AS June 2024 Paper 1 Q3
3
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \times \begin{pmatrix} 3 \\ 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\) | B1 | 1.1 |
| [1] | ||
| (ii) \(\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\) is perpendicular to [both] \(\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) and [also] \(\begin{pmatrix} 3 \\ 5 \\ -2 \end{pmatrix}\) | B1FT | 1.2 |
| [1] | ||
| (iii) If the dot product is zero then the vectors are perpendicular | M1 | 2.1 |
| \(\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}.\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = 1 - 2 + 1 = 0\) \(\begin{pmatrix} 3 \\ 5 \\ -2 \end{pmatrix}.\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = 3 - 5 + 2 = 0\) (so the answer to (a)(i) is perpendicular to both as claimed.) | A1 | 2.2a |
| [2] |
Notes
(a)(ii)
B1FT: This can be stated as a generality (eg \(\mathbf{a} \times \mathbf{b}\) is perpendicular to both \(\mathbf{a}\) and also \(\mathbf{b}\)).
FT their answer to (a)(i) (ie the mark can be awarded if the property is stated for their vector which is not, in fact, perpendicular).
Accept “perpendicular to both vectors” or even just “they are perpendicular”
But if vectors stated must be correct vectors (or their vectors if MR in part (i)). If cross product vector stated must be their cross product vector.
(a)(iii)
M1: M1 can be implied by Attempt to find a relevant dot product and show that this is 0
A1: Correct calculation of \(\mathbf{a}.(\mathbf{a} \times \mathbf{b})\) and \(\mathbf{b}.(\mathbf{a} \times \mathbf{b})\), showing some details of calculation (ie not simply stating “\(= 0\)” without justification).
NB Both these marks are still available to candidates whose answer to (a)(i) is \(\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\).
| Scheme | Marks | AO |
|---|---|---|
| \((2\mathbf{i} - 2\mathbf{j} + \mathbf{k}).(4\mathbf{i} - \mathbf{j} + 8\mathbf{k}) = 8 + 2 + 8 = 18\) | B1 | 1.1 |
| \(\cos\theta = \dfrac{a.b}{|a||b|}\) \(= \dfrac{18}{\sqrt{2^2 + (-2)^2 + 1^2}\sqrt{4^2 + (-1)^2 + (-8)^2}}\) | M1 | 1.1 |
| \(= \dfrac{18}{\sqrt{9}\sqrt{81}} = \dfrac{2}{3}\) \(\therefore \theta = \cos^{-1}\left(\dfrac{2}{3}\right) = 48.2^\circ\) (1 dp) | A1 | 1.1 |
| [3] |
Notes
B1: Can be implied by correct answer
M1: Correct method for evaluation of cosine of required angle, including correct form for both moduli.
M1 can be awarded after correct rearrangement of \(\mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta\) once correct form for moduli seen even if subsequent (calculation) error.
A1: awrt \(48.2^\circ\) or 0.841 rads.
\(48.1896851\ldots\) or \(0.8410686706\)