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a) Let $\mathbf{v}:=(a, b, c)$ and $\mathbf{x}:=(x, y, z)$ be any vectors in $\mathbb{R}^{3}$. Viewed as column vectors, find a $3 \times 3$ matrix $A_{\mathrm{v}}$ so that the cross product $\mathbf{v} \times \mathbf{x}=A_{\mathrm{v}} \mathbf{x}$.
$\mathbf{v} \times \mathbf{x}=A_{\mathbf{v}} \mathbf{x}=\left(\begin{array}{ccc} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)$
where the anti-symmetric matrix $A_{\mathrm{v}}$ is defined by the above formula.
b) From this, one has $\mathbf{v} \times(\mathbf{v} \times \mathbf{x})=A_{\mathbf{v}}(\mathbf{v} \times \mathbf{x})=A_{\mathbf{v}}^{2} \mathbf{x}$ (why?). Combined with the cross product identity $\mathbf{u} \times(\mathbf{v} \times \mathbf{w})=\langle\mathbf{u}, \mathbf{w}\rangle \mathbf{v}-\langle\mathbf{u}, \mathbf{v}\rangle \mathbf{w}$, show that
$A_{\mathrm{v}}^{2} \mathrm{x}=\langle\mathrm{v}, \mathrm{x}\rangle \mathrm{v}-\|\mathrm{v}\|^{2} \mathrm{x}$
c) If $\mathbf{n}=(a, b, c)$ is a unit vector, use this formula to show that (perhaps surprisingly) the orthogonal projection of $\mathbf{x}$ into the plane perpendicular to $\mathbf{n}$ is given by
$\mathbf{x}-(\mathbf{x} \cdot \mathbf{n}) \mathbf{n}=-A_{\mathbf{n}}^{2} \mathbf{x}=-\left(\begin{array}{ccc} -b^{2}-c^{2} & a b & a c \\ a b & -a^{2}-c^{2} & b c \\ a c & b c & -a^{2}-b^{2} \end{array}\right) \mathbf{x}$
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