It is often useful to geometrically represent random variables $X_{1}, \ldots, X_{p}$ (theoretical or empirical data) as vectors $\bf{x}_{1}, \ldots, \bf{x}_{p}$ such that their standard deviations $\sigma(X_{i})$ equal their lengths $||\bf{x}_{i}||$, and their correlations $\rho(X_{i}, X_{j})$ equal the cosine of their angles $\angle(\bf{x}_{i}, \bf{x}_{j})$. One can then use graphical illustrations and geometric intuitions to gain statistical insight.
To this end, let $\bf{S}$ be the $(p \times p)$-covariance matrix of $X_{1}, \ldots, X_{p}$ with rank $k$. Since $\bf{S}$ is positive semidefinite, we can find a decomposition $\bf{S} = \bf{B} \bf{B}'$ by defining the $(p \times k)$-matrix $\bf{B} := \bf{G} \bf{D}^{1/2}$, where $\bf{G}$ is the $(p \times k)$-matrix of eigenvectors of $\bf{S}$ and $\bf{D}$ is the $(k \times k)$-diagonal matrix of corresponding positive eigenvalues.
$\bf{B} \bf{B}'$ is the matrix of dot products of the rows of $\bf{B}$, i.e., $\bf{B}_{ij} = <\bf{B}_{i}, \bf{B}_{j}>$. Now we get the desired representation in $k$-dimensional space by defining $\bf{x}_{i} := \bf{B}_{i}$. Because then we have $$ ||\bf{x}_{i}|| = \sqrt{<\bf{x}_{i}, \bf{x}_{i}>} = \sqrt{\bf{S}_{ii}} = \sigma(X_{i}) $$
And we also have $$ \begin{array}{rcl} \cos(\angle(\bf{x}_{i}, \bf{x}_{j})) &=& \frac{<\bf{x}_{i}, \bf{x}_{j}>}{||\bf{x}_{i}|| \, ||\bf{x}_{j}||} = \frac{<\bf{x}_{i}, \bf{x}_{j}>}{\sqrt{<\bf{x}_{i}, \bf{x}_{i}>} \, \sqrt{<\bf{x}_{j}, \bf{x}_{j}>}}\\ &=& \frac{\bf{S}_{ij}}{\sqrt{\bf{S}_{ii}} \, \sqrt{\bf{S}_{jj}}} = \frac{Cov(X_{i}, X_{j})}{\sigma(X_{i}) \, \sigma(X_{j})}\\ &=& \rho(X_{i}, X_{j}) \end{array} $$
Since $\cos(0) = 1$, maximizing the correlation between variables can be viewed as minimizing the angle between their corresponding vectors.
If we have empirical data $n$-vectors $\bf{x}_{i}$ with means $\bar{\bf{x}}_{i}$, then the representation immediately follows for the corresponding centered variables $\dot{\bf{x}}_{i}$ since $<\dot{\bf{x}}_{i}, \dot{\bf{x}}_{j}> = \sum\limits_{r=1}^{n}(\bf{x}_{ir} - \bar{\bf{x}}_{i})(\bf{x}_{jr} - \bar{\bf{x}}_{j}) = n \, Cov(X_{i}, X_{j})$.
So in this case, $\bf{x}_{i} / \sqrt{n}$ already is the desired representation.
For applications, see e.g. this answer or this answer.