Translation Equations

Usually when we are talking about vectors, we talk about one of the standard planes. In 2-space, we generally talk about the $xy$-plane, $xz$-plane, or $yz$-plane. In 3-space we usually talk about the $xyz$-plane. However, it is important to recognize the the position of vectors are relative to the plane they are in.

Translation of Axes

Let's say that we have some new coordinate system that is the $x'y'$-plane. The diagram below constructs a possible $x'y'$-plane on top of the standard $xy$-plane:


We can analyze the position of the new coordinate system as follows in this diagram with relation to some vector $\vec{u} = (u_1, u_2)$. Note that the position of $\vec{u}$ is different depending on it being placed in the $xy$-plane or the $x'y'$-plane.


We should notice that the origin in the $xy$-plane is $O = (0,0)$, and that the origin in the $x'y'$-plane is $O' = (k, l)$. Hence we can derive translation equations for equivalency.

\begin{align} x' = x - k \quad y' = y - l \end{align}

We should acknowledge this is relatable in 3-space too such that the translation equations are as follows:

\begin{align} \quad x' = x - k \quad y' = y - l \quad z' = z - m \end{align}

Example 1

Suppose we have an $xy$-coordinate system be translated to obtain an $x'y'$-coordinate system that has it's origin at $xy$-coordinates $(2, 3)$. Determine the $x'y'$-coordinates of the point with the $xy$-coordinates $(-2, 1)$.

First let's set up our translation equations:

\begin{align} x' = x - 2 \quad y' = y - 3 \end{align}

We can now substitute appropriate to solve for $x'$ and $y'$:

\begin{align} x' = -2 -2 \quad y' = 1 - 3 \end{align}
\begin{align} x' = -4 \quad, y' = -2 \end{align}

Thus the $x'y'$-coordinates that correspond to the axes translation for the $xy$-coordinates $(-2, 1)$ are $(-4, -2)$.

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