Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 11

Let $\Delta =[\mathbf{a},\mathbf{b},\mathbf{c}]\ne 0$. These three vectors are linearly independent, so write $\mathbf{d}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. Taking the dot product with $\mathbf{b}\times \mathbf{c}$ gives $\alpha =[\mathbf{b},\mathbf{c},\mathbf{d}]/\Delta$. Likewise, the dot products with $\mathbf{c}\times \mathbf{a}$ and $\mathbf{a}\times \mathbf{b}$ giveHere the cyclic invariance and sign change under an interchange of the scalar triple product fix every sign. Substituting the three coefficients provesThis is precisely coordinate extraction using the reciprocal basis $(\mathbf{b}\times \mathbf{c},\mathbf{c}\times \mathbf{a},\mathbf{a}\times \mathbf{b})/\Delta$.

Put $\mathbf{v}={\mathbf{r}}_0-\mathbf{a}$. The squared distance to a point on the line is $|\mathbf{v}-\lambda \hat{\mathbf{t}}{|}^2$. Since $|\hat{\mathbf{t}}|=1$, completing the square in $\lambda$ gives its minimum at $\lambda =\hat{\mathbf{t}}\cdot \mathbf{v}$, with value $|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$. The cross product identity $|\hat{\mathbf{t}}\times \mathbf{v}{|}^2=|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$ therefore provesFor the given line, $\hat{\mathbf{t}}=(1,1,0)/\sqrt{2}$, so its squared distance to $(x,y,z)$ is $(x-y{)}^2/2+z^2$. Substituting the parabola givesThe global minimum occurs at $\mu =\pm 1$, henceThe corresponding points on the parabola are $\pm (1,1+2\sqrt{3},0)+(0,0,-3)$; their nearest points on the line are $\pm (1+\sqrt{3},1+\sqrt{3},0)$. The connecting vectors are orthogonal to the line, as the projection calculation requires.

Write $s=\hat{\mathbf{n}}\cdot \mathbf{x}$. Taking the dot product of the equation with the unit vector $\hat{\mathbf{n}}$ gives $2s+s^2=-1$, so $(s+1{)}^2=0$ and $s=-1$.

Now take the cross product with $\hat{\mathbf{n}}$. The vector triple product gives $\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{x})=s\hat{\mathbf{n}}-\mathbf{x}=-\hat{\mathbf{n}}-\mathbf{x}$, soThe original equation, now with $s^2=1$, also gives $\hat{\mathbf{n}}\times \mathbf{x}=\mathbf{b}-\hat{\mathbf{n}}-2\mathbf{x}$. Substitution eliminates the remaining cross product of the unknown and yieldsIts dot product with $\hat{\mathbf{n}}$ is $-1$, and direct substitution using the vector triple product verifies the original equation. The elimination gives a unique solution without choosing components. It is also an application of the inverse of a scalar plus a cross-product operator after the scalar component has been fixed.