﻿ Solutions: Vector Identities & Plane Trig with GA | Primer on Geometric Algebra | David Hestenes

## Solutions: Vector Identities & Plane Trig with GA

Exercise 1: Prove the following vector identities and show that they are equivalent to trig identities for unit vectors
in a common plane.

(a)
Expand abba to derive the trig Pythagorean Identity:

$\Large{\begin{array}{l}{a^2}{b^2}\,\,\, = \,\,\,{\bf{abba}}\,\,\, = \,\,\,\left( {{\bf{a}} \cdot {\bf{b}}\,\, + \,\,{\bf{a}} \wedge {\bf{b}}} \right)\,\,\left( {{\bf{b}} \cdot {\bf{a}}\,\, + \,\,{\bf{b}} \wedge {\bf{a}}} \right)\,\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,\left( {{\bf{a}} \cdot {\bf{b}}\,\, + \,\,{\bf{a}} \wedge {\bf{b}}} \right)\,\,\left( {{\bf{a}} \cdot {\bf{b}}\,\, - \,\,{\bf{a}} \wedge {\bf{b}}} \right)\,\,\, = \,\,\,{\left( {{\bf{a}} \cdot {\bf{b}}} \right)^2}\,\, - \,\,{\left( {{\bf{a}} \wedge {\bf{b}}} \right)^2}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,{\left( {{\bf{a}} \cdot {\bf{b}}} \right)^2}\,\, + \,\,{\left| {{\bf{a}} \wedge {\bf{b}}} \right|^2}\,\,\, = \,\,\,{a^2}{b^2}\,{\cos ^2}\theta \,\, + \,\,{a^2}{b^2}\,{\sin ^2}\theta \end{array}}$

or, pulling out the magnitudes (leaving unit vectors) and writing these as rotors:

$\Large{\begin{array}{l}{\bf{abba}}\,\,\, = \,\,\,{a^2}{b^2}\,\left( {{\bf{\hat a\hat b}}} \right)\left( {{\bf{\hat b\hat a}}} \right)\,\,\, = \,\,\,{a^2}{b^2}\,{e^{{\bf{i}}\theta }}{e^{ - {\bf{i}}\theta }}\,\,\, = \,\,\,{a^2}{b^2}\\\,\,\,\,\,~~~\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,{a^2}{b^2}\,\left( {\cos \theta \,\, + \,\,{\bf{i}}\,\,\sin \theta } \right)\left( {\cos \theta \,\, - \,\,{\bf{i}}\,\,\sin \theta } \right)\,\,\, = \,\,\,{a^2}{b^2}\,\left( {{{\cos }^2}\theta \,\, + \,\,{{\sin }^2}\theta } \right)\end{array}}$

(b)  Looking at the equation we are asked to prove and at part (a) above, we'll save a little typing
by factoring out all the magnitudes (
divide both sides by ab2c) and using unit vectors. These
magnitudes can always be put back in later, if desired.
The hint was to expand abbc two different ways:

$\Large{\begin{array}{l}{\bf{\hat a}}\left( {{\bf{\hat b\hat b}}} \right){\bf{\hat c}}\,\,\, = \,\,\,{\bf{\hat a\hat c}}\,\,\, = \,\,\,{\bf{\hat a}} \cdot {\bf{\hat c}}\,\, + \,\,{\bf{\hat a}} \wedge {\bf{\hat c}}\,\,\, = \,\,\,\left( {{\bf{\hat a\hat b}}} \right)\left( {{\bf{\hat b\hat c}}} \right)\,\,\, = \,\,\,\left( {{\bf{\hat a}} \cdot {\bf{\hat b}}\,\, + \,\,{\bf{\hat a}} \wedge {\bf{\hat b}}} \right)\left( {{\bf{\hat b}} \cdot {\bf{\hat c}}\,\, + \,\,{\bf{\hat b}} \wedge {\bf{\hat c}}} \right)\\\,\,\,\,\,\,\,\,\,~~~~~~~~\,\,\,\,\,\,\,\,\,\,\, = \,\,\,\left( {{\bf{\hat a}} \cdot {\bf{\hat b}}} \right)\,{\bf{\hat b}} \cdot {\bf{\hat c}}\,\, + \,\,\left( {{\bf{\hat a}} \wedge {\bf{\hat b}}} \right)\left( {{\bf{\hat b}} \wedge {\bf{\hat c}}} \right)\,\, + \,\,\left( {{\bf{\hat a}} \cdot {\bf{\hat b}}} \right){\bf{\hat b}} \wedge {\bf{\hat c}}\,\, + \,\,\left( {{\bf{\hat b}} \cdot {\bf{\hat c}}} \right){\bf{\hat a}} \wedge {\bf{\hat b}}\\\\\\\left( {{\bf{\hat a}} \wedge {\bf{\hat b}}} \right)\left( {{\bf{\hat b}} \wedge {\bf{\hat c}}} \right)\,\,\, = \,\,\,{\bf{\hat a}} \cdot {\bf{\hat c}}\,\, - \,\,\left( {{\bf{\hat a}} \cdot {\bf{\hat b}}} \right)\,{\bf{\hat b}} \cdot {\bf{\hat c}}\,\, - \,\,\left[ {\left( {{\bf{\hat b}} \cdot {\bf{\hat a}}} \right){\bf{\hat b}} \wedge {\bf{\hat c}}\,\, - \,\,\left( {{\bf{\hat b}} \cdot {\bf{\hat b}}} \right){\bf{\hat a}} \wedge {\bf{\hat c}}\,\, + \,\,\left( {{\bf{\hat b}} \cdot {\bf{\hat c}}} \right){\bf{\hat a}} \wedge {\bf{\hat b}}} \right]\end{array}}$

We will have completed the proof if we can show the last expression inside square brackets is zero.
Remember we were also supposed to consider only the case where
the three vectors are coplanar.

This Primer has been concentrating on two dimensional descriptions and examples of GA, so we
haven't yet learned that the expression in square brackets can be written as
b(abc), which is
more clearly zero if the three vectors are coplanar. Since our toolbox of tricks is still a bit empty,
how can we show the expression in square brackets is zero? You should be able to convince yourself
that if
a is parallel or antiparallel to c, the expression is zero. Yet, if a and c are not colinear, we can
write
b in terms of them and see what happens to the expression:

$\Large{\begin{array}{l}{\bf{\hat b}}\,\,\, \equiv \,\,\,\alpha {\bf{\hat a}}\,\, + \,\,\beta {\bf{\hat c}}\,,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{\hat b\hat b}}\,\,\, = \,\,\,1\,\,\, = \,\,\,{\alpha ^2}\,\, + \,\,{\beta ^2}\,\, + \,\,2\alpha \beta \,{\bf{\hat a}} \cdot {\bf{\hat c}}\\\\\left( {{\bf{\hat b}} \cdot {\bf{\hat a}}} \right){\bf{\hat b}} \wedge {\bf{\hat c}}\,\, - \,\,\left( {{\bf{\hat b}} \cdot {\bf{\hat b}}} \right){\bf{\hat a}} \wedge {\bf{\hat c}}\,\, + \,\,\left( {{\bf{\hat b}} \cdot {\bf{\hat c}}} \right){\bf{\hat a}} \wedge {\bf{\hat b}}\\\,\,\,\,\,\,\,\,\,\,\,~~~~\, = \,\,\,\left( {\alpha + \beta \,{\bf{\hat a}} \cdot {\bf{\hat c}}} \right)\alpha \,{\bf{\hat a}} \wedge {\bf{\hat c}}\,\, - \,\,{\bf{\hat a}} \wedge {\bf{\hat c}}\,\, + \,\,\left( {\alpha \,{\bf{\hat a}} \cdot {\bf{\hat c}} + \beta } \right)\beta \,{\bf{\hat a}} \wedge {\bf{\hat c}}\\\,\,\,\,\,\,\,\,\,~~~~\,\,\, = \,\,\,\left( {{\alpha ^2}\,\, + \,\,\alpha \beta \,{\bf{\hat a}} \cdot {\bf{\hat c}}\,\, - \,\,1\,\, + \,\,\alpha \beta \,{\bf{\hat a}} \cdot {\bf{\hat c}}\,\, + \,\,{\beta ^2}} \right)\,\,{\bf{\hat a}} \wedge {\bf{\hat c}}\,\,\, = \,\,\,0\end{array}}$

That was a lot of work. Is it equivalent to some trig identity? As in part (a) above, rotor notation will help:

$\Large{\begin{array}{l}{\bf{\hat a\hat b}}\,\,\, \equiv \,\,\,{e^{{\bf{i}}\theta }}\,\,\, = \,\,\,\cos \theta \,\, + \,\,{\bf{i}}\,\sin \theta \,,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{\hat b\hat c}}\,\,\, \equiv \,\,\,{e^{{\bf{i}}\phi }}\,\,\, = \,\,\,\cos \phi \,\, + \,\,{\bf{i}}\,\sin \phi \\\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\bf{\hat a\hat b}}} \right)\left( {{\bf{\hat b\hat c}}} \right)\,\,\, = \,\,\,{e^{{\bf{i}}\theta }}{e^{{\bf{i}}\phi }}\,\,\, = \,\,\,{e^{{\bf{i}}\left( {\theta + \phi } \right)}}\,\,\, = \,\,\,{\bf{\hat a\hat c}}\,\,\, = \,\,\,\cos \left( {\theta + \phi } \right)\,\, + \,\,{\bf{i}}\,\sin \left( {\theta + \phi } \right)\\\\\\\left( {{\bf{\hat a}} \wedge {\bf{\hat b}}} \right)\left( {{\bf{\hat b}} \wedge {\bf{\hat c}}} \right)\,\,\, = \,\,\,{\bf{\hat a}} \cdot {\bf{\hat c}}\,\, - \,\,\left( {{\bf{\hat a}} \cdot {\bf{\hat b}}} \right)\,{\bf{\hat b}} \cdot {\bf{\hat c}}\\\,\,\,\,\,\,\,\,~~~~~~~~\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\,\left( {{\bf{i}}\,\sin \theta } \right)\left( {{\bf{i}}\,\sin \phi } \right)\,\,\, = \,\,\,\cos \left( {\theta + \phi } \right)\,\, - \,\,\cos \theta \,\,\cos \phi \\\cos \theta \,\,\cos \phi \,\, - \,\,\sin \theta \,\,\sin \phi \,\,\, = \,\,\,\cos \left( {\theta + \phi } \right)\end{array}}$

It turns out our GA identity is equivalent to the trig identity for the cosine of the sum of two angles.
Notice, even here, we require the three vectors to be coplanar, so that the unit bivector
i represents
the same plane in all of these rotor expressions.

Exercise 2: Use rotor products to derive the trigonometric "sum of two angles" formulas.

We've begun this derivation in the last group of expressions above.

$\Large{\begin{array}{l}{\bf{\hat a\hat c}}\,\,\, = \,\,\,{e^{{\bf{i}}\left( {\theta + \phi } \right)}}\,\,\, = \,\,\,\cos \left( {\theta + \phi } \right)\,\, + \,\,{\bf{i}}\,\sin \left( {\theta + \phi } \right)\,\,\, = \,\,\,\left( {{\bf{\hat a\hat b}}} \right)\left( {{\bf{\hat b\hat c}}} \right)\\\,\,\,\,\,\,\,\, = \,\,\,\left( {\cos \theta \,\, + \,\,{\bf{i}}\,\sin \theta } \right)\left( {\cos \phi \,\, + \,\,{\bf{i}}\,\sin \phi } \right)\\\,\,\,\,\,\,\,\, = \,\,\,\left( {\cos \theta \cos \phi \,\, - \,\,\sin \theta \sin \phi } \right)\,\, + \,\,{\bf{i}}\,\left( {\sin \theta \cos \phi \,\, + \,\,\cos \theta \sin \phi } \right)\end{array}}$
$\Large{\begin{array}{l}\left( a \right)\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {\theta + \phi } \right)\,\,\, = \,\,\,\cos \theta \cos \phi \,\, - \,\,\sin \theta \sin \phi \\\\\\\left( b \right)\,\,\,\,\,\,\,\,\,\,\,\,\sin \left( {\theta + \phi } \right)\,\,\, = \,\,\,\sin \theta \cos \phi \,\, + \,\,\cos \theta \sin \phi \end{array}}$

Exercise 3: Prove the following triangle theorems.

(a)  Derive the law of cosines. Let θ be the angle between the directions of a and b. Then γ is its
supplementary angle.

$\Large{\begin{array}{l}{c^2}\,\,\, = \,\,\,{{\bf{c}}^2}\,\,\, = \,\,\,\left( {{\bf{a}} + {\bf{b}}} \right)\left( {{\bf{a}} + {\bf{b}}} \right)\,\,\, = \,\,\,{{\bf{a}}^2}\,\, + \,\,{{\bf{b}}^2}\,\, + \,\,\left( {{\bf{ab}}\,\, + \,\,{\bf{ba}}} \right)\\\,\,\,\,\,\,\,\, = \,\,\,{a^2}\,\, + \,\,{b^2}\,\, + \,\,2\,{\bf{a}} \cdot {\bf{b}}\,\,\, = \,\,\,{a^2}\,\, + \,\,{b^2}\,\, + \,\,2\,ab\,\cos \theta \,\,\, = \,\,\,{a^2}\,\, + \,\,{b^2}\,\, - \,\,2\,ab\,\cos \gamma \end{array}}$

(b)  Derive the law of sines.

$\Large{\begin{array}{l}{\bf{a}} \wedge {\bf{c}}\,\,\, = \,\,\,{\bf{a}} \wedge \left( {{\bf{a}} + {\bf{b}}} \right)\,\,\, = \,\,\,{\bf{a}} \wedge {\bf{b}}\,\,\, = \,\,\,\left( {{\bf{a}} + {\bf{b}}} \right) \wedge {\bf{b}}\,\,\, = \,\,\,{\bf{c}} \wedge {\bf{b}}\\{\bf{i}}\,\,ac\,\,\sin \beta \,\,\, = \,\,\,{\bf{i}}\,\,ab\,\,\sin \left( {\pi - \gamma } \right)\,\,\, = \,\,\,{\bf{i}}\,\,ab\,\,\sin \gamma \,\,\, = \,\,\,{\bf{i}}\,\,cb\,\,\sin \alpha \end{array}}$

These results should be clear with the help of the previous two diagrams. Now, if we divide all terms
by
abc, we have the law of sines.

(c)  Derive triangle laws for supplementary angles (interior/exterior), sum of interior angles, and
sum of exterior angles.

Above, we've assumed previous knowledge that θ and γ are supplementary angles. Let i be the unit
CCW bivector in the plane of the paper. Then
θ is the rotor angle taking the direction of a into the
direction of
b, while γ is the rotor angle taking the direction of b into the direction of a.

$\Large{\begin{array}{l}{\bf{\hat a\hat b}}\left[ {{\bf{\hat b}}\left( { - {\bf{\hat a}}} \right)} \right]\,\,\, = \,\,\,{e^{{\bf{i}}\theta }}{e^{{\bf{i}}\gamma }}\,\,\, = \,\,\,{e^{{\bf{i}}\left( {\theta + \gamma } \right)}}\,\,\, = \,\,\,{\bf{\hat a}}\left( { - {\bf{\hat a}}} \right)\,\,\, = \,\,\,{e^{{\bf{i}}\pi }}\\{\rm{Therefore:}}\,~~~~~~~~\,\theta + \gamma \,\,\, = \,\,\,\pi \end{array}}$

For considering sums of angles, let's look at general n-sided simple (sides intersect only at corners)

polygons. Let all the edge vectors connect tip-of-one-to-tail-of-next, tracing around the polygon in a
general CCW direction. An
exterior angle is defined as the turning angle to get from the direction of
one edge vector to the direction of the next edge vector. If the next edge vector makes a CW turn, that
will count as a negative exterior angle. We have already shown above that each corresponding
interior
angle
will be supplementary to its exterior angle. We'll use rotors to represent turning from one edge
to the next.

$\Large{\begin{array}{l}{e^{{\bf{i}}{\theta _1}}}{e^{{\bf{i}}{\theta _2}}} \cdot \cdot \cdot {e^{{\bf{i}}{\theta _n}}}\,\,\, = \,\,\,{e^{{\bf{i}}2\pi }}\\{\rm{Sum}}\,\,{\rm{of}}\,\,{\rm{exterior}}\,\,{\rm{angles}}\,\,\, = \,\,\,{\theta _1}\,\, + \,\,{\theta _2}\,\, + \,\, \cdot \cdot \cdot \,\, + \,\,{\theta _n}\,\,\, = \,\,\,2\pi \\\\\\\left( {\pi - {\theta _1}} \right)\,\, + \,\,\left( {\pi - {\theta _2}} \right)\,\, + \,\, \cdot \cdot \cdot \,\, + \,\,\left( {\pi - {\theta _n}} \right)\,\,\, = \,\,\,n\pi \,\, - \,\,2\pi \\{\rm{Sum}}\,\,{\rm{of}}\,\,{\rm{interior}}\,\,{\rm{angles}}\,\,\, = \,\,\,\left( {n\,\, - \,\,2} \right)\,\pi \end{array}}$