| 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 +0.3 This is a standard Further Maths question testing routine application of cross product formulas. Part (i) uses the standard formula area = ½|AB × AC|, part (ii) applies the area-base-height relationship, and part (iii) uses the cross product to find the normal vector. All three parts follow textbook procedures with no novel insight required, though it's slightly above average difficulty due to being Further Maths content and requiring careful vector arithmetic across multiple parts. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04g Vector product: a x b perpendicular vector4.04h Shortest distances: between parallel lines and between skew lines |
| 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: Use scalar product to find angle | (M1A1) | |
| Find area using Area \(= \frac{1}{2}ab\sin C\) or equivalent | (M1A1) | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(d = \dfrac{ | \overrightarrow{AB}\times\overrightarrow{BC} | }{ |
| \(= 4.15\left(\frac{1}{5}\sqrt{430}\right)\) | A1 | Area triangle \(= \sin C \times |
| Alt: 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) | |
| Total | 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 | |
| Total | 2 |
## Question 6:
**Part 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: Use scalar product to find angle | (M1A1) | |
| Find area using Area $= \frac{1}{2}ab\sin C$ or equivalent | (M1A1) | |
| Total | **4** | |
**Part 6(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $d = \dfrac{|\overrightarrow{AB}\times\overrightarrow{BC}|}{|\overrightarrow{BC}|} = \dfrac{\sqrt{21^2+3^2+18^2}}{\sqrt{4^2+2^2+5^2}}$ | M1A1 | Alt method: Find angle at $C$ |
| $= 4.15\left(\frac{1}{5}\sqrt{430}\right)$ | A1 | Area triangle $= \sin C \times |AC|$ |
| Alt: 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) | |
| Total | **3** | |
**Part 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 | |
| Total | **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 $A B C$.\\
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(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]}}