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

Question

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 \(\mathbf{v} \times \mathbf{x}=A_{\mathrm{v}} \mathbf{x}\).
ANSWER:
\[
\mathbf{v} \times \mathbf{x}=A_{\mathbf{v}} \mathbf{x}=\left(\begin{array}{ccc}
0 & -c & b \\
c & 0 & -a \\
-b & a & 0
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)
\]
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
\[
\mathbf{x}-(\mathbf{x} \cdot \mathbf{n}) \mathbf{n}=-A_{\mathbf{n}}^{2} \mathbf{x}=-\left(\begin{array}{ccc}
-b^{2}-c^{2} & a b & a c \\
a b & -a^{2}-c^{2} & b c \\
a c & b c & -a^{2}-b^{2}
\end{array}\right) \mathbf{x}
\]