A2 June 2022 Paper 1 Q10
10 In this question all measurements are in centimetres.
A small, thin laser pen is set up with one end at \(A(7, 2, -3)\) and the other end at \(B(9, -3, -2)\)
A laser beam travels from \(A\) to \(B\) and continues in a straight line towards a large thin sheet of glass.
The sheet of glass lies within a plane \(\Pi_1\) which is modelled by the equation
\[4x + py + 5z = 9\]where \(p\) is an integer.
Find the value of \(p\) [6 marks]
This second sheet lies within a plane \(\Pi_2\) which is modelled by the equation
\[4x + py + 5z = -5\]Calculate the distance between the sheets of glass. [2 marks]
Find the coordinates of the image of \(A\) after reflection in \(\Pi_1\) [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains correct normal vector to the plane | B1 | 2.2a |
| Obtains correct expression for \(\overrightarrow{AB}\) or \(\overrightarrow{BA}\) | B1 | 1.2 |
| Obtains their correct scalar (or vector) product | B1F | 1.1b |
| Uses scalar (or vector) product to obtain an equation in \(p\) | M1 | 3.1a |
| Forms an equation in \(p\), by squaring and removing any rational functions Eg \(82 + 2p^2 = 5^2(13 - 5p)^2\) | M1 | 1.1a |
| Solves quadratic and selects correct answer, discarding the other root Condone lack of modulus sign in the working | R1 | 2.1 |
| (6) |
Typical solution
Normal to plane \(\mathbf{n} = \begin{bmatrix}4 \\ p \\ 5\end{bmatrix}\)
Let \(\overrightarrow{AB} = \mathbf{c}\) then
\[\mathbf{c} = \begin{bmatrix}2 \\ -5 \\ 1\end{bmatrix}\]\[\mathbf{n} \cdot \mathbf{c} = 13 - 5p\]\[|\mathbf{c}| = \sqrt{30} \quad \text{and} \quad |\mathbf{n}| = \sqrt{41 + p^2}\]As \(\alpha\) is acute, \(\sin\alpha = \cos\theta = \dfrac{\sqrt{15}}{75}\)
Hence
\[\left|\frac{13 - 5p}{\sqrt{30}\sqrt{41 + p^2}}\right| = \frac{\sqrt{15}}{75}\]\[82 + 2p^2 = 5^2(13 - 5p)^2\]\[623p^2 - 3250p + 4143 = 0\]\[p = 3 \text{ or } p = \frac{1381}{623}\]\[p \in \mathbb{Z}\]\[\therefore p = 3\]| Scheme | Marks | AO |
|---|---|---|
| Recognises need to divide constant term of the plane equation by \(|\mathbf{n}|\) | M1 | 1.1a |
| Finds correct distance for their \(p\), exact or decimal at least 2sf (condone 2) Condone missing units | A1F | 1.1b |
| (2) |
Typical solution
\[|\mathbf{n}| = 5\sqrt{2}\]Distance is
\[\frac{9}{5\sqrt{2}} + \frac{5}{5\sqrt{2}} = \frac{7\sqrt{2}}{5} = 1.98\text{ cm}\]| Scheme | Marks | AO |
|---|---|---|
| Obtains correct equation of the line through \(A\) & \(A^{\prime}\) for their \(p\) Condone lack of \(\mathbf{r} =\) | B1F | 3.1a |
| Forms an equation to find the value of \(\mu\) for their line Condone use of \(\Pi_2\) | M1 | 3.1a |
| Doubles their value of \(\mu\) and uses it to find image point for their line Condone use of \(\Pi_2\) | M1 | 3.2a |
| Obtains correct coordinates for their \(p\) Do not accept position vector Do not condone use of \(\Pi_2\) | A1F | 1.1b |
| (4) | ||
| (12 marks) |
Typical solution
\[\mathbf{r} = \begin{bmatrix}7 \\ 2 \\ -3\end{bmatrix} + \mu\begin{bmatrix}4 \\ 3 \\ 5\end{bmatrix}\]At \(\Pi_1\)
\[4(7 + 4\mu) + 3(2 + 3\mu) + 5(-3 + 5\mu) = 9\]\[\mu = \frac{-1}{5}\]At image point
\[\mu = \frac{-2}{5}\]\[\left(\frac{27}{5}, \frac{4}{5}, -5\right)\]