0 like 0 dislike
407 views
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
$\mathbf{x}-(\mathbf{x} \cdot \mathbf{n}) \mathbf{n}=-A_{\mathbf{n}}^{2} \mathbf{x}=-\left(\begin{array}{ccc} -b^{2}-c^{2} & a b & a c \\ a b & -a^{2}-c^{2} & b c \\ a c & b c & -a^{2}-b^{2} \end{array}\right) \mathbf{x}$
| 407 views

0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
2 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
2 like 0 dislike