Edexcel FP1 AS 2018 June — Question 4 9 marks

Exam BoardEdexcel
ModuleFP1 AS (Further Pure 1 AS)
Year2018
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeVolume of tetrahedron using scalar triple product
DifficultyStandard +0.8 This is a multi-step Further Maths question requiring vector formation, cross product calculation for triangle area, then scalar triple product for tetrahedron volume, and finally unit conversion. While the techniques are standard for FP1, the extended calculation chain and applied context (density calculation with unit conversion) elevate it above routine exercises. The computational demands and multiple stages make it moderately challenging for Further Maths students.
Spec4.04g Vector product: a x b perpendicular vector6.01c Dimensional analysis: error checking
  1. A scientist is investigating the properties of a crystal. The crystal is modelled as a tetrahedron whose vertices are \(A ( 12,4 , - 1 ) , B ( 10,15 , - 3 ) , C ( 5,8,5 )\) and \(D ( 2,2 , - 6 )\), where the length of unit is the millimetre. The mass of the crystal is 0.5 grams.
    1. Show that, to one decimal place, the area of the triangular face \(A B C\) is \(52.2 \mathrm {~mm} ^ { 2 }\)
    2. Find the density of the crystal, giving your answer in \(\mathrm { g } \mathrm { cm } ^ { - 3 }\)
    V349 SIHI NI IMIMM ION OCVJYV SIHIL NI LIIIM ION OOVJYV SIHIL NI JIIYM ION OC
Question 4:
\(A(12,4,-1),\ B(10,15,-3),\ C(10,15,-3),\ D(2,2,-6)\)
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\overrightarrow{AB}=\begin{pmatrix}-2\\11\\-2\end{pmatrix},\ \overrightarrow{AC}=\begin{pmatrix}-7\\4\\6\end{pmatrix},\ \left\{\overrightarrow{BC}=\begin{pmatrix}-5\\-7\\8\end{pmatrix}\right\}\)M1 1.1b — Uses correct method to find any 2 edges of triangle \(ABC\)
\(\text{Area}=\frac{1}{2}\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-2&11&-2\\-7&4&6\end{vmatrix}=\frac{1}{2}\left\\begin{pmatrix}74\\26\\69\end{pmatrix}\right\ =\frac{1}{2}\sqrt{74^2+26^2+69^2}\)
\(= 52.2\ \text{mm}^2\) (1 dp)A1* 2.2a — Correct area 52.2; condone awrt 52.2
Total: 3 marks Note: \(\frac{1}{2}\
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
Finds appropriate vectors for volume of \(ABCD\); complete attempt at tetrahedron volumeM1 3.1a
\(\begin{pmatrix}-10\\-2\\-5\end{pmatrix}\cdot\begin{pmatrix}74\\26\\69\end{pmatrix}=\ldots\) or \(\begin{vmatrix}-2&11&-2\\-7&4&6\\-10&-2&-5\end{vmatrix}=\ldots\)M1 1.1b — Uses appropriate vectors in scalar triple product
\(=\-740-52-345\ \) or \(\
\(V=\dfrac{1137}{6}\ \text{mm}^3\ \left\{\text{or}\ \dfrac{379}{2}\ \text{or}\ 189.5\right\}\)A1 1.1b — Correct volume in mm\(^3\) (allow \(\pm\))
\(\text{Density}=\dfrac{0.5}{189.5}\times 1000\ \text{g cm}^{-3}\)M1 2.1 — Correct method for unit conversion and finding density
\(=2.638522427\ldots \approx 2.6\ \text{g cm}^{-3}\)A1 1.1b — awrt 2.6
Total: 6 marks 
## Question 4:

$A(12,4,-1),\ B(10,15,-3),\ C(10,15,-3),\ D(2,2,-6)$

### Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB}=\begin{pmatrix}-2\\11\\-2\end{pmatrix},\ \overrightarrow{AC}=\begin{pmatrix}-7\\4\\6\end{pmatrix},\ \left\{\overrightarrow{BC}=\begin{pmatrix}-5\\-7\\8\end{pmatrix}\right\}$ | M1 | 1.1b — Uses correct method to find any 2 edges of triangle $ABC$ |
| $\text{Area}=\frac{1}{2}\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-2&11&-2\\-7&4&6\end{vmatrix}=\frac{1}{2}\left\|\begin{pmatrix}74\\26\\69\end{pmatrix}\right\|=\frac{1}{2}\sqrt{74^2+26^2+69^2}$ | M1 | 1.1b — Complete process: vector product of 2 edges, Pythagoras, multiply by 0.5 |
| $= 52.2\ \text{mm}^2$ (1 dp) | A1* | 2.2a — Correct area 52.2; condone awrt 52.2 |
| **Total: 3 marks** | | Note: $\frac{1}{2}\|74\mathbf{i}-26\mathbf{j}+69\mathbf{k}\|=52.2$ condoned for M1M1A1 |

### Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Finds appropriate vectors for volume of $ABCD$; complete attempt at tetrahedron volume | M1 | 3.1a |
| $\begin{pmatrix}-10\\-2\\-5\end{pmatrix}\cdot\begin{pmatrix}74\\26\\69\end{pmatrix}=\ldots$ **or** $\begin{vmatrix}-2&11&-2\\-7&4&6\\-10&-2&-5\end{vmatrix}=\ldots$ | M1 | 1.1b — Uses appropriate vectors in scalar triple product |
| $=\|-740-52-345\|$ **or** $\|-2(-8)-11(95)-2(54)\|\ \{=1137\}$ | A1 | 1.1b — Correct numerical expression (allow $\pm$) |
| $V=\dfrac{1137}{6}\ \text{mm}^3\ \left\{\text{or}\ \dfrac{379}{2}\ \text{or}\ 189.5\right\}$ | A1 | 1.1b — Correct volume in mm$^3$ (allow $\pm$) |
| $\text{Density}=\dfrac{0.5}{189.5}\times 1000\ \text{g cm}^{-3}$ | M1 | 2.1 — Correct method for unit conversion and finding density |
| $=2.638522427\ldots \approx 2.6\ \text{g cm}^{-3}$ | A1 | 1.1b — awrt 2.6 |
| **Total: 6 marks** | | |

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\begin{enumerate}
  \item A scientist is investigating the properties of a crystal. The crystal is modelled as a tetrahedron whose vertices are $A ( 12,4 , - 1 ) , B ( 10,15 , - 3 ) , C ( 5,8,5 )$ and $D ( 2,2 , - 6 )$, where the length of unit is the millimetre. The mass of the crystal is 0.5 grams.\\
(a) Show that, to one decimal place, the area of the triangular face $A B C$ is $52.2 \mathrm {~mm} ^ { 2 }$\\
(b) Find the density of the crystal, giving your answer in $\mathrm { g } \mathrm { cm } ^ { - 3 }$
\end{enumerate}

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\hfill \mbox{\textit{Edexcel FP1 AS 2018 Q4 [9]}}
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