| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Acute angle between two planes |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question requiring routine techniques: finding a normal vector via cross product, converting to Cartesian form, calculating angle between planes using dot product, and finding line of intersection. While it involves multiple steps and Further Maths content, each part follows well-practiced procedures without requiring novel insight or complex problem-solving. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 1 & 3 \\ 0 & 1 & 2 \end{vmatrix} = -\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}\) | M1 | Finds normal to \(\Pi_1\) |
| \(-x + 8y - 4z = 3\) | M1 A1 | Uses point on plane to find Cartesian equation, AEF |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}-1\\8\\-4\end{pmatrix} \cdot \begin{pmatrix}3\\1\\-1\end{pmatrix} = 9\sqrt{11}\cos\theta\) | M1 | Uses scalar product |
| \(\Rightarrow \theta = 72.5°\) | A1 | Accept 1.26 rad |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & -1 \\ -1 & 8 & -4\end{vmatrix} = 4\mathbf{i} + 13\mathbf{j} + 25\mathbf{k}\) | M1 A1 | Finds direction of line of intersection |
| Point on both planes is, e.g. \((1,1,1)\) | M1 A1 | Finds point common to both planes |
| \(\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(4\mathbf{i} + 13\mathbf{j} + 25\mathbf{k})\) | A1 | States vector equation of line |
| 5 |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 1 & 3 \\ 0 & 1 & 2 \end{vmatrix} = -\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}$ | M1 | Finds normal to $\Pi_1$ |
| $-x + 8y - 4z = 3$ | M1 A1 | Uses point on plane to find Cartesian equation, AEF |
| | **3** | |
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-1\\8\\-4\end{pmatrix} \cdot \begin{pmatrix}3\\1\\-1\end{pmatrix} = 9\sqrt{11}\cos\theta$ | M1 | Uses scalar product |
| $\Rightarrow \theta = 72.5°$ | A1 | Accept 1.26 rad |
| | **2** | |
## Question 8(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & -1 \\ -1 & 8 & -4\end{vmatrix} = 4\mathbf{i} + 13\mathbf{j} + 25\mathbf{k}$ | M1 A1 | Finds direction of line of intersection |
| Point on both planes is, e.g. $(1,1,1)$ | M1 A1 | Finds point common to both planes |
| $\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(4\mathbf{i} + 13\mathbf{j} + 25\mathbf{k})$ | A1 | States vector equation of line |
| | **5** | |8 The plane $\Pi _ { 1 }$ has equation
$$\mathbf { r } = \left( \begin{array} { l }
5 \\
1 \\
0
\end{array} \right) + s \left( \begin{array} { r }
- 4 \\
1 \\
3
\end{array} \right) + t \left( \begin{array} { l }
0 \\
1 \\
2
\end{array} \right)$$
(i) Find a cartesian equation of $\Pi _ { 1 }$.\\
The plane $\Pi _ { 2 }$ has equation $3 x + y - z = 3$.\\
(ii) Find the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$, giving your answer in degrees.\\
(iii) Find an equation of the line of intersection of $\Pi _ { 1 }$ and $\Pi _ { 2 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + \lambda \mathbf { b }$. [5]\\
\hfill \mbox{\textit{CAIE FP1 2018 Q8 [5]}}