A2 October 2020 Paper 1 Q4
4. The plane \(\Pi_1\) has equation
\[\mathbf{r} = 2\mathbf{i} + 4\mathbf{j} - \mathbf{k} + \lambda(\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) + \mu(-\mathbf{i} + 2\mathbf{j} + \mathbf{k})\]where \(\lambda\) and \(\mu\) are scalar parameters.
The line \(l\) has equation
\[\frac{x - 1}{5} = \frac{y - 3}{-3} = \frac{z + 2}{4}\]The plane \(\Pi_2\) has equation
\[\mathbf{r}.(2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) = 5\]| Scheme | Marks | AO |
|---|---|---|
| Attempts normal vector: E.g. let \(\mathbf{n} = a\mathbf{i} + b\mathbf{j} + \mathbf{k}\) then \(a + 2b - 3 = 0,\ -a + 2b + 1 = 0\) \(\Rightarrow a = \ldots,\ b = \ldots\) or \(\mathbf{n} = (\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) \times (-\mathbf{i} + 2\mathbf{j} + \mathbf{k})\) | M1 | 3.1a |
| \(\mathbf{n} = k(4\mathbf{i} + \mathbf{j} + 2\mathbf{k})\) | A1 | 1.1b |
| \((4\mathbf{i} + \mathbf{j} + 2\mathbf{k}).(2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) = \ldots\) | M1 | 1.1b |
| \(4x + y + 2z = 10\) | A1 | 2.5 |
| (4) |
Notes
Accept equivalent vector notation throughout.
(a)
M1: Starts by attempting to find a normal vector using a correct method. Allow if there are sign errors in attempts at the cross product.
A1: Obtains a correct normal vector
M1: Attempts scalar product between their normal and a point in the plane
A1: Correct Cartesian form (accept any equivalent Cartesian equation)
Alternative:
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{aligned} x &= 2 + \lambda - \mu\\ y &= 4 + 2\lambda + 2\mu\\ z &= -1 - 3\lambda + \mu \end{aligned} \Rightarrow \begin{aligned} 2x + y &= 8 + 4\lambda\\ y - 2z &= 6 + 8\lambda \end{aligned}\) | M1 A1 | 3.1a 1.1b |
| \(2(2x + y - 8) = y - 2z - 6\) \((4x + y + 2z = 10)\) | M1 A1 | 1.1b 2.5 |
| (4) |
Alternative
M1: Uses the component form to eliminate one of the scalar parameters
A1: Two correct equations with one parameter eliminated OR a correct equation for each parameter in terms of \(x\), \(y\) and \(z\)
M1: Forms a Cartesian equation
A1: Correct Cartesian equation (accept any equivalent form)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{x - 1}{5} = \dfrac{y - 3}{-3} = \dfrac{z + 2}{4} \Rightarrow \mathbf{r} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} + \lambda(5\mathbf{i} - 3\mathbf{j} + 4\mathbf{k})\) \(4(1 + 5\lambda) + 3 - 3\lambda + 2(4\lambda - 2) = 10 \Rightarrow \lambda = \ldots\) | M1 | 3.1a |
| \(\lambda = \dfrac{7}{25} \Rightarrow \mathbf{r} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} + \dfrac{7}{25}(5\mathbf{i} - 3\mathbf{j} + 4\mathbf{k})\) | dM1 | 1.1b |
| \(\left(\dfrac{12}{5}, \dfrac{54}{25}, -\dfrac{22}{25}\right)\) | A1 | 1.1b |
| (3) |
Notes
(b)
M1: Interprets the Cartesian form to give a parametric form (allow sign slips) and substitutes this into their Cartesian equation and proceeds to find a value for their parameter.
NB: Attempts at \(\begin{pmatrix}2 + \lambda - \mu\\ 4 + 2\lambda + 2\mu\\ -1 - 3\lambda + \mu\end{pmatrix} = \begin{pmatrix}1 + 5\lambda\\ 3 - 3\lambda\\ -2 + 4\lambda\end{pmatrix}\) will score M0 as there are only two parameters, but \(\begin{pmatrix}2 + \lambda - \mu\\ 4 + 2\lambda + 2\mu\\ -1 - 3\lambda + \mu\end{pmatrix} = \begin{pmatrix}1 + 5\gamma\\ 3 - 3\gamma\\ -2 + 4\gamma\end{pmatrix}\) leading to a value for \(\gamma\) from solving three equations in three unknowns in M1.
dM1: Substitutes their parameter value back into the parametric form of the line. The parameter must have come from a correct attempt to find the value at intersection.
A1: Correct coordinates. Accept as \(x = \ldots\), \(y=\), … \(z = \ldots\) or as a vector.
Alternative:
| Scheme | Marks | AO |
|---|---|---|
| \(4x + \left(-\dfrac{3}{5}(x - 1) + 3\right) + 2\left(\dfrac{4}{5}(x - 1) - 2\right) = 10 \Rightarrow x = \ldots\) | M1 | 3.1a |
| \(\Rightarrow y = \ldots,\ z = \ldots\) | M1 | 1.1b |
| \(\left(\dfrac{12}{5}, \dfrac{54}{25}, -\dfrac{22}{25}\right)\) | A1 | 1.1b |
| (3) |
Alternative:
M1: Eliminates two of the variables from the equation of plane using the Cartesian equation of the line and solves the linear equation.
dM1: Finds the other two coordinates.
A1: Correct coordinates, as above.
| Scheme | Marks | AO |
|---|---|---|
| \((4\mathbf{i} + \mathbf{j} + 2\mathbf{k}).(2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) = 8 - 1 + 6 = 13\) \(13 = \sqrt{14}\sqrt{21}\cos\theta \Rightarrow \theta = \ldots\) | M1 | 1.1b |
| \(\theta = 41^\circ\) | A1 | 1.1b |
| (2) | ||
| (9 marks) |
Notes
(c)
M1: Complete and correct scalar product method leading to a value for \(\theta\). Note that if \(\sin\theta\) is used instead of \(\cos\theta\) then they must also apply \(90 - \theta\) to access the method.
A1: Correct angle, accept awrt 41. as their final answer (do not isw if they go on to give e.g. \((180 - 41)^\circ\))