| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Volume of tetrahedron using scalar triple product |
| Difficulty | Standard +0.3 This is a standard Further Maths FP1 question testing routine application of vector cross product for area and scalar triple product for volume. The calculations are straightforward with given position vectors, requiring only formula recall and arithmetic. Part (c) adds a simple real-world context but is just multiplication. Slightly easier than average A-level due to being a textbook application with no problem-solving insight 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 | Mark | Guidance |
| \(\overrightarrow{AB}=\begin{pmatrix}-25\\9\\5\end{pmatrix},\ \overrightarrow{AC}=\begin{pmatrix}-20\\5\\-4\end{pmatrix}\) | M1 | Attempts to find 2 edges of the required triangle |
| \(\left | \overrightarrow{AB}\times\overrightarrow{AC}\right | =\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-25&9&5\\-20&5&-4\end{vmatrix}=...\) |
| Area \(=\frac{1}{2}\begin{vmatrix}-61\\-200\\55\end{vmatrix}=\frac{1}{2}\sqrt{61^2+200^2+55^2}=108\)* | A1* | Deduces correct area with no errors; condone sign slips on components |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(AB=\sqrt{25^2+9^2+5^2},\ AC=\sqrt{20^2+5^2+4^2},\ BC=\sqrt{5^2+4^2+9^2}\) | M1 | Attempts lengths of all 3 sides |
| \(\cos ABC=\frac{731+122-441}{2\times\sqrt{731}\times\sqrt{122}}\) via cosine rule | M1 | Applies cosine rule to find one angle |
| Area \(=\frac{1}{2}\sqrt{731}\sqrt{122}\sin ABC=108\)* | A1 | Deduces correct area with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| A complete attempt to find the volume of the tetrahedron | M1 | See scheme |
| \(\begin{vmatrix}18&-14&-2\\-7&-5&3\\-2&-9&-6\end{vmatrix}=...\) | M1 | Uses appropriate vectors in scalar triple product |
| \(=1592\) | A1 | Correct numerical expression for scalar triple product (allow \(\pm\)) |
| \(V=\frac{1592}{6}\) or e.g. \(V=\frac{796}{3}\) | A1 | Correct volume |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Mass \(=\frac{1592}{6}\times 0.85\div 1000\ \text{(kg)}\) | M1 | Correct method for changing units and finding mass in kg |
| \(\{=0.2255333...\}=\) awrt \(0.226\) (kg) | A1 | Correct answer |
## Question 4:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB}=\begin{pmatrix}-25\\9\\5\end{pmatrix},\ \overrightarrow{AC}=\begin{pmatrix}-20\\5\\-4\end{pmatrix}$ | M1 | Attempts to find 2 edges of the required triangle |
| $\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-25&9&5\\-20&5&-4\end{vmatrix}=...$ | M1 | Uses correct process of vector product for 2 appropriate vectors |
| Area $=\frac{1}{2}\begin{vmatrix}-61\\-200\\55\end{vmatrix}=\frac{1}{2}\sqrt{61^2+200^2+55^2}=108$* | A1* | Deduces correct area with no errors; condone sign slips on components |
**Alternative:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $AB=\sqrt{25^2+9^2+5^2},\ AC=\sqrt{20^2+5^2+4^2},\ BC=\sqrt{5^2+4^2+9^2}$ | M1 | Attempts lengths of all 3 sides |
| $\cos ABC=\frac{731+122-441}{2\times\sqrt{731}\times\sqrt{122}}$ via cosine rule | M1 | Applies cosine rule to find one angle |
| Area $=\frac{1}{2}\sqrt{731}\sqrt{122}\sin ABC=108$* | A1 | Deduces correct area with no errors |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| A complete attempt to find the volume of the tetrahedron | M1 | See scheme |
| $\begin{vmatrix}18&-14&-2\\-7&-5&3\\-2&-9&-6\end{vmatrix}=...$ | M1 | Uses appropriate vectors in scalar triple product |
| $=1592$ | A1 | Correct numerical expression for scalar triple product (allow $\pm$) |
| $V=\frac{1592}{6}$ or e.g. $V=\frac{796}{3}$ | A1 | Correct volume |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Mass $=\frac{1592}{6}\times 0.85\div 1000\ \text{(kg)}$ | M1 | Correct method for changing units and finding mass in kg |
| $\{=0.2255333...\}=$ awrt $0.226$ (kg) | A1 | Correct answer |
---\begin{enumerate}
\item With respect to a fixed origin $O$, the points $A$, $B$ and $C$ have position vectors given by
\end{enumerate}
$$\overrightarrow { O A } = 18 \mathbf { i } - 14 \mathbf { j } - 2 \mathbf { k } \quad \overrightarrow { O B } = - 7 \mathbf { i } - 5 \mathbf { j } + 3 \mathbf { k } \quad \overrightarrow { O C } = - 2 \mathbf { i } - 9 \mathbf { j } - 6 \mathbf { k }$$
The points $O , A , B$ and $C$ form the vertices of a tetrahedron.\\
(a) Show that the area of the triangular face $A B C$ of the tetrahedron is 108 to 3 significant figures.\\
(b) Find the volume of the tetrahedron.
An oak wood block is made in the shape of the tetrahedron, with centimetres taken for the units.
The density of oak is $0.85 \mathrm {~g} \mathrm {~cm} ^ { - 3 }$\\
(c) Determine the mass of the block, giving your answer in kg.
\hfill \mbox{\textit{Edexcel FP1 AS 2021 Q4 [9]}}