Past exam of the mathematical methods I course of the University of Oxford / 2021 resit / 3

The normal vectors are ${\mathbf{n}}_A=(1,2,-4)$ and ${\mathbf{n}}_B=(2,-1,3)$. Their dot product is $-12$, and their norms are $\sqrt{21}$ and $\sqrt{14}$. The usual angle between two planes is the smaller, unoriented angle, soFor the particular signed normal vectors printed above, the angle is the supplement $\pi -\theta \approx 134.4^{\circ }$. Reversing either normal vector exchanges these two angles without changing either plane.

A parallel plane has the same normal vector, so write $C:x+2y-4z=d$. Substitution of the given point gives $d=4+4-4=4$. ThusA displacement between the two planes has a component of magnitude $|5-4|/\Vert {\mathbf{n}}_A\Vert$ along their common unit vector normal. The distance between parallel planes is consequently