Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 11 / d

Take the cross product with $\hat{\mathbf{q}}$:Thus one choice isConversely, this cross product equation implies $(\mathbf{r}-\mathbf{p})\times \hat{\mathbf{q}}=0$, so $\mathbf{r}-\mathbf{p}$ is parallel to $\hat{\mathbf{q}}$. It therefore describes precisely the original straight line.

The unit vector $\hat{\mathbf{a}}$ specifies the direction of the straight line. Reversing its sign, together with the sign of $\mathbf{c}$, leaves the same straight line.

Decompose $\mathbf{p}$ into its orthogonal projections parallel and perpendicular to $\hat{\mathbf{a}}$. Since $\mathbf{p}\times \hat{\mathbf{a}}$ removes the parallel component,The last vector is the displacement to the point of the straight line nearest the origin. HenceThe original PDF has $c=|\mathbf{c}|$, not the squared magnitude inserted by the OCR.

The vector $\mathbf{c}=\mathbf{p}\times \hat{\mathbf{a}}$ is a normal vector to the span of $\mathbf{p}$ and $\hat{\mathbf{a}}$. Since $\mathbf{c}\ne 0$, these vectors determine a plane. Thus the missing words are

the line and the origin.

Indeed this plane consists of all linear combinations of $\mathbf{p}$ and the line's direction vector.