= Solution
{c}
The two-vector <Gram matrix> is
$$
A=\begin{pmatrix}\|v_1\|^2&\langle v_1,v_2\rangle\\
\langle v_1,v_2\rangle&\|v_2\|^2\end{pmatrix},
\qquad
\det A=\|v_1\|^2\|v_2\|^2-\langle v_1,v_2\rangle^2.
$$
The upper bound follows because the subtracted square is nonnegative. If either <vector> is zero, the <determinant> is zero. Otherwise apply the non-strict version of the calculation in <4/a/i>: $\|v_1-tv_2\|^2\geq0$ with $t=\langle v_1,v_2\rangle/\|v_2\|^2$ gives $\langle v_1,v_2\rangle^2\leq\|v_1\|^2\|v_2\|^2$. Therefore
$$
\boxed{0\leq\det A\leq\|v_1\|^2\|v_2\|^2.}
$$
Equality on the left holds exactly when the two <vectors> are <linearly dependent>; equality on the right holds exactly when they are <orthogonal>. The <Gram determinant> measures the squared <area> of their <parallelogram>: in the nonzero case, $\det A=\|v_2\|^2\|v_1-tv_2\|^2$, the square of base times perpendicular height.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-i-gram-determinant.png]
{title=The Gram determinant as squared parallelogram area, and its dependence on the angle between unit vectors}
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