| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Volume of tetrahedron using scalar triple product |
| Difficulty | Standard +0.8 This is a Further Maths question requiring knowledge of vector cross product for area and scalar triple product for volume. While the techniques are standard for FM students, it involves multiple steps (finding vectors, computing cross product, then using volume formula with scalar triple product) and requires careful algebraic manipulation. The conceptual demand is moderate for FM but above typical A-level maths. |
| Spec | 4.04g Vector product: a x b perpendicular vector6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete correct process: vector product of 2 edges of triangle \(ABC\), applying Pythagoras and multiplying by 0.5 | M1 | 3.1b |
| \(\overrightarrow{AB} = \begin{pmatrix}4\\0\\-2\end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix}2\\9\\-1\end{pmatrix}\), \(\overrightarrow{BC} = \begin{pmatrix}-2\\9\\1\end{pmatrix}\) | M1 | 1.1b — Uses correct method to find any 2 edges of triangle \(ABC\) |
| \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\4&0&-2\\2&9&-1\end{vmatrix} = \ldots\) or \(\begin{pmatrix}4\\0\\-2\end{pmatrix}\times\begin{pmatrix}2\\9\\-1\end{pmatrix} = \ldots\) | M1 | 1.1b — Attempts vector cross product of 2 edges |
| \(= 18\mathbf{i} + 0\mathbf{j} + 36\mathbf{k}\) | ||
| Area \(ABC = \frac{1}{2}\sqrt{(18)^2+(0)^2+(36)^2}\) | ||
| \(\{= 20.1246\ldots\} = 9\sqrt{5}\) (cm²) or awrt 20.1 (cm²) | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses appropriate vectors to find an angle or perpendicular height in triangle \(ABC\) and uses correct method for complete attempt to find area | M1 | 3.1b |
| \(\overrightarrow{AB} = \begin{pmatrix}4\\0\\-2\end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix}2\\9\\-1\end{pmatrix}\), \(\overrightarrow{BC} = \begin{pmatrix}-2\\9\\1\end{pmatrix}\) | M1 | 1.1b — Uses correct method to find any 2 edges |
| Uses correct method to find an angle or perpendicular height in triangle \(ABC\) | M1 | 1.1b |
| \(\hat{BAC} = 27.905\ldots\), \(\hat{ABC} = 76.047\ldots\), \(\hat{BCA} = 76.047\ldots\) or perpendicular height \(= 9\) | ||
| Area \(ABC = \frac{1}{2}\sqrt{86}\sqrt{20}\sin 76.047\ldots\) or \(\frac{1}{2}\sqrt{86}\sqrt{86}\sin 27.905\ldots\) or \(\frac{1}{2}\sqrt{20}(9)\) | ||
| \(\{= 20.1246\ldots\} = 9\sqrt{5}\) (cm²) or awrt 20.1 (cm²) | A1 | 2.2a |
| Note: Using any of \(\overrightarrow{OA}\), \(\overrightarrow{OB}\) or \(\overrightarrow{OC}\) in vector product is M0 M0 A0 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete attempt to find area by applying \(\frac{1}{2}\ | \overrightarrow{OA}\times\overrightarrow{OB} + \overrightarrow{OB}\times\overrightarrow{OC} + \overrightarrow{OC}\times\overrightarrow{OA}\ | \) or equivalent |
| \(\overrightarrow{OA}\times\overrightarrow{OB} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&1&4\\6&1&2\end{vmatrix} = \ldots\) and \(\overrightarrow{OB}\times\overrightarrow{OC} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\6&1&2\\4&10&3\end{vmatrix} = \ldots\) and \(\overrightarrow{OC}\times\overrightarrow{OA} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\4&10&3\\2&1&4\end{vmatrix} = \ldots\) | M1 | 1.1b — Attempts to apply all three cross products |
| \(\{\overrightarrow{OA}\times\overrightarrow{OB}+\overrightarrow{OB}\times\overrightarrow{OC}+\overrightarrow{OC}\times\overrightarrow{OA}\} = \begin{pmatrix}-2\\20\\-4\end{pmatrix}+\begin{pmatrix}-17\\-10\\56\end{pmatrix}+\begin{pmatrix}37\\-10\\-16\end{pmatrix}\) | M1 | 1.1b — Attempts to add (as vectors) the results |
| Area \(ABC = \frac{1}{2}\sqrt{(18)^2+(0)^2+(36)^2}\) | ||
| \(\{= 20.1246\ldots\} = 9\sqrt{5}\) (cm²) or awrt 20.1 (cm²) | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Finds appropriate vectors to form equation: volume tetrahedron \(ABCD = 21\), giving a linear equation in \(d\). Note: volume must include \(\frac{1}{6}\) | M1 | 3.1a |
| e.g. \(\begin{pmatrix}3\\7\\d-4\end{pmatrix}\cdot\begin{pmatrix}18\\0\\36\end{pmatrix} = \ldots\) or \(\begin{vmatrix}4&0&-2\\2&9&-1\\3&7&d-4\end{vmatrix} = \ldots\) | M1 | 1.1b — Uses appropriate vectors in attempt at scalar triple product |
| \(= | 54+36d-144 | \) or \( |
| \(\frac{1}{6} | 36d-90 | = 21 \Rightarrow |
| Note: Using any of \(\overrightarrow{OA}\), \(\overrightarrow{OB}\), \(\overrightarrow{OC}\) or \(\overrightarrow{OD}\) in scalar triple product is M0 M0 A0 A0 |
## Question 4:
### Part (a) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete correct process: vector product of 2 edges of triangle $ABC$, applying Pythagoras and multiplying by 0.5 | M1 | 3.1b |
| $\overrightarrow{AB} = \begin{pmatrix}4\\0\\-2\end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix}2\\9\\-1\end{pmatrix}$, $\overrightarrow{BC} = \begin{pmatrix}-2\\9\\1\end{pmatrix}$ | M1 | 1.1b — Uses correct method to find any 2 edges of triangle $ABC$ |
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\4&0&-2\\2&9&-1\end{vmatrix} = \ldots$ or $\begin{pmatrix}4\\0\\-2\end{pmatrix}\times\begin{pmatrix}2\\9\\-1\end{pmatrix} = \ldots$ | M1 | 1.1b — Attempts vector cross product of 2 edges |
| $= 18\mathbf{i} + 0\mathbf{j} + 36\mathbf{k}$ | | |
| Area $ABC = \frac{1}{2}\sqrt{(18)^2+(0)^2+(36)^2}$ | | |
| $\{= 20.1246\ldots\} = 9\sqrt{5}$ (cm²) or awrt 20.1 (cm²) | A1 | 2.2a |
### Part (a) — Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses appropriate vectors to find an angle or perpendicular height in triangle $ABC$ and uses correct method for complete attempt to find area | M1 | 3.1b |
| $\overrightarrow{AB} = \begin{pmatrix}4\\0\\-2\end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix}2\\9\\-1\end{pmatrix}$, $\overrightarrow{BC} = \begin{pmatrix}-2\\9\\1\end{pmatrix}$ | M1 | 1.1b — Uses correct method to find any 2 edges |
| Uses correct method to find an angle or perpendicular height in triangle $ABC$ | M1 | 1.1b |
| $\hat{BAC} = 27.905\ldots$, $\hat{ABC} = 76.047\ldots$, $\hat{BCA} = 76.047\ldots$ or perpendicular height $= 9$ | | |
| Area $ABC = \frac{1}{2}\sqrt{86}\sqrt{20}\sin 76.047\ldots$ or $\frac{1}{2}\sqrt{86}\sqrt{86}\sin 27.905\ldots$ or $\frac{1}{2}\sqrt{20}(9)$ | | |
| $\{= 20.1246\ldots\} = 9\sqrt{5}$ (cm²) or awrt 20.1 (cm²) | A1 | 2.2a |
| Note: Using any of $\overrightarrow{OA}$, $\overrightarrow{OB}$ or $\overrightarrow{OC}$ in vector product is M0 M0 A0 A0 | | |
### Part (a) — Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete attempt to find area by applying $\frac{1}{2}\|\overrightarrow{OA}\times\overrightarrow{OB} + \overrightarrow{OB}\times\overrightarrow{OC} + \overrightarrow{OC}\times\overrightarrow{OA}\|$ or equivalent | M1 | 3.1b |
| $\overrightarrow{OA}\times\overrightarrow{OB} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&1&4\\6&1&2\end{vmatrix} = \ldots$ and $\overrightarrow{OB}\times\overrightarrow{OC} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\6&1&2\\4&10&3\end{vmatrix} = \ldots$ and $\overrightarrow{OC}\times\overrightarrow{OA} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\4&10&3\\2&1&4\end{vmatrix} = \ldots$ | M1 | 1.1b — Attempts to apply all three cross products |
| $\{\overrightarrow{OA}\times\overrightarrow{OB}+\overrightarrow{OB}\times\overrightarrow{OC}+\overrightarrow{OC}\times\overrightarrow{OA}\} = \begin{pmatrix}-2\\20\\-4\end{pmatrix}+\begin{pmatrix}-17\\-10\\56\end{pmatrix}+\begin{pmatrix}37\\-10\\-16\end{pmatrix}$ | M1 | 1.1b — Attempts to add (as vectors) the results |
| Area $ABC = \frac{1}{2}\sqrt{(18)^2+(0)^2+(36)^2}$ | | |
| $\{= 20.1246\ldots\} = 9\sqrt{5}$ (cm²) or awrt 20.1 (cm²) | A1 | 2.2a |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds appropriate vectors to form equation: volume tetrahedron $ABCD = 21$, giving a linear equation in $d$. Note: volume must include $\frac{1}{6}$ | M1 | 3.1a |
| e.g. $\begin{pmatrix}3\\7\\d-4\end{pmatrix}\cdot\begin{pmatrix}18\\0\\36\end{pmatrix} = \ldots$ or $\begin{vmatrix}4&0&-2\\2&9&-1\\3&7&d-4\end{vmatrix} = \ldots$ | M1 | 1.1b — Uses appropriate vectors in attempt at scalar triple product |
| $= |54+36d-144|$ or $|4(9d-36+7)-2(14-27)|$ $\{=|36d-90|\}$ | A1 | 1.1b — Correct applied expression for scalar triple product (allow $\pm$, ignore modulus) |
| $\frac{1}{6}|36d-90| = 21 \Rightarrow |36d-90| = 126 \Rightarrow d = 6$ | A1 | 1.1b — Correct solution leading to $d=6$ |
| Note: Using any of $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$ or $\overrightarrow{OD}$ in scalar triple product is M0 M0 A0 A0 | | |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6b9c61ac-23ec-4346-933f-cf00a2e63695-08_435_807_285_630}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of a solid doorstop made of wood. The doorstop is modelled as a tetrahedron.
Relative to a fixed origin $O$, the vertices of the tetrahedron are $A ( 2,1,4 )$, $B ( 6,1,2 ) , C ( 4,10,3 )$ and $D ( 5,8 , d )$, where $d$ is a positive constant and the units are in centimetres.
\begin{enumerate}[label=(\alph*)]
\item Find the area of the triangle $A B C$.
Given that the volume of the doorstop is $21 \mathrm {~cm} ^ { 3 }$
\item find the value of the constant $d$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 AS 2019 Q4 [8]}}