##### Exercise8

A theme that will later unfold concerns the use of coordinate systems. We can identify the point $$(x,y)$$ with the tip of the vector $$\left[\begin{array}{r}x\\y\end{array}\right]\text{,}$$ drawn emanating from the origin. We can then think of the usual Cartesian coordinate system in terms of linear combinations of the vectors

\begin{equation*} \evec_1 = \left[\begin{array}{r} 1 \\ 0 \end{array}\right], \evec_2 = \left[\begin{array}{r} 0 \\ 1 \end{array}\right] \text{.} \end{equation*}

The point $$(2,-3)$$ is identified with the vector

\begin{equation*} \left[\begin{array}{r} 2 \\ -3 \end{array}\right] = 2\evec_1 - 3\evec_2 \text{.} \end{equation*}

If we have vectors

\begin{equation*} \vvec_1 = \left[\begin{array}{r} 2 \\ 1 \end{array}\right], \vvec_2 = \left[\begin{array}{r} 1 \\ 2 \end{array}\right] \text{,} \end{equation*}

we may define a new coordinate system, such that a point $$\{x,y\}$$ will correspond to the vector

\begin{equation*} x\vvec_1 + y\vvec_2 \text{.} \end{equation*}

For instance, the point $$\{2,-3\}$$ is shown on the right side of FigureĀ 8

1. Write the point $$\{2,-3\}$$ in standard coordinates; that is, find $$x$$ and $$y$$ such that

\begin{equation*} (x,y) = \{2,-3\} \text{.} \end{equation*}
2. Write the point $$(2,-3)$$ in the new coordinate system; that is, find $$a$$ and $$b$$ such that

\begin{equation*} \{a,b\} = (2,-3) \text{.} \end{equation*}
3. Convert a general point $$\{a,b\}\text{,}$$ expressed in the new coordinate system, into standard Cartesian coordinates $$(x,y)\text{.}$$

4. What is the general strategy for converting a point from standard Cartesian coordinates $$(x,y)$$ to the new coordinates $$\{a,b\}\text{?}$$ Actually implementing this strategy in general may take a bit of work so just describe the strategy. We will study this in more detail later.

in-context