Hans Georg Schaathun
NTNU, Noregs Teknisk-Naturvitskaplege Universitet
Thu 25 Aug 10:11:51 UTC 2022
Questions - either in class or in discussion fora
... or email if it is personal
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
$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} $$
$$ \begin{bmatrix} a_{11} & a_{21} & a_{23} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} a_{11}x_1 & a_{21}x_2 & a_{23}x_3 \\ a_{21}x_1 & a_{22}x_2 & a_{23}x_3 \\ a_{31}x_1 & a_{32}x_2 & a_{33}x_3 \\ \end{bmatrix} $$