A2 October 2020 Paper 2 Q4
4 The equations of two intersecting lines \(l_1\) and \(l_2\) are
\[l_1 : \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ a \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} \qquad l_2 : \mathbf{r} = \begin{pmatrix} 7 \\ 9 \\ -2 \end{pmatrix} + \mu\begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}\]where \(a\) is a constant.
The equation of the plane \(\Pi\) is
\[\mathbf{r}.\begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} = -14.\]\(l_1\) and \(\Pi\) intersect at \(Q\).
\(l_2\) and \(\Pi\) intersect at \(R\).
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 13 \\ 3 \\ -14 \end{pmatrix}.\begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} = 13 + 15 - 42 = -14\) (so \(R\) is on \(\Pi\)) | B1 | 1.1 |
| eg \(7 - \mu = 13 \Rightarrow \mu = -6 \Rightarrow\) \(\mathbf{r} = \begin{pmatrix} 7 \\ 9 \\ -2 \end{pmatrix} - 6\begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 13 \\ 3 \\ -14 \end{pmatrix}\) (so \(R\) is also on \(l_2\)) | B1 | 1.1 |
| [2] |
Notes
B1: AG. Intermediate working must be seen
B1: AG. Or \(9 + \mu = 3\) or \(-2 + 2\mu = -14\) but must be checked in other two equations.
Alternate method
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}.\left(\begin{pmatrix} 7 \\ 9 \\ -2 \end{pmatrix} + \mu\begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}\right) = 46 + 10\mu = -14 \Rightarrow \mu = -6\) | M1 |
| \(\mu = -6 \Rightarrow\) \(\mathbf{r} = \begin{pmatrix} 7 \\ 9 \\ -2 \end{pmatrix} - 6\begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 13 \\ 3 \\ -14 \end{pmatrix}\) so \(R\) is \((13, 3, -14)\) | A1 |
| [2] |
M1: AG. Substituting in expression of the point into the equation of the plane to find a value for \(\mu\)
A1: AG. Answer in vector form is acceptable.
| Scheme | Marks | AO |
|---|---|---|
| Since lines intersect \(\begin{pmatrix} 1 \\ 0 \\ a \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 7 \\ 9 \\ -2 \end{pmatrix} + \mu\begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}\) for some \(\lambda\) and \(\mu\) so \(1 + 2\lambda = 7 - \mu\) \(\lambda = 9 + \mu\) \((a - 3\lambda = -2 + 2\mu)\) | M1 | 3.1a |
| \(\Rightarrow \lambda = 5,\ \mu = -4\) | A1 | 1.1 |
| so \(a + 5 \times (-3) = -2 + (-4) \times 2 \Rightarrow a = 5\) | A1ft | 1.1 |
| At \(Q\), \(\left(\begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}\right).\begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} = -14\) for some \(\lambda\) so \(1 + 5 \times 3 + \lambda(2 + 5 + (-3) \times 3) = -14\) | *M1 | 2.1 |
| \(-2\lambda = -30 \Rightarrow \lambda = 15 \Rightarrow Q(31, 15, -40)\) | A1 | 1.1 |
| \(\sqrt{(31 - 13)^2 + (15 - 3)^2 + (-40 - -14)^2}\) | dep*M1 | 2.1 |
| \(= \sqrt{18^2 + 12^2 + 26^2} = \sqrt{1144}\) | A1 | 3.2a |
| [7] |
Notes
M1: Equating the lines and deriving 2 useful equations. Ignore attempts at \(z\) coefficient equation
A1: Can be BC
*M1: Substituting general \(\mathbf{r}\) from \(l_1\) into the \(\Pi\) equation and dotting out to form an equation in \(\lambda\). Accept algebraic expressions in \(a\) until final A mark
A1: \(\lambda = 7.5 + 1.5a\)
\(Q(16 + 3a,\ 7.5 + 1.5a,\ -22.5 - 3.5a)\)
dep*M1: Method fully shown or at least 2 of 3 squared terms correct. Depends on correct method shown to find \(Q\)
\(\sqrt{(3 + 3a)^2 + (4.5 + 1.5a)^2 + (-8.5 - 3.5a)^2}\)
A1: \(2\sqrt{286}\)