AS June 2024 Q3
3. Vectors \(\mathbf{u}\) and \(\mathbf{v}\) are given by
\[\mathbf{u} = 5\mathbf{i} + 4\mathbf{j} - 3\mathbf{k} \quad \text{and} \quad \mathbf{v} = a\mathbf{i} - 6\mathbf{j} + 2\mathbf{k}\]where \(a\) is a constant.
Given that
- \(\overrightarrow{AB} = 2\mathbf{u}\)
- \(\overrightarrow{AC} = \mathbf{v}\)
- the area of triangle \(ABC\) is 15
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 5 & 4 & -3\\ a & -6 & 2\end{vmatrix} = \ldots\mathbf{i} - \ldots\mathbf{j} + \ldots\mathbf{k}\) | M1 | 1.1b |
| \(= -10\mathbf{i} - (10 + 3a)\mathbf{j} - (30 + 4a)\mathbf{k}\) or e.g. \(\begin{pmatrix}-10\\ -10 - 3a\\ -30 - 4a\end{pmatrix}\) | A1 | 1.1b |
| (2) |
Notes
M1: Evidence of a correct method for the vector product. May be implied by two out of three correct components if no method shown or by correct work for 1 component.
Attempting \(\mathbf{v} \times \mathbf{u}\) scores M0
A1: Correct answer.
Allow equivalent expressions and condone e.g. \(-10\mathbf{I} + (-10 - 3a)\mathbf{J} + (-30 - 4a)\mathbf{K}\)
Award the mark once a correct vector is seen and isw if necessary.
Must be a vector not coordinates.
| Scheme | Marks | AO |
|---|---|---|
| \(\text{Area} = \dfrac{1}{2}\left|2\mathbf{u} \times \mathbf{v}\right| = 15 \Rightarrow \left|-10\mathbf{i} - (10 + 3a)\mathbf{j} - (30 + 4a)\mathbf{k}\right| = 15\) Or e.g. \(\text{Area} = \dfrac{1}{2}\left|2\mathbf{u} \times \mathbf{v}\right| = \dfrac{1}{2}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 10 & 8 & -6\\ a & -6 & 2\end{vmatrix} = \dfrac{1}{2}\left|\begin{pmatrix}-20\\ -20 - 6a\\ -60 - 8a\end{pmatrix}\right| = 15\) | M1 | 3.1a |
| \(\Rightarrow 10^2 + (10 + 3a)^2 + (30 + 4a)^2 = 15^2\) \(\Rightarrow 25a^2 + 300a + 875 = 0\ (\Rightarrow a^2 + 12a + 35 = 0)\) or e.g. \(20^2 + (20 + 6a)^2 + (60 + 8a)^2 = 30^2\) \(\Rightarrow 400 + 400 + 240a + 36a^2 + 3600 + 960a + 64a^2 = 30^2\) \(\Rightarrow 100a^2 + 1200a + 3500 = 0\ (\Rightarrow a^2 + 12a + 35 = 0)\) | M1 | 1.1b |
| \(\Rightarrow (a + 5)(a + 7) = 0 \Rightarrow a = \ldots\) | ddM1 | 1.1b |
| \(a = -7\) or \(-5\) | A1 cso | 2.2a |
| (4) | ||
| (6 marks) |
Notes
M1: Applies a correct method for the area of the triangle and scalar multiple property of vector product to set up a vector equation in \(a\) using the vector product.
The “1/2” must be seen or implied.
This may be implied by later work e.g. \(\dfrac{1}{2}\sqrt{20^2 + (20 + 6a)^2 + (60 + 8a)^2} = 15\)
M1: Applies the modulus correctly and expands correctly to reach a quadratic in \(a\).
Note that \(\dfrac{1}{4}\left(20^2 + (20 + 6a)^2 + (60 + 8a)^2\right) = 15\) scores M0 (must square both sides)
ddM1: Solves their quadratic by any suitable means including a calculator to obtain at least one real root.
Depends on both previous method marks.
A1cso: Both correct values following correct work e.g. do not condone sign errors in the vector product if they fortuitously lead to the correct answers.
Note there may be more convoluted methods for the area e.g. using “\(\frac{1}{2}ab\sin C\)” and the scalar product:
\(\dfrac{1}{2}\left|\overrightarrow{AB}\right|\left|\overrightarrow{AC}\right|\sin A = 15 \Rightarrow \dfrac{1}{2}10\sqrt{2}\sqrt{a^2 + 40}\sin A = 15\)
\(\mathbf{u}.\mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos A \Rightarrow 5a - 30 = 5\sqrt{2}\sqrt{a^2 + 40}\cos A\)
\(\Rightarrow \sin A = \sqrt{\dfrac{a^2 + 12a + 44}{2(a^2 + 40)}} \Rightarrow 5\sqrt{2}\sqrt{a^2 + 40}\sqrt{\dfrac{a^2 + 12a + 44}{2(a^2 + 40)}} = 15\)
Score M1 for a complete correct method to obtain an equation in \(a\) only then e.g.
\(5\sqrt{a^2 + 12a + 44} = 15 \Rightarrow a^2 + 12a + 35 = 0\) etc.
M2 for simplifying to obtain a quadratic in \(a\).
Then as above. Use review if necessary.