The interpretation of the inner product betweeen two vectors

Question

I am accustomed to Euclidean vector spaces where $x.a$ (both $x$ and $a$ being in $R^2$, say) can be interpreted as the magnitude of the projection of $x$ onto $a$ (or vice versa). Recently I have been reading some notes on geometric algebra and it appears here$^{[1]}$ that $x.a$ is "the projected component of $x$ onto the vector $a^{-1}$" where $a^{-1}=a/||a||^2$. So it seems that only in the case $||a||^2=1$ can $x.a$ be regarded as the projection of $x$ onto $a$.

Can someone please provide a little "geometric intuition" for this interpretation?

[1] Geometric Algebra for Computer Science by Dorst, Fontijne and Mann, p.232

Answer 1

The formula might be better written as

$$\underline P(x) = (x \cdot a^{-1})a$$

So that, if $x = \alpha a + \beta a_\perp$, where $a_\perp$ is some vector perpendicular to $a$, you get

$$\underline P(x) = (\alpha a \cdot a^{-1}) a + (\beta a_\perp \cdot a^{-1})a$$

But $a_\perp$ is perpendicular to $a$, and since $a^{-1}$ is just some scalar multiple of $a$, the second term vanishes. The result is $P(x) = \alpha a$, as you'd expect.

The formula can be written as $(x \cdot a^{-1})a$ or $(x \cdot a)a^{-1}$. They're entirely equivalent by just shifting some factors of $a^2$ around.

Answer 2

As you said the projection is defined to be $$(x\cdot a)/a=(x \cdot a)a^{-1}$$ which is $$(x \cdot a)a^{-1}= (x \cdot a) \frac{a}{ ||a||^2 } = \frac{x \cdot a}{||a||^2}a = \frac{||a||||x||cos(\theta)}{||a||^2} a = ||x||cos(\theta)\frac{a}{||a||} $$ which, if you draw it, is the projection of x onto a.