FP3 June 2018 Q6
6. The line \(l_1\) has equation \[\mathbf{r} = \mathbf{i} + 2\mathbf{k} + \lambda(2\mathbf{i} + 3\mathbf{j} - \mathbf{k})\] where \(\lambda\) is a scalar parameter.
The line \(l_2\) has equation \[\frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 1}{3}\]
The plane \(\Pi\) contains the line \(l_1\) and intersects the line \(l_2\) at the point \((3, 8, 13)\).
| Scheme | Marks |
|---|---|
| \(\mathbf{r} = \begin{pmatrix}1\\ 0\\ 2\end{pmatrix} + \lambda\begin{pmatrix}2\\ 3\\ -1\end{pmatrix},\quad \mathbf{r} = \begin{pmatrix}-1\\ 4\\ 1\end{pmatrix} + \mu\begin{pmatrix}1\\ 1\\ 3\end{pmatrix}\) | |
| \(\begin{pmatrix}2\\ 3\\ -1\end{pmatrix} \neq k\begin{pmatrix}1\\ 1\\ 3\end{pmatrix}\) | B1 |
| \(\mathbf{i}:\ 1 + 2\lambda = -1 + \mu\ \ (1)\) \(\mathbf{j}:\ 3\lambda = 4 + \mu\ \ (2)\) \(\mathbf{k}:\ 2 - \lambda = 1 + 3\mu\ \ (3)\) | |
| (1) and (2) yields \(\lambda = 6,\ \mu = 14\) (1) and (3) yields \(\lambda = -\dfrac{5}{7},\ \mu = \dfrac{4}{7}\) (2) and (3) yields \(\lambda = \dfrac{13}{10},\ \mu = -\dfrac{1}{10}\) | M1 |
| Checking (3): \(-4 \neq 43\) Checking (2): \(-\dfrac{15}{7} \neq \dfrac{32}{7}\) Checking (1): \(3.6 \neq -1.1\) | M1 |
| So the lines are not parallel and do not intersect so the lines are skew | A1 |
| (4) |
Notes
B1: Shows lines are not parallel. If they say “different direction vectors”, the direction vectors must be identified.
Examples:
\(\dfrac{2}{1} \neq \dfrac{3}{1}\)
\(\begin{pmatrix}2\\ 3\\ -1\end{pmatrix} \times \begin{pmatrix}1\\ 1\\ 3\end{pmatrix} = \begin{pmatrix}10\\ -7\\ -1\end{pmatrix} \neq \begin{pmatrix}0\\ 0\\ 0\end{pmatrix}\) (allow \(\neq 0\))
\(\begin{pmatrix}2\\ 3\\ -1\end{pmatrix}\bullet\begin{pmatrix}1\\ 1\\ 3\end{pmatrix} = 2 \neq \sqrt{14}\sqrt{11}\)
\(\begin{pmatrix}2\\ 3\\ -1\end{pmatrix}\bullet\begin{pmatrix}1\\ 1\\ 3\end{pmatrix} = 2 = \sqrt{14}\sqrt{11}\cos\theta \Rightarrow \theta = 80.7^\circ\)
M1: Attempts to solve a pair of equations to find at least one of either \(\lambda = \ldots\) or \(\mu = \ldots\)
M1: Attempts to show a contradiction
A1: All complete and with no errors and conclusion. If they have already stated “not parallel” there is no need to repeat this.
Alternative for the M marks
| Scheme | Marks |
|---|---|
| (1) and (2) yields \(\lambda = 6,\ \mu = 14\) (1) and (3) yields \(\lambda = -\dfrac{5}{7},\ \mu = \dfrac{4}{7}\) (2) and (3) yields \(\lambda = \dfrac{13}{10},\ \mu = -\dfrac{1}{10}\) | M1 |
| Shows any two of (1) and (2) yielding \(\lambda = 6\), (1) and (3) yielding \(\lambda = -\dfrac{5}{7}\), (2) and (3) yielding \(\lambda = \dfrac{13}{10}\) or shows any two of (1) and (2) yielding \(\mu = 14\), (1) and (3) yielding \(\mu = \dfrac{4}{7}\), (2) and (3) yielding \(\mu = -\dfrac{1}{10}\) | M1 |
M1: Attempts to solve a pair of equations to find at least one of either \(\lambda = \ldots\) or \(\mu = \ldots\)
M1: Attempts to show a contradiction
Note that for (b) the only misinterpretations for Position we are allowing are: \(\begin{pmatrix}1\\ 2\\ 0\end{pmatrix}\) for \(\begin{pmatrix}1\\ 0\\ 2\end{pmatrix}\) for the position of \(l_1\) and \(\begin{pmatrix}1\\ -4\\ -1\end{pmatrix}\) for \(\begin{pmatrix}-1\\ 4\\ 1\end{pmatrix}\) for the position of \(l_2\)
But allow obvious slips or mis-copies of e.g. signs or elements if the intention is clear.
Way 1
| Scheme | Marks |
|---|---|
| \(\pm\begin{pmatrix}2\\ 3\\ -1\end{pmatrix} \times \begin{pmatrix}1\\ 1\\ 3\end{pmatrix} = \pm\begin{pmatrix}10\\ -7\\ -1\end{pmatrix}\) | M1 A1 |
| \(\pm\left(\begin{pmatrix}1\\ 0\\ 2\end{pmatrix} - \begin{pmatrix}-1\\ 4\\ 1\end{pmatrix}\right)\bullet\pm\begin{pmatrix}10\\ -7\\ -1\end{pmatrix} = \pm(20 + 28 - 1) = \pm 47\) | M1 |
| \(d = \left|\dfrac{\pm 47}{\sqrt{10^2 + 7^2 + 1^2}}\right| = \dfrac{47}{\sqrt{150}}\) | M1 A1 |
| (5) |
Notes
M1: Attempt cross product of direction vectors. If no method is shown, 2 components should be correct.
A1: Correct vector
M1: Attempt scalar product between the difference of the position vectors and their normal vector.
M1: Correct completion. Divides their scalar product between the difference of the position vectors and their normal vector by the modulus of their vector product.
A1: Any equivalent or awrt 3.84
Way 2
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}2\\ 3\\ -1\end{pmatrix} \times \begin{pmatrix}1\\ 1\\ 3\end{pmatrix} = \begin{pmatrix}10\\ -7\\ -1\end{pmatrix}\) | M1 A1 |
| \(\begin{pmatrix}10\\ -7\\ -1\end{pmatrix}\bullet\begin{pmatrix}1\\ 0\\ 2\end{pmatrix} = 8,\quad \begin{pmatrix}10\\ -7\\ -1\end{pmatrix}\bullet\begin{pmatrix}-1\\ 4\\ 1\end{pmatrix} = -39\) | M1 |
| \(d = \dfrac{8}{\sqrt{10^2 + 7^2 + 1^2}} - \dfrac{-39}{\sqrt{10^2 + 7^2 + 1^2}} = \dfrac{47}{\sqrt{150}}\) | M1 A1 |
| (5) |
M1: Attempt cross product of direction vectors
A1: Correct vector
M1: Attempt equation of both planes
M1: Correct completion
A1: Any equivalent e.g. \(\dfrac{47\sqrt{6}}{30}\) or awrt 3.84 but must be positive.
Way 3
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}1\\ 0\\ 2\end{pmatrix} + \lambda\begin{pmatrix}2\\ 3\\ -1\end{pmatrix} - \left[\begin{pmatrix}-1\\ 4\\ 1\end{pmatrix} + \mu\begin{pmatrix}1\\ 1\\ 3\end{pmatrix}\right] = \begin{pmatrix}2 + 2\lambda - \mu\\ -4 + 3\lambda - \mu\\ 1 - \lambda - 3\mu\end{pmatrix}\) \(\begin{pmatrix}2 + 2\lambda - \mu\\ -4 + 3\lambda - \mu\\ 1 - \lambda - 3\mu\end{pmatrix}\bullet\begin{pmatrix}1\\ 1\\ 3\end{pmatrix} = 0,\quad \begin{pmatrix}2 + 2\lambda - \mu\\ -4 + 3\lambda - \mu\\ 1 - \lambda - 3\mu\end{pmatrix}\bullet\begin{pmatrix}2\\ 3\\ -1\end{pmatrix} = 0\) \(2\lambda - 11\mu = -1\) \(14\lambda - 2\mu = 9\) | M1 |
| \(\lambda = \dfrac{101}{150},\ \mu = \dfrac{16}{75}\) | A1 |
| \(\left(-\dfrac{59}{75}, \dfrac{316}{75}, \dfrac{41}{25}\right), \left(\dfrac{176}{75}, \dfrac{303}{150}, \dfrac{199}{150}\right)\) Or \(\begin{pmatrix}2 + 2\lambda - \mu\\ -4 + 3\lambda - \mu\\ 1 - \lambda - 3\mu\end{pmatrix} = \begin{pmatrix}\frac{47}{15}\\ -\frac{329}{150}\\ -\frac{47}{150}\end{pmatrix}\) | M1 |
| \(d = \sqrt{\left(\dfrac{47}{15}\right)^2 + \left(\dfrac{329}{150}\right)^2 + \left(\dfrac{47}{150}\right)^2} = \dfrac{47\sqrt{6}}{30}\) | M1 A1 |
| (5) |
M1: Finds a general chord between the 2 lines and attempts the scalar product between this and the directions, sets = 0 to give 2 equations in 2 unknowns
A1: Correct values
M1: Uses their values to find the ends of the chord or substitutes into their chord vector
M1: Correct completion by finding the distance between their 2 points
A1: Any equivalent e.g. \(\dfrac{47\sqrt{6}}{30}\) or awrt 3.84
Way 4
| Scheme | Marks |
|---|---|
| \(\pm\begin{pmatrix}2\\ 3\\ -1\end{pmatrix} \times \begin{pmatrix}1\\ 1\\ 3\end{pmatrix} = \pm\begin{pmatrix}10\\ -7\\ -1\end{pmatrix}\) | M1 A1 |
| \(\begin{pmatrix}1\\ 0\\ 2\end{pmatrix} + \lambda\begin{pmatrix}2\\ 3\\ -1\end{pmatrix} - \left[\begin{pmatrix}-1\\ 4\\ 1\end{pmatrix} + \mu\begin{pmatrix}1\\ 1\\ 3\end{pmatrix}\right] = \begin{pmatrix}2 + 2\lambda - \mu\\ -4 + 3\lambda - \mu\\ 1 - \lambda - 3\mu\end{pmatrix}\) \(\begin{pmatrix}2 + 2\lambda - \mu\\ -4 + 3\lambda - \mu\\ 1 - \lambda - 3\mu\end{pmatrix} = k\begin{pmatrix}10\\ -7\\ -1\end{pmatrix} \Rightarrow \begin{aligned}2 + 2\lambda - \mu &= 10k\\ -4 + 3\lambda - \mu &= -7k\\ 1 - \lambda - 3\mu &= -k\end{aligned}\) \(\Rightarrow k = \dfrac{47}{150}\) | M1 |
| \(d = \sqrt{\left(\dfrac{47}{15}\right)^2 + \left(\dfrac{329}{150}\right)^2 + \left(\dfrac{47}{150}\right)^2} = \dfrac{47\sqrt{6}}{30}\) | M1 A1 |
| (5) |
M1: Attempt cross product of direction vectors
A1: Correct vector
M1: Finds a common chord between the 2 lines and sets equal to a multiple of the normal vector to give 3 equations in 3 unknowns and solves to find a value for \(k\)
M1: Correct completion by finding the length of their vector
A1: Any equivalent e.g. \(\dfrac{47\sqrt{6}}{30}\) or awrt 3.84
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}3\\ 8\\ 13\end{pmatrix} - \begin{pmatrix}1\\ 0\\ 2\end{pmatrix} = \begin{pmatrix}2\\ 8\\ 11\end{pmatrix}\) | M1 |
| \(\begin{pmatrix}2\\ 3\\ -1\end{pmatrix} \times \begin{pmatrix}2\\ 8\\ 11\end{pmatrix} = \begin{pmatrix}41\\ -24\\ 10\end{pmatrix}\) | dM1 |
| \(\begin{pmatrix}41\\ -24\\ 10\end{pmatrix}\bullet\begin{pmatrix}1\\ 0\\ 2\end{pmatrix} = \ldots\) or \(\begin{pmatrix}41\\ -24\\ 10\end{pmatrix}\bullet\begin{pmatrix}3\\ 8\\ 13\end{pmatrix} = \ldots\) | ddM1 |
| \(41x - 24y + 10z = 61\) | A1 |
| (4) | |
| (13 marks) |
Notes
M1: Attempt another non-parallel vector in \(\Pi\)
dM1: Attempt cross product of two non-parallel vectors in the plane. If the method is not shown, at least 2 components should be correct. Dependent on the first M mark.
ddM1: Attempt scalar product with a point in the plane. Dependent on both previous method marks.
A1: Any multiple but must be a Cartesian equation.
(c) Way 2
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}3\\ 8\\ 13\end{pmatrix} - \begin{pmatrix}1\\ 0\\ 2\end{pmatrix} = \begin{pmatrix}2\\ 8\\ 11\end{pmatrix}\) | M1 |
| \(\mathbf{r} = \begin{pmatrix}1\\ 0\\ 2\end{pmatrix} + \lambda\begin{pmatrix}2\\ 3\\ -1\end{pmatrix} + \mu\begin{pmatrix}2\\ 8\\ 11\end{pmatrix}\) or \(\begin{aligned}x &= 1 + 2\lambda + 2\mu\ \ (1)\\ y &= 3\lambda + 8\mu\ \ (2)\\ z &= 2 - \lambda + 11\mu\ \ (3)\end{aligned}\) | dM1 |
| \((1) + 2(3):\ x + 2z = 5 + 24\mu\) \((2) + 3(3):\ 3z + y = 6 + 41\mu\) | ddM1 |
| \(\dfrac{3z + y - 6}{41} = \dfrac{x + 2z - 5}{24}\) | A1 |
| (4) |
M1: Attempt another vector in \(\Pi\)
dM1: Forms the vector equation of the plane. Dependent on the first M mark.
ddM1: Eliminates \(\lambda\) or \(\mu\). Dependent on both previous method marks.
A1: Any correct equation but must be a correct Cartesian equation. Isw