| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Session | Specimen |
| Marks | 10 |
| 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 scalar triple product for volume calculation, identification of which face is glued (requiring spatial reasoning about parallelepiped geometry), and area calculation using cross product. While the techniques are standard for FM students, the multi-step nature, need to identify the correct face, and applied context make it moderately challenging but still within typical FM scope. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identifies glued face is triangle \(ABC\) and attempts to find the area, e.g. evidenced by use of \(\frac{1}{2} | \mathbf{AB}\times\mathbf{AC} | \) |
| \(\frac{1}{2} | \mathbf{AB}\times\mathbf{AC} | = \frac{1}{2} |
| \(= \frac{1}{2} | 5\mathbf{i}+3\mathbf{j}+\mathbf{k} | \) |
| \(= \frac{1}{2}\sqrt{35}\ (\text{m}^2)\) | A1 | \(\frac{1}{2}\sqrt{35}\) or exact equivalent |
| Alternative: \( | \mathbf{AB} | ^2=14,\ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Volume of parallelepiped taken up by concrete: \(\frac{1}{6}(\mathbf{OC}\cdot(\mathbf{OA}\times\mathbf{OB}))\) | M1 | Attempts volume of concrete by finding volume of tetrahedron with appropriate method |
| \(= \frac{1}{6}(\mathbf{i}+\mathbf{j}+2\mathbf{k})\cdot(2\mathbf{i}\times(3\mathbf{j}+\mathbf{k}))\) | M1 | Uses formula with correct set of vectors substituted and vector product attempted |
| \(= \frac{10}{6} = \frac{5}{3}\) | A1 | Correct value for the volume of concrete |
| Volume of parallelepiped is \(6\times\) volume of tetrahedron \((= 10)\), so volume of glass is \(10 - \frac{5}{3} = \ldots\) | M1 | Attempt to find total volume of glass by multiplying concrete volume by 6 and subtracting concrete volume |
| Volume of glass \(= \frac{25}{3}\ (\text{m}^3)\) | A1 | \(\frac{25}{3}\) only, ignore reference to units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. The glued surfaces may distort the shapes/reduce the volume of concrete; Measurements in m may not be accurate; The surface of the concrete tetrahedron may not be smooth; Pockets of air may form when the concrete is being poured | B1 | Any acceptable reason in context |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identifies glued face is triangle $ABC$ and attempts to find the area, e.g. evidenced by use of $\frac{1}{2}|\mathbf{AB}\times\mathbf{AC}|$ | M1 | Shows understanding of what is required via attempt at finding area of triangle $ABC$ |
| $\frac{1}{2}|\mathbf{AB}\times\mathbf{AC}| = \frac{1}{2}|(-2\mathbf{i}+3\mathbf{j}+\mathbf{k})\times(-\mathbf{i}+\mathbf{j}+2\mathbf{k})|$ | M1 | Any correct method for triangle area; finds $\mathbf{AB}$ and $\mathbf{AC}$ or appropriate pair of vectors and attempts to use them |
| $= \frac{1}{2}|5\mathbf{i}+3\mathbf{j}+\mathbf{k}|$ | M1 | Correct procedure for vector product with at least 1 correct term |
| $= \frac{1}{2}\sqrt{35}\ (\text{m}^2)$ | A1 | $\frac{1}{2}\sqrt{35}$ or exact equivalent |
| **Alternative:** $|\mathbf{AB}|^2=14,\ |\mathbf{AC}|^2=6,\ \mathbf{AB}\cdot\mathbf{AC}=7$; Area $= \frac{1}{2}\sqrt{(14)(6)-(7)^2} = \frac{1}{2}\sqrt{35}\ (\text{m}^2)$ | M1,M1,M1,A1 | Finds two appropriate sides, attempts scalar product and magnitudes; correct full method |
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## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Volume of parallelepiped taken up by concrete: $\frac{1}{6}(\mathbf{OC}\cdot(\mathbf{OA}\times\mathbf{OB}))$ | M1 | Attempts volume of concrete by finding volume of tetrahedron with appropriate method |
| $= \frac{1}{6}(\mathbf{i}+\mathbf{j}+2\mathbf{k})\cdot(2\mathbf{i}\times(3\mathbf{j}+\mathbf{k}))$ | M1 | Uses formula with correct set of vectors substituted and vector product attempted |
| $= \frac{10}{6} = \frac{5}{3}$ | A1 | Correct value for the volume of concrete |
| Volume of parallelepiped is $6\times$ volume of tetrahedron $(= 10)$, so volume of glass is $10 - \frac{5}{3} = \ldots$ | M1 | Attempt to find total volume of glass by multiplying concrete volume by 6 and subtracting concrete volume |
| Volume of glass $= \frac{25}{3}\ (\text{m}^3)$ | A1 | $\frac{25}{3}$ only, ignore reference to units |
---
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. The glued surfaces may distort the shapes/reduce the volume of concrete; Measurements in m may not be accurate; The surface of the concrete tetrahedron may not be smooth; Pockets of air may form when the concrete is being poured | B1 | Any acceptable reason in context |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ff1fc9b0-6514-44e0-a2a3-46aa6411ce10-08_538_807_251_630}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of a solid sculpture made of glass and concrete. The sculpture is modelled as a parallelepiped.
The sculpture is made up of a concrete solid in the shape of a tetrahedron, shown shaded in Figure 1, whose vertices are $\mathrm { O } ( 0,0,0 ) , \mathrm { A } ( 2,0,0 ) , \mathrm { B } ( 0,3,1 )$ and $\mathrm { C } ( 1,1,2 )$, where the units are in metres. The rest of the solid parallelepiped is made of glass which is glued to the concrete tetrahedron.
\begin{enumerate}[label=(\alph*)]
\item Find the surface area of the glued face of the tetrahedron.
\item Find the volume of glass contained in this parallelepiped.
\item Give a reason why the volume of concrete predicted by this model may not be an accurate value for the volume of concrete that was used to make the sculpture.
\section*{Q uestion 4 continued}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 AS Q4 [10]}}