This article is part 3 in the series about transformation matrices:

2D shearing

In 2D we can skew points towards the $x$ axis by making $x’ = x + sy$, if $s > 0$ then points will skew towards the positive $x$-axis, if $s < 0$ points will move towards the negative $x$-axis

The transformation matrix that skews points towards the $x$ axis is

$$ \begin{equation} \label{2d-shear-x} \mathbf{H_x}(s) = \begin{bmatrix} 1 & s \\ 0 & 1 \end{bmatrix} \end{equation} $$

Towards the $y$ axis is

$$ \begin{equation} \label{2d-shear-y} \mathbf{H_y}(s) = \begin{bmatrix} 1 & 0 \\ s & 1 \end{bmatrix} \end{equation} $$

For example a vector $\mathbf{v}$ multiplied by \eqref{2d-shear-x} results in

$$ \mathbf{v'} = \mathbf{H_x}(s)\mathbf{v} = \begin{bmatrix} 1 & s \\ 0 & 1 \end{bmatrix} \begin{bmatrix} v_x \\ v_y \end{bmatrix} = \begin{bmatrix} v_x + sv_y \\ v_y \end{bmatrix} $$

3D shearing

The notation $\mathbf{H_{xy}}$ indicates that the $x$ and $y$ coordinates are shifted by the other coordinate $z$ i.e.

$$ \begin{align*} x' &= x + sz \\ y' &= y + tz \\ z' &= z \end{align*} $$

The shearing matrices in 3D are

$$ \begin{equation} \label{shear-xy} \mathbf{H_{xy}}(s,t) = \begin{bmatrix} 1 & 0 & s \\ 0 & 1 & t \\ 0 & 0 & 1 \end{bmatrix} \end{equation} $$
$$ \begin{equation} \label{shear-xz} \mathbf{H_{xz}}(s,t) = \begin{bmatrix} 1 & s & 0 \\ 0 & 1 & 0 \\ 0 & t & 1 \end{bmatrix} \end{equation} $$
$$ \begin{equation} \label{shear-yz} \mathbf{H_{yz}}(s,t) = \begin{bmatrix} 1 & 0 & 0 \\ s & 1 & 0 \\ t & 0 & 1 \end{bmatrix} \end{equation} $$

For example a vector $\mathbf{v}$ multiplied by \eqref{shear-xy} results in

$$ \mathbf{v'} = \mathbf{H_{xy}}(s,t) \mathbf{v} = \begin{bmatrix} 1 & 0 & s \\ 0 & 1 & t \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_x \\ v_y \\ v_z \end{bmatrix} = \begin{bmatrix} v_x + sv_z \\ v_y + tv_z \\ v_z \end{bmatrix} $$