| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Area of triangle using vector product |
| Difficulty | Standard +0.3 This is a straightforward application of the vector product formula to find area and perpendicular distance. Students must compute two vectors, perform a standard cross product calculation, then apply the direct formula that area = ½|AB × AC| and use the relationship between area, base, and height. While it requires multiple steps and is from Further Maths, the techniques are routine and mechanical with no conceptual challenges or novel insights required. |
| Spec | 4.04g Vector product: a x b perpendicular vector4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}1\\2\\3\end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix}1\\1\\2\end{pmatrix}\) | B1 | Finds vectors |
| \(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 2 & 3\\1 & 1 & 2\end{vmatrix} = \begin{pmatrix}1\\1\\-1\end{pmatrix}\) | M1A1 | Finds vector product |
| Area of triangle \(ABC = \frac{1}{2}\left\ | \begin{pmatrix}1\\1\\-1\end{pmatrix}\right\ | = \frac{1}{2}\sqrt{3}\) |
| \(\frac{1}{2}\sqrt{1^2+2^2+3^2}\, d = \frac{1}{2}\sqrt{3} \Rightarrow d = \sqrt{\frac{3}{14}}\) | A1 | Finds length of perpendicular |
## Question 4:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix}1\\2\\3\end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix}1\\1\\2\end{pmatrix}$ | B1 | Finds vectors |
| $\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 2 & 3\\1 & 1 & 2\end{vmatrix} = \begin{pmatrix}1\\1\\-1\end{pmatrix}$ | M1A1 | Finds vector product |
| Area of triangle $ABC = \frac{1}{2}\left\|\begin{pmatrix}1\\1\\-1\end{pmatrix}\right\| = \frac{1}{2}\sqrt{3}$ | M1A1 | Finds area of triangle |
| $\frac{1}{2}\sqrt{1^2+2^2+3^2}\, d = \frac{1}{2}\sqrt{3} \Rightarrow d = \sqrt{\frac{3}{14}}$ | A1 | Finds length of perpendicular |
---4 The points $A , B$ and $C$ have position vectors $\mathbf { i } + 2 \mathbf { j } + 2 \mathbf { k } , 2 \mathbf { i } + 4 \mathbf { j } + 5 \mathbf { k }$ and $2 \mathbf { i } + 3 \mathbf { j } + 4 \mathbf { k }$ respectively. Find $\overrightarrow { A B } \times \overrightarrow { A C }$.
Deduce, in either order, the exact value of\\
(i) the area of the triangle $A B C$,\\
(ii) the perpendicular distance from $C$ to $A B$.
\hfill \mbox{\textit{CAIE FP1 2012 Q4 [6]}}