AS June 2019 Paper 1 Q4
4. The line \(l\) has equation
\[\frac{x + 2}{1} = \frac{y - 5}{-1} = \frac{z - 4}{-3}\]The plane \(\mathit{\Pi}\) has equation
\[\mathbf{r}.(\mathbf{i} - 2\mathbf{j} + \mathbf{k}) = -7\]Determine whether the line \(l\) intersects \(\mathit{\Pi}\) at a single point, or lies in \(\mathit{\Pi}\), or is parallel to \(\mathit{\Pi}\) without intersecting it. (5)
| Scheme | Marks | AO |
|---|---|---|
| \((\mathbf{r} =)\begin{pmatrix}-2 + \lambda\\ 5 - \lambda\\ 4 - 3\lambda\end{pmatrix}\) or \(\begin{pmatrix}-2\\ 5\\ 4\end{pmatrix} + \lambda\begin{pmatrix}1\\ -1\\ -3\end{pmatrix}\) (oe) | M1 | 1.1b |
| So meet if \(\begin{pmatrix}-2 + \lambda\\ 5 - \lambda\\ 4 - 3\lambda\end{pmatrix}\bullet\begin{pmatrix}1\\ -2\\ 1\end{pmatrix} = -7 \Rightarrow (-2 + \lambda) \times 1 + (5 - \lambda) \times {-2} + (4 - 3\lambda) \times 1 = -7\) | M1 A1 | 3.1a 1.1b |
| \(\Rightarrow 0\lambda - 8 = -7 \Rightarrow -8 = -7\) a contradiction so no intersection | A1ft | 2.3 |
| Hence \(l\) is parallel to \(\mathit{\Pi}\) but not in it. | A1cso | 3.2a |
| (5) | ||
| (5 marks) |
Notes
M1: Forms a parametric form for the line. Allow one slip.
M1: Substitutes into the equation of the plane to an equation in \(\lambda\). May use Cartesian form of plane to substitute into.
A1: Correct equation in \(\lambda\)
A1ft: Simplifies and derives a contradiction and deduces line and plane do not meet. Follow through in their initial equation in \(\lambda\) so
- contradiction so no intersection if \(\lambda\) disappears and constants unequal
- line lies in plane if a tautology is arrived at
- meet in a point if a solution for \(\lambda\) is found.
But do not allow for incorrect simplification from a correct initial equation in \(\boldsymbol{\lambda}\)
Note that a miscopy/misread of 7 instead of \(-7\) can therefore score a maximum of M1M1A0A1A0.
A1cso: Correct deduction from correct working. This may be seen two separate statements in their working. You may see attempts at showing the line is parallel before/after deducing there is no intersection.
Alternative (Alt 1)
Note that some may attempt a mix of the main scheme and Alt 1. Mark under main scheme unless Alt 1 would score higher.
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}1\\ -1\\ -3\end{pmatrix}\bullet\begin{pmatrix}1\\ -2\\ 1\end{pmatrix} = 1 \times 1 + (-1) \times (-2) + (-3) \times 1 = 0\) | M1 | 3.1a |
| Hence \(l\) is parallel to \(\mathit{\Pi}\) | A1 | 1.1b |
| \((-2, 5, 4)\) on \(l\), but \((1)(-2) + (-2)(5) + 1(4) = -8\) | M1 | 1.1b |
| \(-8 \ne -7\) so \((-2, 5, 4)\) is not on the plane. | A1ft | 2.3 |
| Hence \(l\) is (parallel to \(\mathit{\Pi}\) but) not in the plane. | A1cso | 3.2a |
| (5) |
M1: Attempts the dot product between the two direction vectors.
A1: Shows dot product is zero and makes the correct deduction that line is parallel to plane.
M1: Finds a point on \(l\) and substitutes into the equation of \(\mathit{\Pi}\) (vector or Cartesian)
A1ft: Simplifies and derives a contradiction – follow through their equation, so if arrive at a tautology, they should deduce the line is in the plane.
A1cso: Correct deduction from correct working but may be split across working.
Alternative (Alt 2)
| Scheme | Marks | AO |
|---|---|---|
| Attempts to solve \(\dfrac{x + 2}{1} = \dfrac{y - 5}{-1} = \dfrac{z - 4}{-3}\) and \(x - 2y + z = -7\) simultaneously – eliminates one variable for M mark. | M1 | 3.1a |
| e.g. \(y = -(x + 2) + 5 = -x + 3 \Rightarrow x - 2(-x + 3) + z = -7 \Rightarrow 3x + z = -1\) (oe) | A1 | 1.1b |
| Solves reduced equations, e.g. \(-3(x + 2) = z - 4 \Rightarrow 3x + z = -2\) and \(3x + z = -1 \Rightarrow (3x + z) - (3x + z) = -2 - (-1)\) | M1 | 1.1b |
| \(\Rightarrow 0 = -1\) a contradiction so no intersection | A1ft | 2.3 |
| Hence \(l\) is parallel to \(\mathit{\Pi}\) but not in it. | A1cso | 3.2a |
| (5) |
M1: Attempts to solve the Cartesian equation of the line and plane, using the plane equation to eliminate one variable for the M.
A1: Correct elimination of their chosen variable. (E.g. may see \(3 - 3y + z = -7\) or \(-2x - 2y - 2 = -7\) etc)
M1: Solves the reduced equations in two variables…
A1ft: … and derives a contradiction/line and plane do not meet. Follow through their result, so may reach a tautology and deduce lies in plane, or find single solution and deduce meet in a point.
A1cso: Correct deduction from correct working.