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# System prepended metadata

title: Homogeneous Coordinates

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# Homogeneous Coordinates
#### Created By: Orion Sinacori

Have you ever ran into the problem of needing a linear combination on a plane to not start at the origin? What if I told you that there is a way to translate this linear combination? This issue is typically seen in computer graphics where you have designs that are being mapped on a grid in a computer program. These graphics can't all start at the origin as they would overlap and cause the graphics of the computer to be very unorganized or difficult to follow. This is why Homogeneous Coordinates exist. This idea of translating or shifting your linear combination can be done through homogeneous coordinates which is the idea that each point $(x,y)$ can be identified with the point $(x,y,1)$ on a plane in $\mathbb{R}^3$. The translation that would map $(x,y) \mapsto (x+h,y+k)$ is written as $(x,y,1) \mapsto (x+h,y+k,1)$ and is given by the matrix

$\begin{bmatrix}
    1&0&h\\
    0&1&k\\
    0&0&1
\end{bmatrix}*
\begin{bmatrix}
    x\\
    y\\
    1
\end{bmatrix} = \begin{bmatrix}
x+h\\
y+k\\
1\\
\end{bmatrix}$

An example of ths would be shifting the matrix $A$ up 4 units and to the left 2 units. If we let matrix $A = \begin{bmatrix}
    3&0\\
    0&2\\
\end{bmatrix}$ Then our new shifted matrix in $\mathbb{R}^3$ would be $\begin{bmatrix}
    3&0&-2\\
    0&2&4\\
    0&0&1
\end{bmatrix}$. Now lets find the homogeneous coordinates that rotate $\mathbb{R}^2$ $45^{\circ}$ and shifts it up 3 units. First need to find the augmented matix for our shift then our rotation. After we find our matrices, we will multiply our shift by our rotation since we are rotating first and shifting second. Notice:

$\begin{bmatrix}
    1&0&0\\
    0&1&3\\
    0&0&1
\end{bmatrix} * \begin{bmatrix}
    \frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}&0\\
    \frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}&0\\
    0&0&1
\end{bmatrix} = \begin{bmatrix}
    \frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}&-\frac{3 \sqrt{2}}{2}\\
    \frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}&\frac{3 \sqrt{2}}{2}\\
    0&0&1
\end{bmatrix}$

Now that you have seen Homogeneous Coordinates and got to see how to apply shifts and rotations in Homogeneous Coordinates, it is time to test your knowledge. Try to find the matrix in Homogeneous Coordinates that accomplishes a rotation of $\mathbb{R}^3$ by $90^{\circ}$ about the line $x(t)=3$ $y(t)=2$, $z(t)=t$.