AS June 2022 Paper 1 Q1
1
Vectors \(\mathbf{a}\) and \(\mathbf{b}\) are given by \(\mathbf{a} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} -3 \\ 6 \\ 2 \end{pmatrix}\).
| Scheme | Marks | AO |
|---|---|---|
| \(x\)-coord (or \(y\)-coord): \(4 + 3\lambda = 19\) (or \(-2 + -2\lambda = -12\)) \(\Rightarrow \lambda = 5\) | M1 | 1.1 |
| \(z\)-coord: \(7 + 4\lambda = 17 \Rightarrow \lambda = 2.5\) (or if \(\lambda = 5\) then \(7 + 4\lambda = 27\)) (or \(7 + 4 \times 5 \ne 17\)) | M1 | 1.1 |
| Inconsistency so point does not lie on the line | A1 | 2.2a |
| [3] |
Notes
M1: (1st) Forming and solving an equation in \(\lambda\) for any coordinate
M1: (2nd) Either forming and solving an equation (with a different correct solution) in \(\lambda\) or using the previous value to demonstrate an inconsistency
This second M mark is for considering a coordinate with a different \(\lambda\) (So M2 for \(z\) and one other)
A1: Full marks can be gained for correctly identifying an inconsistency even if there is an error in solving the third equation
Correct conclusion e.g. “point not on line is enough here” as long as with 2 correct different values of \(\lambda\) – no need to see the word “inconsistency”
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\mathbf{a}.\mathbf{b} = 1 \times -3 + -2 \times 6 + 2 \times 2\) | M1 | 1.1 |
| \(\cos\theta = \dfrac{\text{“}{-11}\text{”}}{\sqrt{1^2 + (-2)^2 + 2^2}\sqrt{(-3)^2 + 6^2 + 2^2}}\) | M1 | 1.1 |
| \(\theta = 122^\circ\) (3 sf) | A1 | 1.1 |
| [3] | ||
| (ii) Required vector is \(\mathbf{a} \times \mathbf{b}\) | M1 | 1.1 |
| \(= \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} \times \begin{pmatrix} -3 \\ 6 \\ 2 \end{pmatrix} = \begin{pmatrix} -16 \\ -8 \\ 0 \end{pmatrix}\) | A1 | 1.1 |
| [2] |
Notes
(b)(i)
M1: (1st) Forming the dot product. Can be implied by \(-11\) (but not 11)
M1: (2nd) Their dot product divided by the product of the (correctly formed) moduli. Allow sin/cos confusion here
\(-\dfrac{11}{21}\) but can be awarded for \(\dfrac{11}{21}\)
A1: Do not ISW (so e.g. M2A0 for working leading to final answer of \(58.4^\circ\))
\(121.5881\ldots\)
If more accurate than 3s.f. accept answers in range \([121.4, 121.8]\)
(b)(ii)
M1: This mark can be awarded for any non-zero multiple, even if non-numerical.
Some evidence of method needed for M1. Could be one correct value, or correct calculation for one value.
A1: or any numerical non-zero multiple e.g. \(\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\)
Ignore errors in attempts to simplify. Condone incorrect statements e.g. \(\begin{pmatrix} -16 \\ -8 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\)
(b)(ii) Alternative method
| Scheme | Marks |
|---|---|
| Assume vector is of the form \(\begin{pmatrix} 1 \\ p \\ q \end{pmatrix}\). Then: \(1 - 2p + 2q = 0\) and \(-3 + 6p + 2q = 0\) | M1 |
| \(2p - 2q = 1\) and \(6p + 2q = 3 \Rightarrow p = \tfrac{1}{2}, q = 0\) so required vector is \(\begin{pmatrix} 1 \\ \frac{1}{2} \\ 0 \end{pmatrix}\). | A1 |
M1: Need both ‘dot product’ equations. Ignore lack of consideration of \(\begin{pmatrix} 0 \\ p \\ q \end{pmatrix}\).
Could also be awarded for considering e.g. \(\begin{pmatrix} t \\ p \\ q \end{pmatrix}\) and deriving \(t - 2p + 2q = 0\) and \(-3t + 6p + 2q = 0\)
A1: or any numerical non-zero multiple e.g. \(\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\)
Eliminating e.g. \(q\) leads to \(t = 2p\) and then \(q = 0\) and so to \(\begin{pmatrix} 2p \\ p \\ 0 \end{pmatrix}\) but a (non-zero) value for \(p\) must chosen so that the final answer is a vector and not a family of vectors.