2016 H2 Maths Paper 1, Question 11

With $a = 0$, $p$ has direction vectors $\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 4 \\ - 2 \end{pmatrix}$. A normal vector of $p$ is

Since $l$ has direction vector $\begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix}$, which is parallel to $\mathbf{n}$, $l$ is perpendicular to $p$  (shown).

Using the point $(1, - 3,2)$ on $p$ and normal $\begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix}$, the Cartesian equation of $p$ is

With $a = 0$, $l$ is $x = - 1 - 2t$, $y = t$, $z = 1 + 2t$. Substituting into the equation of $p$:

So the point of intersection has coordinates

To find $\lambda$ and $\mu$, we equate this point to the vector equation of $p$:

From the $x$-component, $1 + \lambda = \frac{1}{9}$, so $\lambda = - \frac{8}{9}$.

From the $z$-component, $2 - 2\mu = - \frac{1}{9}$, so $\mu = \frac{19}{18}$.

$p$ has Cartesian equation $- 2x + y + 2z = - 1$, with normal $\mathbf{n} = \begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix}$, $\left| \mathbf{n} \right| = \sqrt{4 + 1 + 4} = 3$.

A general plane parallel to $p$ has equation $- 2x + y + 2z = k$ for some constant $k$. The perpendicular distance between this plane and $p$ is

For general $a$, a normal vector of $p$ is

$l$ has direction vector $\begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix}$. Requiring this to be perpendicular to the normal of $p$: