Standard +0.3 This is a standard Further Maths vectors question requiring routine techniques: substituting a line equation into a plane equation to find intersection point P, reading off the normal vector, computing a cross product, and forming a line equation. While it involves multiple steps and the cross product calculation requires care, each component is a textbook procedure with no novel insight required. Slightly easier than average A-level due to straightforward application of standard methods.
8 A line, passing through the point \(A ( 3,0,2 )\), has vector equation \(\mathbf { r } = 3 \mathbf { i } + 2 \mathbf { k } + \lambda ( 2 \mathbf { i } + \mathbf { j } - 2 \mathbf { k } )\). It meets the plane \(\Pi\), which has equation \(\mathbf { r } \cdot ( \mathbf { i } + 2 \mathbf { j } + \mathbf { k } ) = 3\), at the point \(P\). Find the coordinates of \(P\).
Write down a vector \(\mathbf { n }\) which is perpendicular to \(\Pi\), and calculate the vector \(\mathbf { w }\), where
$$\mathbf { w } = \mathbf { n } \times ( 2 \mathbf { i } + \mathbf { j } - 2 \mathbf { k } )$$
The point \(Q\) lies in \(\Pi\) and is the foot of the perpendicular from \(A\) to \(\Pi\). Use the vector \(\mathbf { w }\) to determine an equation of the line \(P Q\) in the form \(\mathbf { r } = \mathbf { u } + \mu \mathbf { v }\).
Direction of \(PQ\) is \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-5&4&-3\\1&2&1\end{vmatrix} = \begin{pmatrix}10\\2\\-14\end{pmatrix} \sim \begin{pmatrix}5\\1\\-7\end{pmatrix}\)
M1A1
OE
Equation of \(PQ\) is \(\mathbf{r} = \begin{pmatrix}1\\-1\\4\end{pmatrix} + \mu\begin{pmatrix}5\\1\\-7\end{pmatrix}\)
dM1A1
[N.B. For methods not using \(\mathbf{w}\): at most three B1 marks.]
## Question 8:
| Working/Answer | Marks | Notes |
|---|---|---|
| $\begin{pmatrix}3+2\lambda\\ \lambda\\ 2-2\lambda\end{pmatrix} \cdot \begin{pmatrix}1\\2\\1\end{pmatrix} = 3$ | M1 | |
| $3+2\lambda+2\lambda+2-2\lambda = 3 \Rightarrow \lambda = -1$ | A1 | |
| $P$ is $(1,-1,4)$ | A1 | Accept position vector |
| $\mathbf{n} = \begin{pmatrix}1\\2\\1\end{pmatrix}$ | B1 | |
| $\mathbf{w} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2&1\\2&1&-2\end{vmatrix} = \begin{pmatrix}-5\\4\\-3\end{pmatrix}$ | M1A1 | |
| Direction of $PQ$ is $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-5&4&-3\\1&2&1\end{vmatrix} = \begin{pmatrix}10\\2\\-14\end{pmatrix} \sim \begin{pmatrix}5\\1\\-7\end{pmatrix}$ | M1A1 | OE |
| Equation of $PQ$ is $\mathbf{r} = \begin{pmatrix}1\\-1\\4\end{pmatrix} + \mu\begin{pmatrix}5\\1\\-7\end{pmatrix}$ | dM1A1 | |
| [N.B. For methods not using $\mathbf{w}$: at most three B1 marks.] | | |
8 A line, passing through the point $A ( 3,0,2 )$, has vector equation $\mathbf { r } = 3 \mathbf { i } + 2 \mathbf { k } + \lambda ( 2 \mathbf { i } + \mathbf { j } - 2 \mathbf { k } )$. It meets the plane $\Pi$, which has equation $\mathbf { r } \cdot ( \mathbf { i } + 2 \mathbf { j } + \mathbf { k } ) = 3$, at the point $P$. Find the coordinates of $P$.
Write down a vector $\mathbf { n }$ which is perpendicular to $\Pi$, and calculate the vector $\mathbf { w }$, where
$$\mathbf { w } = \mathbf { n } \times ( 2 \mathbf { i } + \mathbf { j } - 2 \mathbf { k } )$$
The point $Q$ lies in $\Pi$ and is the foot of the perpendicular from $A$ to $\Pi$. Use the vector $\mathbf { w }$ to determine an equation of the line $P Q$ in the form $\mathbf { r } = \mathbf { u } + \mu \mathbf { v }$.
\hfill \mbox{\textit{CAIE FP1 2015 Q8 [10]}}