A2 June 2022 Paper 2 Q8
8. Two birds are flying towards their nest, which is in a tree.
Relative to a fixed origin, the flight path of each bird is modelled by a straight line.
In the model, the equation for the flight path of the first bird is
\[\mathbf{r}_1 = \begin{pmatrix}-1\\ 5\\ 2\end{pmatrix} + \lambda\begin{pmatrix}2\\ a\\ 0\end{pmatrix}\]and the equation for the flight path of the second bird is
\[\mathbf{r}_2 = \begin{pmatrix}4\\ -1\\ 3\end{pmatrix} + \mu\begin{pmatrix}0\\ 1\\ -1\end{pmatrix}\]where \(\lambda\) and \(\mu\) are scalar parameters and \(a\) is a constant.
In the model, the angle between the birds’ flight paths is 120°
The position of the nest is modelled as being at this common point.
The tree containing the nest is in a park.
The ground level of the park is modelled by the plane with equation
\[2x - 3y + z = 2\]| Scheme | Marks | AO |
|---|---|---|
| A complete method to use the scalar product of the direction vectors and the angle 120° to form an equation in \(a\) \(\dfrac{\begin{pmatrix}2\\ a\\ 0\end{pmatrix}\bullet\begin{pmatrix}0\\ 1\\ -1\end{pmatrix}}{\sqrt{2^2 + a^2}\sqrt{1^2 + (-1)^2}} = \cos 120\) | M1 | 3.1b |
| \(\dfrac{a}{\sqrt{4 + a^2}\sqrt{2}} = -\dfrac{1}{2}\) | A1 | 1.1b |
| \(2a = -\sqrt{4 + a^2}\sqrt{2} \Rightarrow 4a^2 = 8 + 2a^2 \Rightarrow a^2 = 4 \Rightarrow a = \ldots\) | M1 | 1.1b |
| \(a = -2\) | A1 | 2.2a |
| (4) |
Notes
M1: See scheme, allow a sign slip and cos 60
A1: Correct simplified equation in \(a\), cos 120 must be evaluated to \(-\frac{1}{2}\) and dot product calculated
Note: If the candidate states either \(\left|\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\right| = \cos\theta\) or \(\left|\dfrac{a}{\sqrt{4 + a^2}\sqrt{2}}\right| = \cos 60\) then has the equation \(\dfrac{a}{\sqrt{4 + a^2}\sqrt{2}} = \dfrac{1}{2}\) award this mark. If the module of the dot product is not seen then award A0 for this equation.
dM1: Solve a quadratic equation for \(a\), by squaring and solving an equation of the form \(a^2 = K\) where \(K \gt 0\)
A1: Deduces the correct value of \(a\) from a correct equation. Must be seen in part (a) using the angle between the lines.
Alternative cross product method
M1: \(\begin{vmatrix}2 & a & 0\\ 0 & 1 & -1\end{vmatrix} = \sqrt{2^2 + a^2}\sqrt{1^2 + (-1)^2}\sin 120\)
A1: \(\sqrt{a^2 + 8} = \sqrt{4 + a^2}\sqrt{2}\dfrac{\sqrt{3}}{2}\)
Then as above
Note If they use the point of intersection to find a value for \(a\) this scores no marks
| Scheme | Marks | AO |
|---|---|---|
| Any two of \(\mathbf{i}\): \(-1 + 2\lambda = 4\quad (1)\) \(\mathbf{j}\): \(5 + \text{`their } {-2}\text{'}\lambda = -1 + \mu\quad (2)\) \(\mathbf{k}\): \(2 = 3 - \mu\quad (3)\) | M1 | 3.4 |
| Solves the equations to find a value of \(\lambda\left\{= \dfrac{5}{2}\right\}\) and \(\mu\{= 1\}\) | M1 | 1.1b |
| \(r_1 = \begin{pmatrix}-1\\ 5\\ 2\end{pmatrix} + \dfrac{5}{2}\begin{pmatrix}2\\ \text{`their } {-2}\text{'}\\ 0\end{pmatrix}\) or \(r_2 = \begin{pmatrix}4\\ -1\\ 3\end{pmatrix} + 1\begin{pmatrix}0\\ 1\\ -1\end{pmatrix}\) | dM1 | 1.1b |
| \((4, 0, 2)\) or \(\begin{pmatrix}4\\ 0\\ 2\end{pmatrix}\) | A1 | 1.1b |
| Checks the third equation e.g. \(\lambda = \dfrac{5}{2}: \text{LHS} = 5 - 2\lambda = 5 - 5 = 0\) \(\mu = 1: \text{RHS} = -1 + \mu = -1 + 1 = 0\) therefore common point/intersect/consistent/tick or substitutes the values of \(\lambda\) and \(\mu\) into the relevant lines and achieves the same coordinate | B1 | 2.1 |
| (5) |
Notes
M1: Uses the model to write down any two correct equations
M1: Solve two equations simultaneously to find a value for \(\mu\) and \(\lambda\)
dM1: Dependent on previous method mark. Substitutes \(\mu\) and \(\lambda\) into a relevant equation. If no method shown two correct ordinates implies this mark.
A1: Correct coordinates. May be seen in part (c)
B1: Shows that the values of \(\mu\) and \(\lambda\) give the same third coordinate or point of intersection and draws the conclusion that the lines intersect/common point/consistent or tick.
Note: If an incorrect value for \(a\) is found in part (a) but in part (b) they find that \(a = -2\) this scores B0 but all other marks are available
| Scheme | Marks | AO |
|---|---|---|
| Full attempt to find the minimum distance from the point of intersection (nest) to the plane (ground) E.g. Minimum distance \(= \dfrac{\left|2 \times \text{`}4\text{'} + (-3) \times \text{`}0\text{'} + 1 \times \text{`}2\text{'} - 2\right|}{\sqrt{2^2 + (-3)^2 + 1^2}} = \ldots\) Alternatively \(\mathbf{r} = \begin{pmatrix}\text{`}4\text{'}\\ \text{`}0\text{'}\\ \text{`}2\text{'}\end{pmatrix} + \lambda\begin{pmatrix}2\\ -3\\ 1\end{pmatrix}\quad 2(\text{`}4\text{'} + 2\lambda) - 3(\text{`}0\text{'} - 3\lambda) + (\text{`}2\text{'} + \lambda) = 2 \Rightarrow \lambda = \ldots\left\{-\dfrac{4}{7}\right\}\) \(\mathbf{r} = \begin{pmatrix}\text{`}4\text{'}\\ \text{`}0\text{'}\\ \text{`}2\text{'}\end{pmatrix} + \text{`}{-\dfrac{4}{7}}\text{'}\begin{pmatrix}2\\ -3\\ 1\end{pmatrix} = \begin{pmatrix}\tfrac{20}{7}\\ \tfrac{12}{7}\\ \tfrac{10}{7}\end{pmatrix}\) Minimum distance \(= \sqrt{\left(2 \times \text{`}{-\dfrac{4}{7}}\text{'}\right)^2 + \left(-3 \times \text{`}{-\dfrac{4}{7}}\text{'}\right)^2 + \left(1 \times -\dfrac{4}{7}\right)^2} = \ldots\) \(= \sqrt{\left(\text{`}4\text{'} - \text{`}\dfrac{20}{7}\text{'}\right)^2 + \left(\text{`}0\text{'} - \text{`}\dfrac{12}{7}\text{'}\right)^2 + \left(\text{`}2\text{'} - \text{`}\dfrac{10}{7}\text{'}\right)^2} = \ldots\) | M1 A1ft | 3.1b 3.4 |
| \(\dfrac{8}{\sqrt{14}}\) or \(\dfrac{4\sqrt{14}}{7}\) or awrt 2.1 | A1 | 2.2b |
| (3) |
Notes
This is M1M1A1 on ePen marking as M1 A1ft A1
M1: Full attempt to find the minimum distance from a point to a plane. Condone a sign slip with the value of \(d\).
A1ft: Following through on their point of intersection. Uses the model to find a correct expression for minimum distance from the nest to the ground
A1: Correct distance
Alternative
| Scheme | Marks | AO |
|---|---|---|
| Find perpendicular distance from plane to the origin \(2x - 3y + z = 2\) \(|n| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{14}\) shortest distance \(= \dfrac{2}{\sqrt{14}}\) Find perpendicular distance from the plane containing the point of intersection to the origin \(2x - 3y + z = \begin{pmatrix}4\\ 0\\ 2\end{pmatrix}\bullet\begin{pmatrix}2\\ -3\\ 1\end{pmatrix} = 10\) shortest distance \(= \dfrac{10}{\sqrt{14}}\) Minimum distance \(= \dfrac{10}{\sqrt{14}} - \dfrac{2}{\sqrt{14}}\) | M1 A1ft | 3.1b 3.4 |
| \(\dfrac{8}{\sqrt{14}}\) or \(\dfrac{4\sqrt{14}}{7}\) or awrt 2.1 | A1 | 2.2b |
| (3) |
M1: Find the shortest distance from a point to plane by finding the perpendicular distance from the given plane to the origin and the perpendicular distance from the plane contacting their point of intersection to the origin and subtracts
A1ft: Following through on their point of intersection. Uses the model to find a correct expression for minimum distance from the nest to the ground
A1: Correct distance
| Scheme | Marks | AO |
|---|---|---|
| For example Not reliable as the birds will not fly in a straight line Not reliable as angle between flights paths will not always be 120° Not reliable/reliable as the ground will not be flat/smooth Not reliable as bird’s nest is not a point | B1 | 3.2b |
| (1) | ||
| (13 marks) |
Notes
B1: Comments on one of the models
- Flight path of the birds modelled as a straight line
- Angle between flight paths modelled as 120°
- The bird’s nest is modelled as a point
- Ground modelled as a plane
Then states unreliable
Any correct answer seen, ignore any other incorrect answers