## Geometric Algebra Worked Exercises, Chapter 7

$\let\oldcdot\cdot \renewcommand{\cdot}{\kern-.2em\oldcdot\kern-.2em} \let\oldwedge\wedge \renewcommand{\wedge}{\kern-.2em\oldwedge\kern-.2em} \renewcommand{\vec}[1]{\mathbf{#1}} \let\oldtimes\times \renewcommand{\times}{\kern-.2em\oldtimes\kern-.2em} \newcommand{\dvec}[1]{\dot{\vec{#1}}} \newcommand{\ip}[2]{\vec{#1}\cdot\vec{#2}} \newcommand{\op}[2]{\vec{#1}\wedge\vec{#2}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\hf}{\frac{1}{2}} \newcommand{\rn}{\rangle^\hf_0} $
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7.1 Unit Vector Derivative $\vec{u} = \vec{u}(t)$ a vector valued function, $u = \lvert \vec{u} \rvert$. Show
$$ \ddt\left(\frac{\vec{u}}{u}\right) = \frac{\vec{u}(\vec{u}\wedge\dvec{u})}{u^3} = \frac{(\vec{u}\times\dvec{u})\times\vec{u}}{u^3} $$
Two useful intermediate results.
Vector Magnitude Squared: $$ u^2 = \vec{u}^2 $$
Proof: $$ u^2 = \left(\langle \vec{u}^\sim\vec{u} \rn\right)^2 $$ $$ = \langle \vec{u}^2 \rangle_0 $$ $$ = \vec{u}^2 $$