| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Area of triangle using cross product |
| Difficulty | Standard +0.8 This is a Further Maths question requiring cross product for triangle area, perpendicular distance calculation, and plane equation derivation. While the techniques are standard for FP1, the multi-part nature, vector manipulation across three parts, and the fact it's Further Maths content places it moderately above average difficulty for A-level overall. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04g Vector product: a x b perpendicular vector4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} = \mathbf{i}+5\mathbf{j}-2\mathbf{k}\), \(\overrightarrow{BC} = -4\mathbf{i}-2\mathbf{j}+5\mathbf{k}\), \(\overrightarrow{AC} = -3\mathbf{i}+\mathbf{j}+3\mathbf{k}\) | B1 | 2 correct required |
| \(\overrightarrow{AB} \times \overrightarrow{BC} = 21\mathbf{i}+3\mathbf{j}+18\mathbf{k}\) | M1A1 | OE |
| Area of triangle \(ABC = \frac{1}{2}\sqrt{21^2+3^2+18^2} = 13.9\left(\frac{3}{2}\sqrt{86}\right)\) | A1 | |
| Alt method: Use scalar product to find angle | (M1A1 | |
| Find area using Area \(= \frac{1}{2}ab\sin C\) or equivalent | M1A1) | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(d = \frac{ | \overrightarrow{AB}\times\overrightarrow{BC} | }{ |
| \(= 4.15\left(\frac{1}{5}\sqrt{430}\right)\) | A1 | Area triangle \(= \sin C \times |
| Alt method: Use equation of BC to find D (foot of perpendicular) in terms of parameter and scalar product to find parameter, \(\lambda = 8/15\). Find length | (M1A1) | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| From (*) Cartesian equation is \(7x + y + 6z = \text{const.}\) | M1 | |
| Through \((2,-1,1)\), hence \(7x + y + 6z = 19\) | A1 | |
| 2 |
## Question 6:
**6(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = \mathbf{i}+5\mathbf{j}-2\mathbf{k}$, $\overrightarrow{BC} = -4\mathbf{i}-2\mathbf{j}+5\mathbf{k}$, $\overrightarrow{AC} = -3\mathbf{i}+\mathbf{j}+3\mathbf{k}$ | B1 | 2 correct required |
| $\overrightarrow{AB} \times \overrightarrow{BC} = 21\mathbf{i}+3\mathbf{j}+18\mathbf{k}$ | M1A1 | OE |
| Area of triangle $ABC = \frac{1}{2}\sqrt{21^2+3^2+18^2} = 13.9\left(\frac{3}{2}\sqrt{86}\right)$ | A1 | |
| Alt method: Use scalar product to find angle | (M1A1 | |
| Find area using Area $= \frac{1}{2}ab\sin C$ or equivalent | M1A1) | |
| | **4** | |
**6(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $d = \frac{|\overrightarrow{AB}\times\overrightarrow{BC}|}{|\overrightarrow{BC}|} = \frac{\sqrt{21^2+3^2+18^2}}{\sqrt{4^2+2^2+5^2}}$ | M1A1 | Alt method: Find angle at $C$ |
| $= 4.15\left(\frac{1}{5}\sqrt{430}\right)$ | A1 | Area triangle $= \sin C \times |AC|$ |
| Alt method: Use equation of BC to find D (foot of perpendicular) in terms of parameter and scalar product to find parameter, $\lambda = 8/15$. Find length | (M1A1) | |
| | **3** | |
**6(iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| From (*) Cartesian equation is $7x + y + 6z = \text{const.}$ | M1 | |
| Through $(2,-1,1)$, hence $7x + y + 6z = 19$ | A1 | |
| | **2** | |
---6 The points $A , B$ and $C$ have position vectors $2 \mathbf { i } - \mathbf { j } + \mathbf { k } , 3 \mathbf { i } + 4 \mathbf { j } - \mathbf { k }$ and $- \mathbf { i } + 2 \mathbf { j } + 4 \mathbf { k }$ respectively.\\
(i) Find the area of the triangle $A B C$.\\
(ii) Find the perpendicular distance of the point $A$ from the line $B C$.\\
(iii) Find the cartesian equation of the plane through $A , B$ and $C$.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q6 [9]}}