Finding the coordinates of a vector with respect to a basis $\beta = \{\mathbf{v}_1, \mathbf{v}_2\}$ normally means solving a linear system or inverting the basis matrix $M_\beta$ (see the
Change of Basis page). However, when the basis vectors are pairwise orthogonal, there is a much quicker way: the coordinates are simply (orthognal) projections, which provide significant computational efficeincy.
\subsection*{Orthogonal sets and bases}
Recall that $\mathbf{u}, \mathbf{v} \in \R^n$ are \textbf{orthogonal} if $\mathbf{u}^\mathsf{T}\mathbf{v} = 0$, and that the norm $\|\mathbf{u}\| = \sqrt{\mathbf{u}^\mathsf{T}\mathbf{u}}$.
\begin{definition}
A set $S = \{\mathbf{v}_1, \ldots, \mathbf{v}_p\} \subset \R^n$ is an \textbf{orthogonal set} if its vectors are pairwise orthogonal, $\mathbf{v}_i^\mathsf{T}\mathbf{v}_j = 0$ for $i \neq j$. It is an \textbf{orthonormal set} if, in addition, $\|\mathbf{v}_i\| = 1$ for all $i$. A basis of a subspace $W$ that is also an orthogonal (orthonormal) set is called an \textbf{orthogonal basis} (\textbf{orthonormal basis}) of $W$.
\end{definition}
Orthogonality is a stronger requirement than linear independence: if the vectors of $S$ are nonzero and pairwise orthogonal, then $S$ is linearly independent. (The proof uses the same trick as the theorem below.) So in $\R^2$, any two nonzero orthogonal vectors form an orthogonal basis (resp. in $\R^m$, any $m$ nonzero orthogonal vectors form one too).
\subsection*{Coordinates by projection}
Recall that the projection of $\mathbf{x}$ onto a nonzero vector $\mathbf{v}$ is
$$P_\mathbf{v}\mathbf{x} = \frac{\mathbf{v}^\mathsf{T}\mathbf{x}}{\|\mathbf{v}\|^2}\,\mathbf{v}.$$
\begin{theorem}
Let $\beta = \{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ be an orthogonal basis of $W$ and $\mathbf{x} \in W$. Then the coordinates of $\mathbf{x}$ with respect to $\beta$ are
$$c_i = \frac{\mathbf{v}_i^\mathsf{T}\mathbf{x}}{\|\mathbf{v}_i\|^2}, \qquad i = 1, \ldots, p.$$
\end{theorem}
\begin{proof}
Write $\mathbf{x} = c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p$ and multiply on the left by $\mathbf{v}_1^\mathsf{T}$. By orthogonality every cross term $\mathbf{v}_1^\mathsf{T}\mathbf{v}_j$ with $j \neq 1$ vanishes, leaving
$$\mathbf{v}_1^\mathsf{T}\mathbf{x} = c_1\,\mathbf{v}_1^\mathsf{T}\mathbf{v}_1 = c_1\|\mathbf{v}_1\|^2.$$
Since $\mathbf{v}_1 \neq \mathbf{0}$ (it belongs to a basis, so it needs to be linearly independent with the rest of the vectors), we may divide by $\|\mathbf{v}_1\|^2$. The same argument holds with $\mathbf{v}_i^\mathsf{T}$, giving each $c_i$.
\end{proof}
Each term $c_i\mathbf{v}_i$ is exactly the projection of $\mathbf{x}$ onto $\mathbf{v}_i$. In other words, for an orthogonal basis,
$$\mathbf{x} = P_{\mathbf{v}_1}\mathbf{x} + P_{\mathbf{v}_2}\mathbf{x} + \cdots + P_{\mathbf{v}_p}\mathbf{x}.$$
This is the purple arrow in the graph landing on $\mathbf{x}$ (when the basis is, in fact, orthogonal).
\begin{remark}
If $\beta$ is orthonormal, then $\|\mathbf{v}_i\| = 1$ and the formula simplifies even further: $c_i = \mathbf{v}_i^\mathsf{T}\mathbf{x}$.
\end{remark}
\begin{example}
Let
$$\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \qquad \mathbf{v}_2 = \begin{bmatrix} -1 \\ \phantom{-}1 \end{bmatrix}, \qquad \mathbf{x} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}.$$
Since $\mathbf{v}_1^\mathsf{T}\mathbf{v}_2 = -1 + 1 = 0$, $\beta = \{\mathbf{v}_1, \mathbf{v}_2\}$ is an orthogonal basis of $\R^2$, with $\|\mathbf{v}_1\|^2 = \|\mathbf{v}_2\|^2 = 2$. We can tehn directly calculate
$$c_1 = \frac{\mathbf{v}_1^\mathsf{T}\mathbf{x}}{\|\mathbf{v}_1\|^2} = \frac{3}{2}, \qquad c_2 = \frac{\mathbf{v}_2^\mathsf{T}\mathbf{x}}{\|\mathbf{v}_2\|^2} = -\frac{1}{2}.$$
As is good practice, we should verify with simple vector scaling and addition
$$\frac{3}{2}\begin{bmatrix} 1 \\ 1 \end{bmatrix} - \frac{1}{2}\begin{bmatrix} -1 \\ \phantom{-}1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \mathbf{x}.$$
\end{example}
\subsection*{Without orthogonality}
When $\mathbf{v}_1^\mathsf{T}\mathbf{v}_2 \neq 0$, the cross terms in the proof no longer vanish, and the projections do not sum to $\mathbf{x}$.
\begin{example}
Let
$$\mathbf{z}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad \mathbf{z}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \qquad \mathbf{x} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}.$$
These form a basis of $\R^2$, but $\mathbf{z}_1^\mathsf{T}\mathbf{z}_2 = 1 \neq 0$. The true coordinates are $c_1 = c_2 = 1$, since $\mathbf{z}_1 + \mathbf{z}_2 = \mathbf{x}$. The projections, however, are
$$P_{\mathbf{z}_1}\mathbf{x} = \frac{2}{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix}, \qquad P_{\mathbf{z}_2}\mathbf{x} = \frac{3}{2}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3/2 \\ 3/2 \end{bmatrix},$$
and their sum is
$$\begin{bmatrix} 7/2 \\ 3/2 \end{bmatrix} \neq \mathbf{x}.$$
\end{example}