Geometric Product
\[\newcommand{\ihat}{\hat{\boldsymbol{\imath}}} \newcommand{\jhat}{\hat{\boldsymbol{\jmath}}} \newcommand{\khat}{\hat{\boldsymbol{k}}} \newcommand{\vc}[1]{\mathbf{#1}} \newcommand{\inner}[2]{ \langle #1, #2 \rangle } \newcommand{\abs}[1]{ \lvert #1 \rvert }\]
Dot Product versus our Criteria for a “Good Product”
Let’s review the dot product against our expectations:
- Useful applications
Yes We showed above that the dot product can be used to determine the angle between vectors, is useful to test perpendicularity and parallelity, find projections onto basis or other vectors, etc.
Also an example from physics: $ W = \vc{F} \cdot \vc{d} $ - Geometric meaning
Yes The dot product can be viewed as a projection-based multiply. - Works the same as regular number multiplication
Yes The dot product obeys the commutative, associative and distributive properties. - Support division that works the same as regular number division
No The dot product loses information in producing it’s scalar result. You can’t start with a scalar dot product result and recover one of the initial vectors. - Regular number multiplication should be just a special case
Yes If a vector has a single dimension, then the dot project becomes identical to - Should be closed under multiplication, just like reals and complex numbers
No Dot product of two vectors produces a scalar. Why doesn’t it produce a vector? Because it is symmetrical, you could think of it as a projection of either vector onto the other. So it would have to choose the dirction of one of the vectors.
So it’s a pretty good multiply but not perfect.
Cross Product versus our Criteria for a “Good Product”
Let’s review the cross product against our expectations:
- Useful applications
TBD - Geometric meaning
TBD - Works the same as regular number multiplication
TBD - Support division that works the same as regular number division
TBD - Regular number multiplication should be just a special case
TBD - Should be closed under multiplication, just like reals and complex numbers
TBD
So it’s a pretty good multiply but not perfect.
Multiplying Two Quarternions
Quarternions are an extension of complex numbers. Here we write $ \vc{u} $ and $ \vc{v} $ as pure vector quarternions:
\[\begin{aligned} \vc{u} &= u_x \ihat + u_y \jhat + u_z \khat \\ \vc{v} &= v_x \ihat + v_y \jhat + v_z \khat \end{aligned}\]Think of i, j and k analogous to x, y, and z.
\[\begin{aligned} \vc{u} \vc{v} = \; &\begin{pmatrix} u_x \ihat + u_y \jhat + u_z \khat \end{pmatrix} \begin{pmatrix} v_x \ihat + v_y \jhat + v_z \khat \end{pmatrix} \\ = \; &u_x v_x \ihat \ihat + u_x v_y \ihat \jhat + u_x v_z \ihat \khat +\\ &u_y v_x \jhat \ihat + u_y v_y \jhat \jhat + u_y v_z \jhat \khat +\\ &u_z v_x \khat \ihat + u_z v_y \khat \jhat + u_z v_z \khat \khat \end{aligned}\]Just like $ \vc{i} $ for imaginary numbers, imagine $ \vc{j}, \vc{k} $ are similar:
\[\ihat \ihat = \jhat \jhat = \khat \khat = \ihat \jhat \khat = -1\]We can now derive $ \ihat \jhat $:
\[\begin{aligned} \ihat \jhat \khat &= -1 \\ \ihat \jhat \khat \khat &= -1 \khat \\ \ihat \jhat (-1) &= -1 \khat \\ \ihat \jhat &= \khat \\ \end{aligned}\]A similar process produces:
\[\ihat \jhat = \khat \;\;\; \jhat \khat = \ihat \;\;\; \khat \ihat = \jhat\]Next:
\[\begin{aligned} (\jhat \khat)(\khat \ihat) &= \ihat \jhat \\ \jhat (\khat \khat) \ihat &= \khat \\ \jhat (-1) \ihat &= \khat \\ \jhat \ihat &= -\khat \\ \end{aligned}\]A similar process produces:
\[\khat \jhat = -\ihat \;\;\; \jhat \ihat = -\khat \;\;\; \ihat \khat = -\jhat\]Now we can substitute to get:
\[\begin{aligned} \vc{u} \vc{v} = \; &-(u_x v_x + u_y v_y + u_z v_z) \\ &+ (u_y v_z - u_z v_y) \ihat \\ &+ (u_z v_x - u_x v_z) \jhat \\ &+ (u_x v_y - u_y v_x) \khat \\ \vc{u} \vc{v} = \; &-(\vc{u} \cdot \vc{v}) + (\vc{u} \times \vc{v}) \end{aligned}\]