| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Equation of plane through three points |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard vector product techniques: computing a cross product from given vectors, using it to find area, and deriving a plane equation. All steps are routine applications of formulas with no conceptual challenges, making it slightly easier than average even for FM students. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AC} = 3\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}\) | B1 | cao, any form |
| \(\overrightarrow{BC} = -3\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\) | B1 | cao, any form |
| \(\overrightarrow{AC} \times \overrightarrow{BC} = 10\mathbf{i} - 15\mathbf{j} + 30\mathbf{k}\) | M1 A1 | Attempt to find cross product, modulus of one term correct; cao, any form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of triangle \(ABC = \frac{1}{2}\ | 10\mathbf{i} - 15\mathbf{j} + 30\mathbf{k}\ | = \frac{1}{2}\sqrt{1225} = 17.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(10x - 15y + 30z = -20\) or \(2x - 3y + 6z = -4\) | M1 | Using answer to (a) to find equation of plane |
| \(\mathbf{r}\cdot(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = -4\) or correct multiple | A1 | cao |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AC} = 3\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}$ | B1 | cao, any form |
| $\overrightarrow{BC} = -3\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}$ | B1 | cao, any form |
| $\overrightarrow{AC} \times \overrightarrow{BC} = 10\mathbf{i} - 15\mathbf{j} + 30\mathbf{k}$ | M1 A1 | Attempt to find cross product, modulus of one term correct; cao, any form |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of triangle $ABC = \frac{1}{2}\|10\mathbf{i} - 15\mathbf{j} + 30\mathbf{k}\| = \frac{1}{2}\sqrt{1225} = 17.5$ | M1 A1 | Modulus of answer to (a), condone missing $\frac{1}{2}$; cao |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $10x - 15y + 30z = -20$ or $2x - 3y + 6z = -4$ | M1 | Using answer to (a) to find equation of plane |
| $\mathbf{r}\cdot(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = -4$ or correct multiple | A1 | cao |
---3. The position vectors of the points $A , B$ and $C$ relative to an origin $O$ are $\mathbf { i } - 2 \mathbf { j } - 2 \mathbf { k } , 7 \mathbf { i } - 3 \mathbf { k }$ and $4 \mathbf { i } + 4 \mathbf { j }$ respectively.
Find
\begin{enumerate}[label=(\alph*)]
\item $\overrightarrow { A C } \times \overrightarrow { B C }$,
\item the area of triangle $A B C$,
\item an equation of the plane $A B C$ in the form $\mathbf { r } . \mathbf { n } = p$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2012 Q3 [8]}}