# 3.II.6E

Derive an expression for the triple scalar product $\left(\mathbf{e}_{1} \times \mathbf{e}_{2}\right) \cdot \mathbf{e}_{3}$ in terms of the determinant of the matrix $E$ whose rows are given by the components of the three vectors $\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$.

Use the geometrical interpretation of the cross product to show that $\mathbf{e}_{a}, a=1,2,3$, will be a not necessarily orthogonal basis for $\mathbb{R}^{3}$ as long as $\operatorname{det} E \neq 0$.

The rows of another matrix $\hat{E}$ are given by the components of three other vectors $\hat{\mathbf{e}}_{b}, b=1,2,3$. By considering the matrix $E \hat{E}^{\mathrm{T}}$, where ${ }^{\mathrm{T}}$ denotes the transpose, show that there is a unique choice of $\hat{E}$ such that $\hat{\mathbf{e}}_{b}$ is also a basis and

$\mathbf{e}_{a} \cdot \hat{\mathbf{e}}_{b}=\delta_{a b}$

Show that the new basis is given by

$\hat{\mathbf{e}}_{1}=\frac{\mathbf{e}_{2} \times \mathbf{e}_{3}}{\left(\mathbf{e}_{1} \times \mathbf{e}_{2}\right) \cdot \mathbf{e}_{3}} \quad \text { etc. }$

Show that if either $\mathbf{e}_{a}$ or $\hat{\mathbf{e}}_{b}$ is an orthonormal basis then $E$ is a rotation matrix.