A2 June 2025 Paper 1 Q7
7 A 3-D coordinate system, whose units are metres, is set up to model a street containing telephone cables \(T_1\) and \(T_2\).
The cables are modelled as straight lines with vector equations
\(T_1: \mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix}\) and \(T_2: \mathbf{r} = \begin{pmatrix} 8 \\ 2 \\ 4 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\).
To access the cables for maintenance, a ladder can be used. The base of the ladder is placed at a fixed point on the ground.
The ladder is modelled as a straight-line segment. The base of the ladder is modelled as being located at the point \((4, 5, 0)\).
| Scheme | Marks | AO |
|---|---|---|
| \([1]: 3 + 5\lambda = 8 + 2\mu\) \([2]: 1 + 4\lambda = 2 - \mu\) \([3]: 1 + \lambda = 4 + \mu\) | M1 | 3.4 |
| e.g. solving [1] and [2] gives: \(\lambda = \dfrac{7}{13}, \mu = -\dfrac{15}{13}\) | A1 | 1.1 |
| Substituting \(\lambda = \dfrac{7}{13}, \mu = -\dfrac{15}{13}\) into [3] gives \(1 + \dfrac{7}{13} = \dfrac{20}{13}\) and \(4 - \dfrac{15}{13} = \dfrac{37}{13}\) so the lines do not intersect (as \(\dfrac{20}{13} \neq \dfrac{37}{13}\)) | A1 | 2.2a |
| [3] |
Notes
M1: Attempt to solve any pair of simultaneous equation for either \(\lambda\) or \(\mu\). So must obtain a value for either \(\lambda\) or \(\mu\). No MR in this part but allow either a single sign error in one equation only or one incorrect value in one equation only
A1: For finding either \(\lambda\) or \(\mu\) correctly for one pair of equations.
For reference, solving [2] and [3] gives: \(\lambda = \frac{4}{5}, \mu = -\frac{11}{5}\) and solving [1] and [3] gives: \(\lambda = -\frac{1}{3}, \mu = -\frac{10}{3}\)
A1: Substitutes correct \(\lambda\) and \(\mu\) into third equation and showing inconsistency + conclusion (e.g. ‘lines do not intersect’, ‘no solutions’, ‘equations not consistent’). Values being compared must be evaluated e.g. \(1 + \frac{7}{13} \neq 4 - \frac{15}{13}\) only is A0.
For reference (if equations not re-arranged):
\(\lambda = \frac{4}{5}, \mu = -\frac{11}{5}\) into [1] gives 7 and \(\frac{18}{5}\)
\(\lambda = -\frac{1}{3}, \mu = -\frac{10}{3}\) into [2] gives \(-\frac{1}{3}\) and \(\frac{16}{3}\)
Alternative method is to find one value of \(\lambda\) or \(\mu\) correctly and then solve a different pair of equations and obtain a different value of either \(\lambda\) or \(\mu\) (values of \(\lambda, \mu\) appear above for checking purposes)
| Scheme | Marks | AO |
|---|---|---|
| \(\left(\begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix} - \begin{pmatrix} 4 \\ 5 \\ 0 \end{pmatrix}\right) \cdot \begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix} \; (= 0)\) | M1* | 3.3 |
| \(\big(5(5\lambda - 1) + 4(4\lambda - 4) + (\lambda + 1) = 0 \Rightarrow\big)\) \(\lambda = \dfrac{10}{21}\) | A1 | 1.1 |
| \(\left|\begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} + \dfrac{10}{21}\begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix} - \begin{pmatrix} 4 \\ 5 \\ 0 \end{pmatrix}\right| = \sqrt{\left(\dfrac{29}{21}\right)^2 + \left(-\dfrac{44}{21}\right)^2 + \left(\dfrac{31}{21}\right)^2}\) | M1dep* | 3.4 |
| \(= 291\) (cm) | A1 | 1.1 |
| [4] |
Notes
M1*: Scalar product of direction component of \(T_1\) with vector from \((4, 5, 0)\) to general point on \(T_1\). No MR in this part but condone a single numerical slip in a value and sign errors only (but must be subtracting the general point from \((4, 5, 0)\)). No evaluation of the scalar product needed for this mark
Alternative for first M mark
| Scheme | Marks |
|---|---|
| \(D^2 = (5\lambda - 1)^2 + (4\lambda - 4)^2 + (\lambda + 1)^2\) \(\Rightarrow 2D \times \dfrac{\mathrm{d}D}{\mathrm{d}\lambda} = 10(5\lambda - 1) + 8(4\lambda - 4) + 2(\lambda + 1)\) | M1* |
Finds distance (or distance squared) from \((4, 5, 0)\) to general point on \(T_1\) and then differentiates with respect to \(\lambda\) - condone a single numerical slip in a value and sign errors only when setting up the expression for the distance (or distance squared) (but must be subtracting the general point from \((4, 5, 0)\))
A1: Solves to correctly find \(\lambda\) at closest point to \((4,5,0)\) on \(T_1\). Need not find coordinates of point
M1dep*: Finding the magnitude from \((4,5,0)\) to the closest point on \(T_1\) using their value of \(\lambda\) following through their \(\begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix} - \begin{pmatrix} 4 \\ 5 \\ 0 \end{pmatrix}\)
(corrected from the printed mark scheme: the third component is printed as \(\left(-\frac{31}{21}\right)^2\); the \(z\)-component is \(1 + \frac{10}{21} = +\frac{31}{21}\), and the distance is unchanged)
A1: 3 sf answer must be seen somewhere. Accept 291 with no units, but not 2.91 m or 2.91
Alternative method 1
| Scheme | Marks |
|---|---|
| \(\dfrac{\left|\left(\begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} 4 \\ 5 \\ 0 \end{pmatrix}\right) \times \begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix}\right|}{\left|\begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix}\right|}\) | M1* |
| \(\begin{pmatrix} -1 \\ -4 \\ 1 \end{pmatrix} \times \begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix} = \begin{pmatrix} -8 \\ 6 \\ 16 \end{pmatrix}\) | A1 |
| \(= \dfrac{\sqrt{(-8)^2 + 6^2 + 16^2}}{\sqrt{5^2 + 4^2 + 1^2}}\) | M1dep* |
| \(= 291\) (cm) | A1 |
M1*: For using the formula \(D = \dfrac{\left|\overrightarrow{AP} \times \mathbf{d}\right|}{|\mathbf{d}|}\) where \(A\) is \((4, 5, 0)\), \(P\) is any point on \(T_1\) and \(\mathbf{d}\) is the direction vector of \(T_1\). No MR in this part but condone a single numerical slip in a value and sign errors only. No evaluation of the vector product needed for this mark
A1: For calculating correct vector product \(\pm\begin{pmatrix} -8 \\ 6 \\ 16 \end{pmatrix}\)
M1dep*: For \(= \dfrac{\sqrt{a^2 + b^2 + c^2}}{\sqrt{5^2 + 4^2 + 1^2}}\) where \(a\), \(b\) and \(c\) are the components of their vector product
A1: 3 sf answer must be seen somewhere. Accept 291 with no units, but not 2.91 m or 2.91
Alternative method 2
| Scheme | Marks |
|---|---|
| \(\left(\begin{pmatrix} 4 \\ 5 \\ 0 \end{pmatrix} - \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}\right) \cdot \begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix} = \left|\begin{pmatrix} 4 \\ 5 \\ 0 \end{pmatrix} - \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}\right| \times \left|\begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix}\right| \times \cos\theta\) | M1* |
| Using \((3, 1, 1)\) gives \(\cos\theta = \pm\dfrac{10\sqrt{21}}{63}\) | A1 |
| \(d = \sqrt{(4 - 3)^2 + (5 - 1)^2 + (0 - 1)^2}\,\sin(43.3317\ldots)\) | M1dep* |
| \(= 291\) (cm) | A1 |
M1*: Using \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta\) with the direction component of \(T_1\) and a vector from \((4, 5, 0)\) to any point on \(T_1\). No MR in this part but condone a single numerical slip in a value and sign errors only. No evaluation is needed for this mark
A1: Correct expression for the cosine of the angle between \(T_1\) and the line from \((4, 5, 0)\) to any point on \(T_1\) – for reference the most common point on \(T_1\) is \((3, 1, 1)\) but any point could have been used. If \((3, 1, 1)\) then accept \(\cos\theta = \pm 0.73\) or better \((\pm 0.72739\ldots)\) - if the angle found then accept an angle of 43 or better (43.3317…) or 137 or better (136.668…)
M1dep*: Correct expression for the required distance using their angle and the magnitude of the distance between \((4, 5, 0)\) and their point on \(T_1\)
A1: 3 sf answer must be seen somewhere. Accept 291 with no units, but not 2.91 m or 2.91
| Scheme | Marks | AO |
|---|---|---|
| e.g. the cables are modelled as lines (so they have zero width) but they will have width (and so may intersect) e.g. the cables are modelled as straight/lines but (they are likely to not be straight, as) they might bend (under gravity, so may intersect) | B1 | 3.5b |
| [1] |
Notes
B1: For identifying a modelling assumption that would affect the answer to part (a). Also, other reasonable answers. B0 for ‘the lines are modelled as being infinite’ only
Allow modelling assumption about how the cable might impact on the answer to part (b) e.g. the cables are modelled as rigid, and the ladder may cause them to bend. However, an answer which does not refer to the cables scores B0