A2 June 2023 Q7
7. With respect to a fixed origin \(O\) the point \(A\) has coordinates \((3, 6, 5)\) and the line \(l\) has equation
\[\left(\mathbf{r} - (12\mathbf{i} + 30\mathbf{j} + 39\mathbf{k})\right) \times (7\mathbf{i} + 13\mathbf{j} + 24\mathbf{k}) = \mathbf{0}\]The points \(B\) and \(C\) lie on \(l\) such that \(AB = AC = 15\)
Given that \(A\) does not lie on \(l\) and that the \(x\) coordinate of \(B\) is negative,
The point \(D\) has coordinates \((-2, 1, \alpha)\), where \(\alpha\) is a constant.
Given that the volume of the tetrahedron \(ABCD\) is 147
Given that \(\alpha \gt 0\)
| Scheme | Marks | AO |
|---|---|---|
| Vector \(A\) to \(l\) is \(\pm\left(\begin{pmatrix}12\\ 30\\ 39\end{pmatrix} + \lambda\begin{pmatrix}7\\ 13\\ 24\end{pmatrix} - \begin{pmatrix}3\\ 6\\ 5\end{pmatrix}\right)\) | B1 | 2.2a |
| \(\left|\begin{matrix}9 + 7\lambda\\ 24 + 13\lambda\\ 34 + 24\lambda\end{matrix}\right| = 15 \Rightarrow (9 + 7\lambda)^2 + (24 + 13\lambda)^2 + (34 + 24\lambda)^2 = 15^2 \Rightarrow \lambda = \ldots\) | M1 | 3.1a |
| \(794\lambda^2 + 2382\lambda + 1588 = 0 \Rightarrow \lambda = -1, -2\) | A1 | 1.1b |
| \(B(-2, 4, -9)\) and \(C(5, 17, 15)\) | A1 | 2.2a |
| (4) |
Notes
Notes: Accept alternative vector forms throughout (including use of row vectors), and accept coordinates as vectors. For cross products with no working shown accept two out three correct entries to imply the method.
B1: Deduces the correct vector for \(A\) to \(l\) in parametric form (either direction).
M1: Makes the key step of attempting the general distance from \(A\) to \(l\), sets equal to 15 and attempts to solve for \(\lambda\)
A1: Correct values for \(\lambda\)
A1: Both coordinates correct
| Scheme | Marks | AO |
|---|---|---|
| E.g. \(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}-5\\ -2\\ -14\end{pmatrix} \times \begin{pmatrix}2\\ 11\\ 10\end{pmatrix} =\) or \(\overrightarrow{AB} \times \overrightarrow{BC} = \begin{pmatrix}-5\\ -2\\ -14\end{pmatrix} \times \begin{pmatrix}7\\ 13\\ 24\end{pmatrix}\ \left(= \pm\begin{pmatrix}134\\ 22\\ -51\end{pmatrix}\right)\) | M1 | 3.1a |
| \(\begin{pmatrix}134\\ 22\\ -51\end{pmatrix} \bullet \begin{pmatrix}3\\ 6\\ 5\end{pmatrix} = 279\) | M1 | 1.1b |
| \(134x + 22y - 51z = 279\) | A1 | 1.1b |
| (3) |
Notes
M1: Realises that the normal to the plane is required and applies the vector product to 2 vectors in the plane. (Many possibilities are possible, the two most common are in the scheme.)
M1: Attempts scalar product using a point in the plane and their normal vector.
A1: Correct equation (allow any multiple).
(b) Alt
| Scheme | Marks | AO |
|---|---|---|
| Plane is e.g. \(\mathbf{r} = \begin{pmatrix}3\\ 6\\ 5\end{pmatrix} + \lambda\begin{pmatrix}7\\ 13\\ 24\end{pmatrix} + \mu\begin{pmatrix}5\\ 2\\ 14\end{pmatrix} \Rightarrow \left\{\begin{aligned}x &= 3 + 7\lambda + 5\mu\\ y &= 6 + 13\lambda + 2\mu\\ z &= 5 + 24\lambda + 14\mu\end{aligned}\right.\) | M1 | 1.1b |
| \(\Rightarrow \left\{\begin{aligned}5y - 2x &= 24 + 51\lambda\\ 7y - z &= 37 + 67\lambda\end{aligned}\right. \Rightarrow 51(7y - z - 37) - 67(5y - 2x - 24) = 0\) | M1 | 3.1a |
| \(134x + 22y - 51z = 279\) | A1 | 1.1b |
| (3) |
Notes
(Corrected from the printed mark scheme: the third equation of the Alt is printed as \(x = 5 + 24\lambda + 14\mu\); it is the \(z\) equation, \(z = 5 + 24\lambda + 14\mu\).)
| Scheme | Marks | AO |
|---|---|---|
| E.g. \(\overrightarrow{DA} = \begin{pmatrix}3\\ 6\\ 5\end{pmatrix} - \begin{pmatrix}-2\\ 1\\ \alpha\end{pmatrix} = \begin{pmatrix}5\\ 5\\ 5 - \alpha\end{pmatrix}\) or \(\overrightarrow{BD} = \begin{pmatrix}-2\\ 1\\ \alpha\end{pmatrix} - \begin{pmatrix}-2\\ 4\\ -9\end{pmatrix} = \begin{pmatrix}0\\ -3\\ 9 + \alpha\end{pmatrix}\) | M1 | 3.1a |
| \(\begin{pmatrix}134\\ 22\\ -51\end{pmatrix} \bullet \begin{pmatrix}5\\ 5\\ 5 - \alpha\end{pmatrix} = \ldots\) or \(\begin{pmatrix}7\\ 13\\ 24\end{pmatrix} \bullet \left(\begin{pmatrix}5\\ 2\\ 14\end{pmatrix} \times \begin{pmatrix}0\\ -3\\ 9 + \alpha\end{pmatrix}\right) = \ldots\) | dM1 | 1.1b |
| \(\dfrac{1}{6}|525 + 51\alpha| = \pm 147 \Rightarrow \alpha = \ldots\) | M1 | 3.1a |
| \(\alpha = 7, -\dfrac{469}{17}\) | A1 | 1.1b |
| (4) |
Notes
M1: Finds an appropriate vector using the coordinates of \(D\) e.g. joining \(D\) to \(A\) or \(B\).
dM1: Forms and evaluates an appropriate scalar triple product.
M1: Realises that \(\pm 147\) is possible for the value of 1/6 of the triple product and attempts to solve to obtain 2 distinct values for \(\alpha\). Allow this mark if one value is later rejected (or crossed out).
A1: Correct values, but A0 if one is later rejected.
| Scheme | Marks | AO |
|---|---|---|
| E.g. Vector connecting \(A\) and \(l\) is \(\pm\begin{pmatrix}9\\ 24\\ 34\end{pmatrix}\) or \(D\) and \(l\) is \(\pm\begin{pmatrix}0\\ -3\\ 16\end{pmatrix}\) | B1ft | 2.2a |
| \(\overrightarrow{AD} \times \overrightarrow{BC} = \begin{pmatrix}-5\\ -5\\ 2\end{pmatrix} \times \begin{pmatrix}7\\ 13\\ 24\end{pmatrix} = \begin{pmatrix}-146\\ 134\\ -30\end{pmatrix}\) | M1 | 1.1b |
| \(d = \dfrac{\left|\begin{pmatrix}-146\\ 134\\ -30\end{pmatrix} \bullet \begin{pmatrix}9\\ 24\\ 34\end{pmatrix}\right|}{\sqrt{146^2 + 134^2 + 30^2}} = \ldots\) | M1 | 3.1a |
| \(d =\) awrt 4.4 | A1 | 1.1b |
| (4) | ||
| (15 marks) |
Notes
B1ft: Deduces a correct vector joining \(A\) or \(D\) and \(l\) (follow through their \(B\) or \(C\) if used). Accept in either direction (do not be concerned about the labelling). Many other vectors are possible here.
M1: Calculates the vector product between the directions.
M1: Fully correct strategy for the shortest distance.
A1: Awrt 4.4
(d) Alt
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{AD} \times \overrightarrow{BC} = \begin{pmatrix}-5\\ -5\\ 2\end{pmatrix} \times \begin{pmatrix}7\\ 13\\ 24\end{pmatrix} = \begin{pmatrix}-146\\ 134\\ -30\end{pmatrix}\) and need distance between planes \(\mathbf{r} \bullet \begin{pmatrix}-146\\ 134\\ -30\end{pmatrix} = \begin{pmatrix}3\\ 6\\ 5\end{pmatrix} \bullet \begin{pmatrix}-146\\ 134\\ -30\end{pmatrix} = 216\) and \(\mathbf{r} \bullet \begin{pmatrix}-146\\ 134\\ -30\end{pmatrix} = \begin{pmatrix}-2\\ 4\\ -9\end{pmatrix} \bullet \begin{pmatrix}-146\\ 134\\ -30\end{pmatrix} = 1098\) (one correct B1ft on their normal vectors, M1 attempt both) | B1ft M1 | 2.2a 1.1b |
| \(d = \dfrac{1098 - 216}{\sqrt{146^2 + 134^2 + 30^2}}\) | M1 | 3.1a |
| \(d =\) awrt 4.4 | A1 | 1.1b |
| (4) |
Notes
(Corrected from the printed mark scheme: in the two plane equations of this Alt the normal vector is printed as \(\begin{pmatrix}-146\\ 134\\ 30\end{pmatrix}\); it is \(\overrightarrow{AD} \times \overrightarrow{BC} = \begin{pmatrix}-146\\ 134\\ -30\end{pmatrix}\), which gives the printed values 216 and 1098.)
(d) Alt 2
| Scheme | Marks | AO |
|---|---|---|
| E.g. \(\mathbf{r} = \begin{pmatrix}9\\ 24\\ 34\end{pmatrix} + \lambda\begin{pmatrix}7\\ 13\\ 24\end{pmatrix} + \mu\begin{pmatrix}5\\ 5\\ -2\end{pmatrix}\) Sets up plane perpendicular to both direction vectors, passing through point represented by vector that joins the two lines. (Translates e.g. point \(A\) to the origin and considers parallel planes.) | B1ft | 2.2a |
| \(\mathbf{r} \bullet \begin{pmatrix}9 + 7\lambda + 5\mu\\ 24 + 13\lambda + 5\mu\\ 34 + 24\lambda - 2\mu\end{pmatrix} \bullet \begin{pmatrix}7\\ 13\\ 24\end{pmatrix} = 0 \Rightarrow a + b\lambda + c\mu = 0\) and \(\mathbf{r} \bullet \begin{pmatrix}9 + 7\lambda + 5\mu\\ 24 + 13\lambda + 5\mu\\ 34 + 24\lambda - 2\mu\end{pmatrix} \bullet \begin{pmatrix}5\\ 5\\ -2\end{pmatrix} = 0 \Rightarrow d + e\lambda + f\mu = 0\) Attempts scalar products of the plane with both direction vectors (finds point on second plane normal the origin). | M1 | 1.1b |
| \(\lambda = \ldots, \mu = \ldots \Rightarrow\) \(d = \sqrt{\left(9 + 7\text{“}\lambda\text{”} + 5\text{“}\mu\text{”}\right)^2 + \left(24 + 13\text{“}\lambda\text{”} + 5\text{“}\mu\text{”}\right)^2 + \left(34 + 24\text{“}\lambda\text{”} - 2\text{“}\mu\text{”}\right)^2} = \ldots\) Solves for \(\lambda\) and \(\mu\) and proceeds to find the distance using their values. | M1 | 3.1a |
| \(d =\) awrt 4.4 | A1 | 1.1b |
| (4) |
Notes
(Corrected from the printed mark scheme: the third component of the vector in Alt 2 is printed as \(34 + 24\lambda = 2\mu\); it is \(34 + 24\lambda - 2\mu\), as in the distance formula below it.)