| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Topic | Vectors: Cross Product & Distances |
| Type | Volume of tetrahedron using scalar triple product |
| Difficulty | Standard +0.3 This is a straightforward Further Maths vectors question testing standard techniques: scalar triple product for volume, cross product to find normal vector, and solving simultaneous plane equations. All parts follow textbook methods with no novel problem-solving required, though the multi-step nature and Further Maths context place it slightly above average A-level difficulty. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Attempt at scalar triple product: \(\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}\) or \(\mathbf{a} \times \mathbf{b} \cdot \mathbf{c}\) | M1 | |
| NB \(\mathbf{b} \times \mathbf{c} = \begin{pmatrix} 36 \\ -41 \\ 145 \end{pmatrix}\) and \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -73 \\ 139 \\ 97 \end{pmatrix}\) | M1 | |
| Use of \(V = \frac{1}{6} | \mathbf{a} \cdot \mathbf{b} \times \mathbf{c} | \) formula |
| Answer \(335\frac{1}{6}\) or \(\frac{2011}{6}\) | A1 | |
| [3] | ||
| (b)(i) \(\mathbf{n} = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 6 \\ 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 7 \\ -21 \\ 0 \end{pmatrix}\) accept \(\pm \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}\) etc. | M1, A1 | Alt approach possible that eliminates and from the original eqn. |
| \(\mathbf{r} \cdot \mathbf{n} = \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 3 \\ 0 \end{pmatrix} = 1 = d\) | M1, A1 | ft incorrect \(\mathbf{n}\) |
| [4] | ||
| (ii) \(-x + 3y = 1\) and \(x + 4y + 7z = 13\) ft 1st eqn. from (i) | M1 | |
| Set \(y = \lambda \Rightarrow\) Parametrisation attempt | M1, A1, A1 | |
| \(x = 3\lambda - 1, z = 2 - \lambda\) | ||
| \(\mathbf{r} = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}\) or any other correct vector line eqn. form ft | B1 | |
| MUST have \(\mathbf{r} =\) at the start (allow \(\mathbf{r} =\)) | ||
| [4] | ||
| ALTERNATIVE | ||
| d.v. \(= \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ 4 \\ 7 \end{pmatrix} = \begin{pmatrix} -21 \\ -7 \\ 7 \end{pmatrix}\) accept \(\pm \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}\) etc. | M1, A1 | |
| ft \(\begin{pmatrix} -3 \\ 1 \\ -1 \end{pmatrix}\) or \(\begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}\) | ||
| Finding any point on the line; e.g. \((-1, 0, 2), (2, 1, 1), (5, 2, 0)\), etc. | B1, A1 | |
| Answer as above ft point and d.v. | [4] |
**(a)** Attempt at scalar triple product: $\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}$ or $\mathbf{a} \times \mathbf{b} \cdot \mathbf{c}$ | M1 |
NB $\mathbf{b} \times \mathbf{c} = \begin{pmatrix} 36 \\ -41 \\ 145 \end{pmatrix}$ and $\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -73 \\ 139 \\ 97 \end{pmatrix}$ | M1 |
Use of $V = \frac{1}{6}|\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}|$ formula | M1 |
Answer $335\frac{1}{6}$ or $\frac{2011}{6}$ | A1 |
| [3] |
**(b)(i)** $\mathbf{n} = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 6 \\ 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 7 \\ -21 \\ 0 \end{pmatrix}$ accept $\pm \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}$ etc. | M1, A1 | Alt approach possible that eliminates and from the original eqn.
$\mathbf{r} \cdot \mathbf{n} = \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 3 \\ 0 \end{pmatrix} = 1 = d$ | M1, A1 | ft incorrect $\mathbf{n}$
| [4] |
**(ii)** $-x + 3y = 1$ and $x + 4y + 7z = 13$ ft 1st eqn. from (i) | M1 |
Set $y = \lambda \Rightarrow$ Parametrisation attempt | M1, A1, A1 |
$x = 3\lambda - 1, z = 2 - \lambda$ | |
$\mathbf{r} = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$ or any other correct vector line eqn. form ft | B1 |
MUST have $\mathbf{r} =$ at the start (allow $\mathbf{r} =$) | |
| [4] |
**ALTERNATIVE** | |
d.v. $= \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ 4 \\ 7 \end{pmatrix} = \begin{pmatrix} -21 \\ -7 \\ 7 \end{pmatrix}$ accept $\pm \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$ etc. | M1, A1 |
ft $\begin{pmatrix} -3 \\ 1 \\ -1 \end{pmatrix}$ or $\begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}$ | |
Finding any point on the line; e.g. $(-1, 0, 2), (2, 1, 1), (5, 2, 0)$, etc. | B1, A1 |
Answer as above ft point and d.v. | [4] |
---\begin{enumerate}[label=(\alph*)]
\item The points $A$, $B$ and $C$ have position vectors
$$\mathbf{a} = \begin{pmatrix} 19 \\ 3 \\ 10 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 12 \\ 7 \\ -1 \end{pmatrix} \quad \text{and} \quad \mathbf{c} = \begin{pmatrix} 5 \\ 15 \\ 3 \end{pmatrix}$$
respectively, and $O$ is the origin. Calculate the volume of the tetrahedron $OABC$. [3]
\item \begin{enumerate}[label=(\roman*)]
\item The plane $\Pi_1$ has equation $\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 6 \\ 2 \\ 5 \end{pmatrix}$. Determine an equation for $\Pi_1$ in the form $\mathbf{r} \cdot \mathbf{n} = d$. [4]
\item A second plane, $\Pi_2$, has equation $\mathbf{r} \cdot \begin{pmatrix} 1 \\ 4 \\ 7 \end{pmatrix} = 13$. Find a vector equation for the line of intersection of $\Pi_1$ and $\Pi_2$. [4]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2011 Q9 [11]}}