AS June 2023 Paper 1 Q6
6. The line \(l_1\) has equation \(\mathbf{r} = \begin{pmatrix}-2\\ 2\\ 0\end{pmatrix} + \lambda\begin{pmatrix}3\\ 0\\ 1\end{pmatrix}\) where \(\lambda\) is a scalar parameter.
The line \(l_2\) is parallel to \(\begin{pmatrix}1\\ 2\\ -3\end{pmatrix}\)
The plane \(\Pi\) contains the line \(l_1\) and is perpendicular to \(\begin{pmatrix}1\\ 2\\ -3\end{pmatrix}\)
Given that
- the point of intersection of \(\Pi\) and \(l_2\) has coordinates \((2, 3, 2)\)
- the point \(B(p, q, r)\) lies on \(l_2\)
- the distance \(AB\) is \(2\sqrt{5}\)
- \(p\), \(q\) and \(r\) are positive integers
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}3\\ 0\\ 1\end{pmatrix} \bullet \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = 3\{+0\} - 3\) | M1 | 1.1b |
| = 0 therefore the lines are perpendicular. | A1 | 2.4 |
| (2) |
Notes
M1: Applies the dot product to the direction vectors. Minimum requirement is 3 – 3
A1: Shows that the dot product = 0 and concludes that the lines are perpendicular.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{r} \bullet \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = \begin{pmatrix}-2\\ 2\\ 0\end{pmatrix} \bullet \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = \ldots\{2\}\) | M1 | 1.1b |
| \(x + 2y - 3z = 2\) o.e. | A1 | 2.5 |
| (2) |
Notes
M1: Applies \(\mathbf{r} \bullet \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = \begin{pmatrix}-2\\ 2\\ 0\end{pmatrix} \bullet \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = \ldots\) or \(\begin{pmatrix}x\\ y\\ z\end{pmatrix} \bullet \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = \begin{pmatrix}-2\\ 2\\ 0\end{pmatrix} \bullet \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = \ldots\)
A1: Correct Cartesian equation \(x + 2y - 3z = 2\) o.e.
Note: \(\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} = 2\) is M1A0
| Scheme | Marks | AO |
|---|---|---|
| \(3 + 2(1) - 3(1) = 2\) (therefore lies on the plane) | B1 | 1.1b |
| (1) |
Notes
B1: See scheme, no conclusion required
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}p\\ q\\ r\end{pmatrix} = \begin{pmatrix}2\\ 3\\ 2\end{pmatrix} + \mu\begin{pmatrix}1\\ 2\\ -3\end{pmatrix}\) or \(\begin{pmatrix}p\\ q\\ r\end{pmatrix} + \mu\begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = \begin{pmatrix}2\\ 3\\ 2\end{pmatrix}\) leading to \(\begin{aligned}p &= 2 - \mu\\ q &= 3 - 2\mu\\ r &= 2 + 3\mu\end{aligned}\) | M1 | 3.1a |
| \((p - 3)^2 + (q - 1)^2 + (r - 1)^2 = \left(2\sqrt{5}\right)^2\) \(((2 + \mu) - 3)^2 + ((3 + 2\mu) - 1)^2 + ((2 - 3\mu) - 1)^2 = \left(2\sqrt{5}\right)^2\) \((-1 + \mu)^2 + (2 + 2\mu)^2 + (1 - 3\mu)^2 = 20\) or \((p - 3)^2 + (q - 1)^2 + (r - 1)^2 = \left(2\sqrt{5}\right)^2\) \(((2 - \mu) - 3)^2 + ((3 - 2\mu) - 1)^2 + ((2 + 3\mu) - 1)^2 = \left(2\sqrt{5}\right)^2\) \((-1 - \mu)^2 + (2 - 2\mu)^2 + (1 + 3\mu)^2 = 20\) | M1 | 3.1a |
| \(14\mu^2 - 14 = 0\) o.e | A1 | 1.1b |
| Solves their quadratic \(\{\mu = -1 \text{ or } \mu = 1\}\) | M1 | 1.1b |
| Uses \(\mu = -1\) \(\begin{aligned}p &= 2 + (-1) = \ldots\\ q &= 3 + 2(-1) = \ldots\\ r &= 2 - 3(-1) = \ldots\end{aligned}\) or Using \(\mu = 1\) \(\begin{aligned}p &= 2 - (1) = \ldots\\ q &= 3 - 2(1) = \ldots\\ r &= 2 + 3(1) = \ldots\end{aligned}\) | ddM1 | 1.1b |
| \((1, 1, 5)\) only | A1 | 3.2a |
| (6) | ||
| (11 marks) |
Notes
M1: Uses the point of intersection to find the coordinates of \(B\) as functions of a parameter
M1: Uses the distance between the point \(A\) and the point \(B\) to form an equation for their parameter only.
A1: Correct simplified quadratic equation
M1: Solves their quadratic equation to find a value for \(\mu\)
ddM1: Dependent on the first two method marks. Uses any one of their values for their parameter to find the coordinates of \(B\), it need not be the correct one.
A1: Correct coordinates for \(B\), condone as a vector, if seen (3, 5, −1) must be disregarded
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(|AX| = \sqrt{(3 - 2)^2 + (1 - 3)^2 + (1 - 2)^2} = \sqrt{6}\) | M1 | 3.1a |
| Correctly uses Pythagoras to find the length of \(XB\) \(|XB| = \sqrt{\left(2\sqrt{5}\right)^2 - 6} = \sqrt{14}\) | M1 | 3.1a |
| Find the magnitude of the direction vector and compares to the length of \(XB\) to find a value for \(\mu\) | M1 | 1.1b |
| \(\mu = -1\) or \(\mu = 1\) | A1 | 1.1b |
| Uses \(\mu = -1\) \(\begin{aligned}p &= 2 + (-1) = \ldots\\ q &= 3 + 2(-1) = \ldots\\ r &= 2 - 3(-1) = \ldots\end{aligned}\) or Using \(\mu = 1\) \(\begin{aligned}p &= 2 - (1) = \ldots\\ q &= 3 - 2(1) = \ldots\\ r &= 2 + 3(1) = \ldots\end{aligned}\) | ddM1 | 1.1b |
| \((1, 1, 5)\) only | A1 | 3.2a |
| (6) |
M1: Finds the length \(AX\)
M1: Uses Pythagoras to find the length of \(XB\)
NOTE the change in order of the M1 and A1
M1: Find the length of the direction vector and compares to find a value for \(\mu\)
A1: A correct values for \(\mu\)
ddM1: Dependent on the first two method marks. Uses any one of their values for their parameter to find the coordinates of \(B\), it need not be the correct one.
A1: Correct coordinates for \(B\), condone as a vector, if seen (3, 5, −1) must be disregarded