A2 October 2020 Q3
3. The points \(A\), \(B\) and \(C\), with position vectors \(\mathbf{a} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}\), \(\mathbf{b} = \mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\) and \(\mathbf{c} = -2\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}\) respectively, lie on the plane \(\Pi\)
The point \(D\) has position vector \(8\mathbf{i} + 7\mathbf{j} + 5\mathbf{k}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{AB} = -2\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}\) and \(\overrightarrow{AC} = -5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\) | B1 | 1.1b |
| \(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix}-2 & 6 & 4\\ -5 & 5 & 2\end{vmatrix}\) \(= (6 \times 2 - 4 \times 5)\mathbf{i} - (-2 \times 2 - 4 \times -5)\mathbf{j} + (-2 \times 5 - 6 \times -5)\mathbf{k}\) | M1 | 1.1b |
| \(= -8\mathbf{i} - 16\mathbf{j} + 20\mathbf{k}\) | A1 | 1.1b |
| (3) |
Notes
B1: Both \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) correct.
M1: Applies the cross product to their \(\overrightarrow{AB}\) and their \(\overrightarrow{AC}\). There must be at least two correct components if no method seen. Method can be implied by \(\mathbf{i}\begin{vmatrix}6 & 4\\ 5 & 2\end{vmatrix} - \mathbf{j}\begin{vmatrix}-2 & 4\\ -5 & 2\end{vmatrix} + \mathbf{k}\begin{vmatrix}-2 & 6\\ -5 & 5\end{vmatrix} = \ldots\) with at least one correct component.
A1: Correct vector
| Scheme | Marks | AO |
|---|---|---|
| E.g. \(\mathbf{n} = -8\mathbf{i} - 16\mathbf{j} + 20\mathbf{k}\) gives \(p = (3\mathbf{i} - 2\mathbf{j} + \mathbf{k}).(-8\mathbf{i} - 16\mathbf{j} + 20\mathbf{k}) = \ldots\) E.g. \(\mathbf{n} = 2\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}\) gives \(p = (3\mathbf{i} - 2\mathbf{j} + \mathbf{k}).(2\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}) = \ldots\) | M1 | 1.1b |
| Equation is \(\mathbf{r}.(-8\mathbf{i} - 16\mathbf{j} + 20\mathbf{k}) = 28\) or \(\mathbf{r}.(2\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}) = -7\) (oe) | A1 | 2.5 |
| (2) |
Notes
M1: Uses their \(\mathbf{n}\) (which may be any multiple of their \(\overrightarrow{AB} \times \overrightarrow{AC}\)) and any point on the plane in an attempt to find \(p\). (Use of \(\overrightarrow{AB}\) or \(\overrightarrow{AC}\) is M0.)
A1: Correct equation in form stated. Accept any multiples, e.g. \(\mathbf{r}.(-8\mathbf{i} - 16\mathbf{j} + 20\mathbf{k}) = 28\)
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{AD}.\left(\overrightarrow{AB} \times \overrightarrow{AC}\right) = \text{“}\overrightarrow{AD}\text{”}.(-8\mathbf{i} - 16\mathbf{j} + 20\mathbf{k}) = \ldots\) | M1 | 1.1b |
| \(\overrightarrow{AD} = 5\mathbf{i} + 9\mathbf{j} + 4\mathbf{k}\) | B1 | 1.1b |
| Volume \(= \dfrac{1}{6}\left|\overrightarrow{AD}.\left(\overrightarrow{AB} \times \overrightarrow{AC}\right)\right| = \dfrac{1}{6}\left|(5\mathbf{i} + 9\mathbf{j} + 4\mathbf{k}).(-8\mathbf{i} - 16\mathbf{j} + 20\mathbf{k})\right| = \ldots\) | M1 | 3.1a |
| \(= \dfrac{52}{3}\) o.e. \(17\dfrac{1}{3}\) | A1 | 1.1b |
| (4) | ||
| (9 marks) |
Notes
M1: Attempts a suitable scalar triple product, e.g. \(\overrightarrow{AD}.\left(\overrightarrow{AB} \times \overrightarrow{AC}\right)\). Must include a complete method to use all necessary vectors.
B1: Correct ‘\(\overrightarrow{AD}\)’ if using \(\overrightarrow{AD}.\left(\overrightarrow{AB} \times \overrightarrow{AC}\right)\), or all vectors correct if using a different product.
M1: Use of volume \(= \dfrac{1}{6}\left|\text{their } \overrightarrow{AD}.\left(\overrightarrow{AB} \times \overrightarrow{AC}\right)\right|\) (oe full method to find the volume).
A1: Correct exact answer.