| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Line of intersection of two planes |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring cross products, plane equations, and finding the line of intersection of two planes. While the individual techniques (cross product, normal vector to plane, parametric line from two planes) are standard FP3 content, the question requires careful execution across multiple steps and the final part demands understanding of the vector form (r-a)×b=0 for a line. This is moderately challenging for Further Maths students but follows established procedures without requiring novel insight. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04g Vector product: a x b perpendicular vector |
| Answer | Marks |
|---|---|
| \(\overrightarrow{PQ} = \begin{pmatrix} -3 \\ 4 \\ -4 \end{pmatrix}\), \(\overrightarrow{QR} = \begin{pmatrix} 2 \\ 2 \\ -2 \end{pmatrix}\) | B1 |
| \(\overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 4 & -4 \\ 2 & 2 & -2 \end{vmatrix} = \begin{pmatrix} 0 \\ -14 \\ -14 \end{pmatrix}\) | M1 A2, 1, 0 (4) |
| Answer | Marks |
|---|---|
| \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = -2 \Rightarrow y + z = -2\) or equivalent | M1 A1 (2) |
| Answer | Marks |
|---|---|
| Let \(z = \lambda \Rightarrow y = -\lambda - 2, x = 2\lambda + 8\) | M1 A1 A1 |
| \(\mathbf{r} = \begin{pmatrix} 8 \\ -2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\) | M1 |
| \(\left(\mathbf{r} - \begin{pmatrix} 8 \\ -2 \\ 0 \end{pmatrix}\right) \times \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = 0\) | A1 (5) |
**(a)**
$\overrightarrow{PQ} = \begin{pmatrix} -3 \\ 4 \\ -4 \end{pmatrix}$, $\overrightarrow{QR} = \begin{pmatrix} 2 \\ 2 \\ -2 \end{pmatrix}$ | B1 |
$\overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 4 & -4 \\ 2 & 2 & -2 \end{vmatrix} = \begin{pmatrix} 0 \\ -14 \\ -14 \end{pmatrix}$ | M1 A2, 1, 0 (4) |
**(b)**
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = -2 \Rightarrow y + z = -2$ or equivalent | M1 A1 (2) |
**(c)**
$y + z = -2$
$x + y - z = 6$
Let $z = \lambda \Rightarrow y = -\lambda - 2, x = 2\lambda + 8$ | M1 A1 A1 |
$\mathbf{r} = \begin{pmatrix} 8 \\ -2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$ | M1 |
$\left(\mathbf{r} - \begin{pmatrix} 8 \\ -2 \\ 0 \end{pmatrix}\right) \times \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = 0$ | A1 (5) |
(11 marks)\begin{enumerate}
\item Referred to a fixed origin O , the points $\mathrm { P } , \mathrm { Q }$ and R have coordinates $( \mathbf { i } - 3 \mathbf { j } + \mathbf { k } ) , ( - 2 \mathbf { i } + \mathbf { j } - 3 \mathbf { k } )$ and $( 3 \mathbf { j } - 5 \mathbf { k } )$ respectively. The plane $\Pi _ { 1 }$ passes through $\mathrm { P } , \mathrm { Q }$ and R . Find\\
(a) $\overrightarrow { \mathrm { PQ } } \times \overrightarrow { \mathrm { QR } }$,\\
(b) a cartesian equation of $\Pi _ { 1 }$.
\end{enumerate}
The plane $\Pi _ { 2 }$ has equation $\mathbf { r }$. ( $\mathbf { i } + \mathbf { j } - \mathbf { k }$ ) $= 6$. The planes $\Pi _ { 1 }$ and $\Pi _ { 2 }$ intersect in the line I .\\
(c) Find a vector equation of I, giving your answer in the form ( $\mathbf { r } - \mathbf { a }$ ) $\times \mathbf { b } = \mathbf { 0 }$.\\
\hfill \mbox{\textit{Edexcel FP3 Q6 [11]}}