AS June 2018 Paper 1 Q1
1
Express the equation of this line in vector form. [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} \times \begin{pmatrix} -3 \\ -6 \\ 4 \end{pmatrix}\) | M1 | 1.1a |
| \(\begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix}\) | A1 | 1.1 |
| [2] |
Notes
M1: Use of cross product
Correct pairs of numbers used together, allow one numerical or sign error.
A1: or any non-zero multiple
Alternative
| Scheme | Marks |
|---|---|
| \(3 - 2a = 0\) | M1 |
| \(\begin{pmatrix} 0 \\ 1 \\ 1.5 \end{pmatrix}\) | A1 |
M1: Assuming a vector of the form \(\begin{pmatrix} 0 \\ 1 \\ a \end{pmatrix}\) and understanding that both dot products are zero. Allow one numerical or sign error
A1: or any non-zero multiple
Alternative
| Scheme | Marks |
|---|---|
| \(a + 3b - 2c = 0\) \(-3a - 6b + 4c = 0\) \(\Rightarrow a = 0, 3b - 2c = 0\) | M1 |
| \(\begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix}\) | A1 |
M1: Use dot product with both vectors to find two simultaneous equations. Either eliminate \(a\) or \(b\) & \(c\) to show that \(a = 0\) or \(3b - 2c = 0\)
A1: or any non-zero multiple
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{x - 0}{2}\) or \(\dfrac{y - 3}{1}\) or \(\dfrac{z - (-2)}{\frac{1}{2}}\) | M1 | 1.1a |
| \(\mathbf{r} = \begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix} + \ldots\) | A1 | 1.1 |
| \(\ldots + \lambda\begin{pmatrix} 2 \\ 1 \\ \frac{1}{2} \end{pmatrix}\) | A1 | 1.1 |
| [3] |
Notes
M1: Or \(x = (0+)2\lambda\) or \(y = 3 + (1)\lambda\) or \(z = \dfrac{\lambda - 4}{2}\)
or form \((\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}\)
Can be implied by correct answer
A1: (1st) Could have any multiple of the direction vector added. e.g. [4, 5, −1]
If M0 then SC1 for \(\begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix} + \ldots\) (i.e. no \(\lambda\) attached)
A1: (2nd) Vector could be any multiple e.g. [4, 2, 1].
Any sensible parameter name (not e.g. \(r\), \(x\), \(y\) or \(z\)).
Not dependent on previous A mark so M1 A0 A1 is possible.
If M0 then SC1 for \(\begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ \frac{1}{2} \end{pmatrix}\) (corrected from the printed mark scheme: the last entry of each vector in this special case is cut off in the printed scheme)