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

The scalar $\mathbf{r}\cdot \hat{\mathbf{u}}$ is the signed scalar projection along the unit vector $\hat{\mathbf{u}}$. Multiplying by that unit vector makes $P_{\hat{\mathbf{u}}}(\mathbf{r})$ the orthogonal projection of $\mathbf{r}$ onto its span.

For the other expression, the vector triple product identity givesThis is the orthogonal projection onto a plane perpendicular to $\hat{\mathbf{u}}$ through the origin. Its dot product with $\hat{\mathbf{u}}$ is zero, while the difference from $\mathbf{r}$ is parallel to $\hat{\mathbf{u}}$.

The two orthogonal projections are complementary, givingThey split a vector into its components parallel and perpendicular to $\hat{\mathbf{u}}$.

The straight line consists of $\mathbf{y}+s(\mathbf{z}-\mathbf{y})$, with $s\in \mathbb{R}$. Since the endpoints are distinct, the direction vector can be normalized. ChoosePutting $t=s|\mathbf{z}-\mathbf{y}|$ gives the required parametric equation of a straight line $\mathbf{r}=\mathbf{p}+t\hat{\mathbf{q}}$. Other choices of $\mathbf{p}$ on the same straight line, or the opposite direction for $\hat{\mathbf{q}}$, give equivalent descriptions.

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.

A necessary compatibility condition is $\mathbf{f}\cdot \hat{\mathbf{g}}=0$, because a cross product with $\hat{\mathbf{g}}$ is perpendicular to it. If this condition fails, there is no solution.

If it holds, take $\hat{\mathbf{g}}\times$ the equation and use the vector triple product identity:Thus the perpendicular orthogonal projection is fixed and the parallel component is free:Conversely, $(\hat{\mathbf{g}}\times \mathbf{f})\times \hat{\mathbf{g}}=\mathbf{f}$ under the compatibility condition, so every displayed vector solves the equation. The solution set is a straight line, including the line through the origin when $\mathbf{f}=0$.