Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV / 1

For vectors $\mathbf{a},\mathbf{b},\mathbf{c}\in {\mathbb{R}}^3$, the scalar product, cross product and scalar triple product are respectivelyExpanding the dot product givesInterchanging $\mathbf{a}$ and $\mathbf{b}$ changes the sign of their cross product, so its dot product with $\mathbf{b}$ is also zero. Thus the cross product is orthogonal to both vectors, including when it is the zero vector.

The displayed scalar triple product is a determinant. A square matrix has nonzero determinant exactly when its rows are linearly independent. Hence

Subtracting the first row from the other two in the determinant givesThis vanishes exactly when the two displacements $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ are linearly dependent. If $\mathbf{a}\ne \mathbf{b}$, that means $\mathbf{c}-\mathbf{a}=t(\mathbf{b}-\mathbf{a})$ for some real number $t$. Thus all three points lie on the lineConversely, putting three points on any one line makes their displacements linearly dependent, so the determinant vanishes.

The equivalence remains true without distinctness. If exactly two points coincide, the two distinct locations determine a line containing all three, and two rows of the original determinant coincide. If all three points coincide, any line through that point works and the determinant is again zero. Existence persists, although uniqueness of the line can fail.

The value $f(\mathbf{a},\mathbf{b},\mathbf{c})$ is twice the signed area of the triangle traversed in that order; it is positive for counterclockwise orientation. Choose points on the parabolaFor $i<j<k$, subtracting the first row in the determinant and factoring givesEach of the three factors is positive, proving the required property for every integer $n\ge 3$.