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

  • $\begingroup$ If $a$ is not a unit vector, then $x\cdot a$ will depend on the magnitude of $a$. Does it make sense for the projection of $x$ to depend on how long $a$ is? The projection is usually defined as $(x\cdot a)/a$. $\endgroup$ – Andrey Sokolov Mar 10 '14 at 1:09
  • $\begingroup$ I guess my question then is why is the projection defined as $(x.a)/a=(x.a)a^{-1}$, since that looks like it would be parallel to the vector $a^{-1}$ and not necessarily to the vector $a$. E.g., if $a^2=-1$ (i.e. $a.a=||a||^2=-1$), then $(x.a)/a=-(x.a)a$ which seems to be in the opposite direction to $a$. $\endgroup$ – njt Mar 10 '14 at 4:19

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.

  • $\begingroup$ Yes! $(x.a^{-1})a=(\alpha a.a^{-1})a=(\alpha a.a/a^2)a=\alpha a$ and $(x.a)a^{-1}=(\alpha a.a)a^{-1}=(\alpha a.a)a/a^2=\alpha a$. I see it now! $\endgroup$ – njt Mar 11 '14 at 16:21

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.

  • $\begingroup$ I believe that the relation $x.a=||a||||x||\cos (\theta)$ is a strictly Euclidean concept. E.g., I think if $a^2=-1$, we would end up with $\cosh(\theta)$ instead. $\endgroup$ – njt Mar 11 '14 at 15:55

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