Hans Georg Schaathun
NTNU, Noregs Teknisk-Naturvitskaplege Universitet
21 August 2023
Questions - either in class or in discussion fora
Eye Model from Introduction to Psychology by University of Minnesota
1912 International Lawn Tennis Challenge
By Adrian Rosebrock , CC BY-SA 4.0,wikimedia commons
Wikimedia
200~By NASA on The Commons No restrictions,Wikimedia Commons
Vision System
i.e. putting it all together
A point in space | $\mathbf{X} = [X_1,X_2,X_3]^\mathrm{T}\in\mathbb{R}^3$ |
A bound vector, from $\mathbf{X}$ to $\mathbf{Y}$ | $\vec{\mathbf{XY}}$ |
A free vector is the same difference, but without any specific anchor point | represented as $\mathbf{Y} - \mathbf{X}$ |
$$x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\quad y=\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}$$
$$\langle x,y\rangle = x^\mathrm{T}y = x_1y_1+x_2y_2+x_3y_3$$
Euclidean Norm: $||x|| = \sqrt{\langle x,x\rangle}$
Orthogonal vectors when $\langle x,y\rangle=0$
$$x\times y = \begin{bmatrix} x_2y_3 - x_3y_2 \\ x_3y_1 - x_1y_3 \\ x_1y_2 - x_2y_1 \end{bmatrix} \in \mathbb{R}^3$$
Observe that
$$x\times y = \hat xy \quad\text{where}\quad \hat x = \begin{bmatrix} 0 & -x_3 & x_2 \\ x_3 & 0 & -x_1 \\ -x_2 & x_1 & 0 \end{bmatrix} \in \mathbb{R}^{3\times3}$$
$\hat x$ is a skew-symmetric matrix because $\hat x=-\hat x^\mathrm{T}$
i.e. the cross-product is a linear operation
By Acdx - Self-made, based on Image:Right_hand_cross_product.png,
Matrices define linear operations on vectors
$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} $$
$$ \begin{bmatrix} a_{11} & a_{21} & a_{23} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} a_{11}x_1 & a_{21}x_2 & a_{23}x_3 \\ a_{21}x_1 & a_{22}x_2 & a_{23}x_3 \\ a_{31}x_1 & a_{32}x_2 & a_{33}x_3 \\ \end{bmatrix} $$