Answer
See solution
Work Step by Step
$||\vec{u}-\vec{v}||^2=\langle \vec{u}-\vec{v},\vec{u}-\vec{v} \rangle=\langle \vec{u},\vec{u}-\vec{v} \rangle+\langle -\vec{v},\vec{u}-\vec{v} \rangle=\langle \vec{u},\vec{u}-\vec{v} \rangle-\langle \vec{v},\vec{u}-\vec{v} \rangle$
$\langle \vec{u},\vec{u}-\vec{v} \rangle=\langle \vec{u}-\vec{v},\vec{u} \rangle=\langle \vec{u},\vec{u}\rangle+\langle -\vec{v},\vec{u} \rangle=\langle \vec{u},\vec{u}\rangle-\langle \vec{v},\vec{u} \rangle$
$\langle \vec{v},\vec{u}-\vec{v} \rangle=\langle \vec{u}-\vec{v},\vec{v} \rangle=\langle \vec{u},\vec{v}\rangle+\langle -\vec{v},\vec{v} \rangle=\langle \vec{u},\vec{v}\rangle-\langle \vec{v},\vec{v} \rangle$
Plugging back into the original equation, we get
$||\vec{u}-\vec{v}||^2=\langle \vec{u},\vec{u}-\vec{v} \rangle-\langle \vec{v},\vec{u}-\vec{v} \rangle=\langle \vec{u},\vec{u}\rangle-\langle \vec{v},\vec{u} \rangle-\langle \vec{u},\vec{v}\rangle+\langle \vec{v},\vec{v} \rangle=\langle \vec{u},\vec{u}\rangle-2\langle \vec{u},\vec{v}\rangle+\langle \vec{v},\vec{v} \rangle$
$\langle \vec{u},\vec{u}\rangle=||\vec{u}||^2=1$
$\langle \vec{v},\vec{v}\rangle=||\vec{v}||^2=1$ because $\vec{u}$ and $\vec{v}$ are normal vectors.
$\langle \vec{u},\vec{v}\rangle=0$ because $\vec{u}$ and $\vec{v}$ and orthogonal.
Therefore, $||\vec{u}-\vec{v}||=\sqrt{2}$
Axioms used by order: 2, 3, 2, 3, 2, 3, 1
