AS June 2023 Q5
5. The points \(A\), \(B\) and \(C\) are the vertices of a triangle.
Given that
- \(\overrightarrow{AB} = \begin{pmatrix}p\\ 4\\ 6\end{pmatrix}\) and \(\overrightarrow{AC} = \begin{pmatrix}q\\ 4\\ 5\end{pmatrix}\) where \(p\) and \(q\) are constants
- \(\overrightarrow{AB} \times \overrightarrow{AC}\) is parallel to \(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\)
| Scheme | Marks | AO |
|---|---|---|
| A complete method to find the values for \(p\) and \(q\). There must be an attempt to find \(\overrightarrow{AB} \times \overrightarrow{AC}\) and set equal to a multiple of \(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\) and forms and solve two simultaneous equations for \(p\) and \(q\) | M1 | 3.1a |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ p & 4 & 6\\ q & 4 & 5\end{vmatrix} = (20 - 24)\mathbf{i} - (5p - 6q)\mathbf{j} + (4p - 4q)\mathbf{k}\) | M1 A1 | 1.1b 1.1b |
| \(\overrightarrow{AB} \times \overrightarrow{AC} = -2(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})\) or \(\begin{pmatrix}2\\ 3\\ 4\end{pmatrix} \times \left(\overrightarrow{AB} \times \overrightarrow{AC}\right) = \mathbf{0}\) | B1 | 2.2a |
| \(-(5p - 6q) = -2 \times 3\) and \((4p - 4q) = -2 \times 4\) \(\{\text{alt}: 32p - 36q = 0, 8p - 8q + 16 = 0, 12q - 10p + 12 = 0\}\) | M1 | 3.1a |
| Solves \(5p - 6q = 6\) and \(4p - 4q = -8\) To find values for \(p\) and \(q\) | dM1 | 1.1b |
| \(p = -18,\ q = -16\) | A1 | 1.1b |
| (7) |
Notes
M1: A complete method to find the values for \(p\) and \(q\). There must be an attempt to find \(\overrightarrow{AB} \times \overrightarrow{AC}\) and sets equal to a multiple (not equal to 1) of \(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\) (allowing minor slips) to form and solve two simultaneous equations for \(p\) and \(q\). This is for the overall approach, so may be scored if the method for the cross product is incorrect, as long as an attempt has been made.
M1: A complete method to find the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\). Form should be correct \((\ldots)\mathbf{i} - (\ldots)\mathbf{j} + (\ldots)\mathbf{k}\) with correct coefficients in each bracket, though allow slips in sign in these.
A1: Correct cross product.
B1: Deduces that \(\overrightarrow{AB} \times \overrightarrow{AC} = -2(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})\) May be implied by their working. Alternatively, a correct cross product statement for the parallel vectors.
M1: Forms simultaneous equations by setting the \(\mathbf{j}\) component of the cross product to \(k\) times 3 and setting the \(\mathbf{k}\) component of the cross product to \(k\) times 4, where \(k \neq 1\) is an attempt at a scale factor for the parallel vectors. Alternatively, attempts the cross product and equates at least two components to 0 for form simultaneous equations - the attempt at the cross product may be incorrect for this mark as long as the intent is clear.
dM1: Solves their simultaneous equations to find a value for \(p\) or \(q\). Accept for any \(p\) and \(q\) following forming simultaneous equations (do not be concerned about the method used). Dependent on the previous method mark
A1: Correct values for \(p\) and \(q\).
Alt
| Scheme | Marks | AO |
|---|---|---|
| A complete method to find the values for \(p\) and \(q\). Attempts scalar product of both \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) with \(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\) sets equal to 0 and solves both equations. | M1 | 3.1a |
| \(\begin{pmatrix}p\\ 4\\ 6\end{pmatrix} \cdot \begin{pmatrix}2\\ 3\\ 4\end{pmatrix} = 2p + 12 + 24\) or \(\begin{pmatrix}q\\ 4\\ 5\end{pmatrix} \cdot \begin{pmatrix}2\\ 3\\ 4\end{pmatrix} = 2q + 12 + 20\) | M1 A1 | 1.1b 1.1b |
| \(\begin{pmatrix}p\\ 4\\ 6\end{pmatrix} \cdot \begin{pmatrix}2\\ 3\\ 4\end{pmatrix} = 0 \Rightarrow 2p + 36 = 0\) or \(\begin{pmatrix}q\\ 4\\ 5\end{pmatrix} \cdot \begin{pmatrix}2\\ 3\\ 4\end{pmatrix} = 0 \Rightarrow 2q + 32 = 0\) | B1 | 2.2a |
| \(\Rightarrow p = \ldots\) or \(\Rightarrow q = \ldots\) | M1 | 1.1b |
| \(\Rightarrow p = \ldots\) and \(\Rightarrow q = \ldots\) | dM1 | 3.1a |
| \(p = -18,\ q = -16\) | A1 | 1.1b |
| (7) |
M1: A complete method to find the values for \(p\) and \(q\). There must be an attempt to apply the scalar product of both \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) with \(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\) then set each equal to 0 and solves both equations. (Corrected from the printed mark scheme: printed as “of both \(\overrightarrow{AB}\) and \(\overrightarrow{AB}\)”.)
M1: A correct method for one of the scalar products.
A1: Correct expression for one of the scalar products.
B1: Forms at least one correct equation for \(p\) or \(q\) by setting the scalar product equal to zero and producing a correct linear equation for \(p\) or \(q\)
M1: Proceeds to find at least one of the two values.
dM1: Solves to find a value for \(p\) and for \(q\). Dependent on the previous method mark
A1: Correct values for \(p\) and \(q\).
(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)
| Scheme | Marks | AO |
|---|---|---|
| E.g. Area \(= \dfrac{1}{2}\left|\begin{pmatrix}-4\\ -6\\ -8\end{pmatrix}\right| = \dfrac{1}{2}\left[\sqrt{(-4)^2 + (-6)^2 + (-8)^2}\right] = \ldots\) | M1 | 1.1b |
| Area \(= \sqrt{29}\) or \(\dfrac{1}{2}\sqrt{116}\) | A1 | 1.1b |
| (2) | ||
| (9 marks) |
Notes
M1: A complete method for the area of the triangle. E.g. may use area \(= \dfrac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|\) with their cross product from (a), or starting from scratch, and correct attempt at the modulus. Alternative may find angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) using scalar product, \(\cos\theta = \dfrac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{\left|\overrightarrow{AB}\right|\left|\overrightarrow{AC}\right|}\), and then Area \(= \dfrac{1}{2}\left|\overrightarrow{AB}\right| \times \left|\overrightarrow{AC}\right|\sin\theta\).
A1: Correct exact area from a correct \(\overrightarrow{AB} \times \overrightarrow{AC}\) (or correct \(p\) and \(q\) in the alt) (oe). There may be variations in method used, but if an incorrect vectors is used in the process score A0.
(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)