AS June 2018 Q4
4. 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.
| Scheme | Marks | AO |
|---|---|---|
| \(A(12, 4, -1),\ B(10, 15, -3),\ C(5, 8, 5),\ D(2, 2, -6)\) | ||
| \(\overrightarrow{AB} = \begin{pmatrix}-2\\ 11\\ -2\end{pmatrix},\ \overrightarrow{AC} = \begin{pmatrix}-7\\ 4\\ 6\end{pmatrix},\ \left\{\overrightarrow{BC} = \begin{pmatrix}-5\\ -7\\ 8\end{pmatrix}\right\}\) | M1 | 1.1b |
| Area \(= \dfrac{1}{2}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ -2 & 11 & -2\\ -7 & 4 & 6\end{vmatrix} = \dfrac{1}{2}\left|\begin{pmatrix}74\\ 26\\ 69\end{pmatrix}\right| = \dfrac{1}{2}\sqrt{(74)^2 + (26)^2 + (69)^2}\) | M1 | 1.1b |
| \(\{= 52.23265\ldots\} = 52.2\ (\text{mm}^2)\ (1\text{ dp})\ *\) | A1* | 2.2a |
| (3) |
Notes
M1: Uses a correct method to find any 2 edges of triangle \(ABC\)
M1: Complete process of taking the vector product between 2 edges of triangle \(ABC\), applying Pythagoras and multiplying the result by 0.5
A1*: Deduces the correct area of \(52.2\ (\text{mm}^2)\). Condone awrt 52.2
Note: Condone \(\dfrac{1}{2}\left|74\mathbf{i} - 26\mathbf{j} + 69\mathbf{k}\right| = \dfrac{1}{2}\sqrt{(74)^2 + (-26)^2 + (69)^2} = 52.2\), o.e. for M1M1A1
Note: As an alternative, \(\dfrac{1}{2}\sqrt{129}\sqrt{101}\sin(66.2343\ldots) = 52.2\ (1\text{ dp})\), where the angle has been found by applying the scalar product between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\)
(Corrected from the printed mark scheme: the first line of the scheme gives \(C\) as \((10, 15, -3)\); \(C\) is \((5, 8, 5)\).)
| Scheme | Marks | AO |
|---|---|---|
| Finds appropriate vectors to find the volume of \(ABCD\) and makes a complete attempt to find the volume of the tetrahedron | M1 | 3.1a |
| e.g. \(\left|\begin{pmatrix}-10\\ -2\\ -5\end{pmatrix} \bullet \begin{pmatrix}74\\ 26\\ 69\end{pmatrix}\right| = \ldots\) or \(\begin{vmatrix}-2 & 11 & -2\\ -7 & 4 & 6\\ -10 & -2 & -5\end{vmatrix} = \ldots\) | M1 | 1.1b |
| \(= |-740 - 52 - 345|\) or \(|-2(-8) - 11(95) - 2(54)|\ \{= 1137\}\) | A1 | 1.1b |
| \(V = \dfrac{1137}{6}\ (\text{mm}^3)\ \left\{\text{or } \dfrac{379}{2} \text{ or } 189.5\right\}\) | A1 | 1.1b |
| Density \(= \dfrac{0.5}{189.5} \times 1000\ (\text{g cm}^{-3})\) | M1 | 2.1 |
| \(\{= 2.638522427\ldots\} = \text{awrt } 2.6\ (\text{g cm}^{-3})\) | A1 | 1.1b |
| (6) | ||
| (9 marks) |
Notes
M1: See scheme
M1: Uses appropriate vectors to in an attempt at the scalar triple product
A1: Correct numerical expression for the scalar triple product (allow \(\pm\))
A1: Correct volume (in \(\text{mm}^3\)) (allow \(\pm\))
M1: A correct method for changing their units for their volume and for finding density
A1: Obtains the correct density in \(\text{g cm}^{-3}\). Allow awrt 2.6
Note: Using any of \(\overrightarrow{OA}\), \(\overrightarrow{OB}\), \(\overrightarrow{OC}\) or \(\overrightarrow{OD}\) in their scalar triple product is M0M0A0A0
Note: Allow M1M1A0A0 for
\[V = \frac{1}{6}\left|\begin{pmatrix}-10\\ -2\\ -5\end{pmatrix} \bullet \begin{pmatrix}74\\ 26\\ 69\end{pmatrix}\right| = \frac{1}{6}\left|-740\mathbf{i} - 52\mathbf{j} - 345\mathbf{k}\right| = \frac{1}{6}\sqrt{(-740)^2 + (-52)^2 + (-345)^2}\]\[= \frac{1}{6}(818.125296\ldots) = 136.354216\ldots\](Corrected from the printed mark scheme: the last value is printed as \(135.354216\ldots\); \(\dfrac{818.125296\ldots}{6} = 136.354216\ldots\))
Note: Some vector product calculations for reference:
\[\left|\overrightarrow{AD}.\left(\overrightarrow{AB} \times \overrightarrow{AC}\right)\right| = \begin{vmatrix}-10 & -2 & -5\\ -2 & 11 & -2\\ -7 & 4 & 6\end{vmatrix} = \left|\begin{pmatrix}-10\\ -2\\ -5\end{pmatrix} \bullet \begin{pmatrix}74\\ 26\\ 69\end{pmatrix}\right| = |-740 - 52 - 345| = 1137\]\[\left|\overrightarrow{AB}.\left(\overrightarrow{AC} \times \overrightarrow{AD}\right)\right| = \begin{vmatrix}-2 & 11 & -2\\ -7 & 4 & 6\\ -10 & -2 & -5\end{vmatrix} = \left|\begin{pmatrix}-2\\ 11\\ -2\end{pmatrix} \bullet \begin{pmatrix}-8\\ -95\\ 54\end{pmatrix}\right| = |16 - 1045 - 108| = 1137\]\[\left|\overrightarrow{AC}.\left(\overrightarrow{AB} \times \overrightarrow{AD}\right)\right| = \begin{vmatrix}-7 & 4 & 6\\ -2 & 11 & -2\\ -10 & -2 & -5\end{vmatrix} = \left|\begin{pmatrix}-7\\ 4\\ 6\end{pmatrix} \bullet \begin{pmatrix}-59\\ 10\\ 114\end{pmatrix}\right| = |413 + 40 + 684| = 1137\]Note: Some candidates apply \(\overrightarrow{AB} \times \overrightarrow{AC}\) incorrectly to give \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ -2 & 11 & -2\\ -7 & 4 & 6\end{vmatrix} = 74\mathbf{i} - 26\mathbf{j} + 69\mathbf{k}\)
This leads to an incorrect \(\left|\overrightarrow{AD}.\left(\overrightarrow{AB} \times \overrightarrow{AC}\right)\right| = \left|\begin{pmatrix}-10\\ -2\\ -5\end{pmatrix} \bullet \begin{pmatrix}74\\ -26\\ 69\end{pmatrix}\right| = |-740 + 52 - 345| = 1033\)