A2 June 2019 Paper 1 Q7
7. The line \(l_1\) has equation
\[\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z - 4}{3}\]The line \(l_2\) has equation
\[\mathbf{r} = \mathbf{i} + 3\mathbf{k} + t(\mathbf{i} - \mathbf{j} + 2\mathbf{k})\]where \(t\) is a scalar parameter.
Way 1
| Scheme | Marks | AO |
|---|---|---|
| \(1 + 2\lambda = 1 + t\) \(-1 - \lambda = -t\) \(4 + 3\lambda = 3 + 2t\) \(\Rightarrow t = \ldots\) or \(\lambda = \ldots\) | M1 | 3.1a |
| Checks the third equation with \(t = 2\) and \(\lambda = 1\) Or shows that the coordinate (3, -2, 7) lies on both lines | A1 | 1.1b |
| As the lines intersect at a point the lines lie in the same plane. | A1 | 2.4 |
| (3) |
Notes
Allow using \(\begin{pmatrix}1\\3\\0\end{pmatrix}\) instead of \(\begin{pmatrix}1\\0\\3\end{pmatrix}\) for the method mark.
Way 1
M1: Starts by attempting to find where the two lines intersect. They must set up a parametric equation for line 1 (allow sign slips and as long as the intention is clear), forms simultaneous equations by equating coordinates and attempts to solve to find a value for \(t = \ldots\) or \(\lambda = \ldots\).
A1: Shows that there is a unique solution by checking the third equation or shows that the coordinate (3, -2, 7) lies on both lines.
A1: Achieves the correct values \(t = 2\) and \(\lambda = 1\), checks the third equation and concludes that either
- a common point,
- the lines intersect
- the equations are consistent
therefore, the lines lie in the same plane
Alternative: Way 2
| Scheme | Marks | AO |
|---|---|---|
| \(1 = 1 + 2\lambda + t\) \(-1 = \quad -\lambda - t\) \(4 = 3 + 3\lambda + 2t\) \(\Rightarrow t = \ldots\) or \(\lambda = \ldots\) or \(1 = 1 + 2\lambda + t\) \(0 = -1 - \lambda - t\) \(3 = 4 + 3\lambda + 2t\) \(\Rightarrow t = \ldots\) or \(\lambda = \ldots\) | M1 | 3.1a |
| Checks the third equation with \(t = 2\) and \(\lambda = -1\) or Checks the third equation with \(t = -2\) and \(\lambda = 1\) | A1 | 1.1b |
| Second coordinates lie on the plane; therefore, the lines lie on the same plane | A1 | 2.4 |
| (3) |
M1: Finds the vector equation of the plane with the both direction vectors and one coordinate (allow a sign slip), sets equal to the other coordinate, forms simultaneous equations and attempts to solve to find a value for \(t = \ldots\) or \(\lambda = \ldots\).
A1: Shows that the other coordinate lies on the plane by checking the third equation.
A1: Achieves the correct values \(t = -2\) and \(\lambda = 1\) or \(t = 2\) and \(\lambda = -1\) and concludes that the second coordinate lie on the plane; therefore, the lines lie on the same plane
Alternative: Way 3
| Scheme | Marks | AO |
|---|---|---|
| \(x = 1 + t,\quad y = -t,\quad z = 3 + 2t\) \(\dfrac{1 + t - 1}{2} = \dfrac{-t + 1}{-1} = \dfrac{3 + 2t - 4}{3}\) Solves a pair of equations \(t = \ldots\) | M1 | 3.1a |
| Solve two pairs of equations to find \(t = 2\) | A1 | 1.1b |
| As the lines intersect at a point the lines lie in the same plane. | A1 | 2.4 |
| (3) |
M1: Substitutes line 2 into line 1 and solves a pair of equations to find a value for \(t\). Allow slip with the position of 0 and sign slips as long as the intention is clear.
A1: Solve two pairs of equations to achieve \(t = 2\) for each.
A1: Achieves the correct value \(t = 2\) and concludes that either
- a common point,
- the lines intersect
- the equations are consistent
therefore, the lines lie in the same plane
Alternative: Way 4 (Using Further Pure 2 knowledge)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\cdot\begin{pmatrix}x\\y\\z\end{pmatrix} \Rightarrow 2x - y + 3z = 0\) and \(\begin{pmatrix}1\\-1\\2\end{pmatrix}\cdot\begin{pmatrix}x\\y\\z\end{pmatrix} \Rightarrow x - y + 2z = 0\) attempts to solve the equations to find a normal vector OR attempts the cross product \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\times\begin{pmatrix}1\\-1\\2\end{pmatrix} = \ldots\) AND either finds the equation of one plane OR finds dot product between the normal and one coordinate \(\mathbf{r}\cdot\begin{pmatrix}1\\-1\\-1\end{pmatrix} = \begin{pmatrix}1\\-1\\4\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\-1\end{pmatrix} = \ldots\) or \(\mathbf{r}\cdot\begin{pmatrix}1\\-1\\-1\end{pmatrix} = \begin{pmatrix}1\\0\\3\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\-1\end{pmatrix} = \ldots\) OR \(\begin{pmatrix}1\\-1\\4\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\-1\end{pmatrix} = \ldots\) or \(\begin{pmatrix}1\\0\\3\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\-1\end{pmatrix} = \ldots\) | M1 | 3.1a |
| Achieves the correct planes containing each line \(\mathbf{r}\cdot\begin{pmatrix}1\\-1\\-1\end{pmatrix} = -2\) or \(x - y - z = -2\) o.e. OR Shows that \(\begin{pmatrix}1\\-1\\4\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\-1\end{pmatrix} = -2\) and \(\begin{pmatrix}1\\0\\3\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\-1\end{pmatrix} = -2\) o.e. | A1 | 1.1b |
| Both planes are the same, therefore the lines lie in the same plane. | A1 | 2.4 |
| (3) |
M1: A complete method to finds a vector which is normal to both lines and attempts to finds the equation of the plane containing one line.
A1: Achieves the correct equation for the plane containing each line.
A1: Conclusion, planes are the same, therefore the lines lie in the same plane.
| Scheme | Marks | AO |
|---|---|---|
| e.g. \(\mathbf{r} = \begin{pmatrix}1\\0\\3\end{pmatrix} + p\begin{pmatrix}2\\-1\\3\end{pmatrix} + q\begin{pmatrix}1\\-1\\2\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}1\\-1\\4\end{pmatrix} + p\begin{pmatrix}2\\-1\\3\end{pmatrix} + q\begin{pmatrix}1\\-1\\2\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}3\\-2\\7\end{pmatrix} + p\begin{pmatrix}2\\-1\\3\end{pmatrix} + q\begin{pmatrix}1\\-1\\2\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}3\\-2\\7\end{pmatrix} + p\begin{pmatrix}0\\-1\\1\end{pmatrix} + q\begin{pmatrix}1\\-1\\2\end{pmatrix}\) or \(\mathbf{r}.k\begin{pmatrix}1\\-1\\-1\end{pmatrix} = -2k\) | B1 | 2.5 |
| (1) |
Notes
This may be seen in part (a)
B1: Correct vector equation allow any letter for the scalers.
Must start with \(\mathbf{r} = \ldots\) and uses two out of the following direction vectors \(\pm\begin{pmatrix}1\\-1\\2\end{pmatrix}\), \(\pm\begin{pmatrix}2\\-1\\3\end{pmatrix}\) or \(\pm\begin{pmatrix}0\\-1\\1\end{pmatrix}\) and one of the following position vectors \(\begin{pmatrix}1\\0\\3\end{pmatrix}\), \(\begin{pmatrix}1\\-1\\4\end{pmatrix}\) or \(\begin{pmatrix}3\\-2\\7\end{pmatrix}\)
Way 1
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\bullet\begin{pmatrix}1\\-1\\2\end{pmatrix} = 2 + 1 + 6\) | M1 | 1.1b |
| \(\sqrt{2^2 + (-1)^2 + 3^2}\sqrt{1^2 + (-1)^2 + 2^2}\cos\theta = 9\) \(\Rightarrow \cos\theta = \dfrac{9}{\sqrt{2^2 + (-1)^2 + 3^2}\sqrt{1^2 + (-1)^2 + 2^2}}\) | dM1 | 2.1 |
| \(\theta = 11\) cao | A1 | 1.1b |
| (3) | ||
| (7 marks) |
Notes
Way 1
M1: Calculates the scalar product between the direction vectors, allow one slip, if the intention is clear
dM1: Dependent on the previous method mark. Applies the scalar product formula with their scalar product to find a value for \(\cos\theta\)
A1: Correct answer only
Alternative: Way 2 (Using Further Pure 2 knowledge)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\times\begin{pmatrix}1\\-1\\2\end{pmatrix} = \begin{pmatrix}1\\-1\\-1\end{pmatrix}\) | M1 | 1.1b |
| \(\sqrt{2^2 + (-1)^2 + 3^2}\sqrt{1^2 + (-1)^2 + 2^2}\sin\theta = \sqrt{1^2 + (-1)^2 + (-1)^2}\) \(\Rightarrow \sin\theta = \dfrac{\sqrt{1^2 + (-1)^2 + (-1)^2}}{\sqrt{2^2 + (-1)^2 + 3^2}\sqrt{1^2 + (-1)^2 + 2^2}}\) | dM1 | 2.1 |
| \(\theta = 11\) cao | A1 | 1.1b |
| (3) |
M1: Calculates the vector product between the direction vectors, allow one slip, if the intention is clear
dM1: Dependent on the previous method mark. Applies the vector product formula with their vector product to find a value for \(\sin\theta\)
A1: Correct answer only