Geometric Algebra Rejection, Projection and reflection rotation, confused on how end result is actually calculated

Question

I recently found a video that claimed to give intuition to Quaternions. And to my suprise it nearly did, but I have a few large hang ups on the lack of definitions of certain operations, it also appears that many other resources on the internet suffer the same issues and don't adequately define operators in a way that I could figure this out independently.

First projection on the projection and rejection operators:

While I understand what both do, I do not understand how either can actually be computed given the Wikipedia definitions of the un-generalized geometric product, the contraction product (generalized dot product), and the wedge product (exterior product). These are the same definitions given in the video.

video definition at 36:28

Projection of $v$ onto $u$ is $Geometric Product($$v \bullet u$, $inv(u))$

This implies would need to take the dot product of $u$ and $v$, a scalar, then take the geometric product of that scalar... I don't understand how I'm supposed to do that (it would require taking the dot product of a scalar and taking the wedge product of a scalar and a vector, which makes me think the un-generalized definition is not what I want here?)

video definition at 42:05

Similarly, Rejection of $v$ onto $u$ is $Geometric Product($$v \land u$, $inv(u))$ which if $u$ is unit length, then $v \land u$ $\bullet$ $u$ should be $v$ using the contraction product because $v,u$ ^ $u$ $= 0$, which also doesn't make any sense (clearly the rejection in the example shown is not going to be $v$, the vector being projected itself...)

video definition at 54:13

rotation of $x$ given vector $\theta$ /2 ccwise from $x$, $a$, and $\theta$ /2 ccwise from $a$, $b$ is given by $GeometricProduct(Geometric Product(GeometricProduct(b, inv(a), x), GeometricProduct(a, inv(b)))$

Finally, while it did make sense why we needed to do two reflections to achieve a rotation, and how this absolutely shows quaternion analogies, I did not understand how the function could actually be calculated with out running into bivector scalar and vector scalar geometric products.

I must be misunderstanding something, but I can't figure out how these functions are supposed to be actually evaluated given the current information I have.

Answer 1

I have not seen the video but maybe can help you.

This implies would need to take the dot product of u and v, a scalar, then take the geometric product of that scalar... I don't understand how I'm supposed to do that (it would require taking the dot product of a scalar and taking the wedge product of a scalar and a vector, which makes me think the un-generalized definition is not what I want here?)

No, in the case of a scalar and a vector, the geometric product (gp) reduces to the familiar vector space product-with-scalar, so gp($\alpha$, $u$) = $\alpha u = \alpha (u\cdot e_1) + \alpha (u\cdot e_2) + \alpha (u\cdot e_3)$. So the projection of $v$ on $u$: $(u \cdot v) u^{-1}$ = $\frac{1}{\|u\|^2}(u \cdot v) u$, just the same as in vector space.

Similarly, Rejection of v onto u is GeometricProduct(GeometricProduct(v ∧ u, inv(u))) which if u is unit length, then v ∧ u ∙ u should be v using the contraction product because v,u ^ u = 0, which also doesn't make any sense (clearly the rejection in the example shown is not going to be v, the vector being projected itself...)

Considering $\|u\| = 1$, the rejection $(v \wedge u) u^{-1}$ is equal to $(v \wedge u) u$. The expansion is $(v \wedge u) \cdot u + (v \wedge u) \wedge u$, since $(v \wedge u) \wedge u = 0$ the rejection is $(v \wedge u) \cdot u = -(v \cdot u) \wedge u + v \wedge (u \cdot u)$. The term $(v \cdot u) \wedge u = (v \cdot u) u$ since the wedge product of a scalar and a vector reduces to the familiar vector space product, and since $u \cdot u = 1$, the rejection is $-(v \cdot u) u + v$.

Finally, while it did make sense why we needed to do two reflections to achieve a rotation, and how this absolutely shows quaternion analogies, I did not understand how the function could actually be calculated with out running into bivector scalar and vector scalar geometric products.

These geometric products reduces to the familiar vector space products, they commute and only can change the weight of a k-vector but never its attitude.