge B|^2. $$ $|\cdot|^2$ is multiplicative on products of vectors, and the Hodge star is an isometry so $$ |A|^2|B|^2 = (A\cdot B)^2 + |A\times B|^2. $$