C4 June 2014 (R) Q6
6. With respect to a fixed origin, the point \(A\) with position vector \(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\) lies on the line \(l_1\) with equation \[\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + \lambda\begin{pmatrix}0\\2\\-1\end{pmatrix}, \qquad \text{where } \lambda \text{ is a scalar parameter,}\] and the point \(B\) with position vector \(4\mathbf{i} + p\mathbf{j} + 3\mathbf{k}\), where \(p\) is a constant, lies on the line \(l_2\) with equation \[\mathbf{r} = \begin{pmatrix}7\\0\\7\end{pmatrix} + \mu\begin{pmatrix}3\\-5\\4\end{pmatrix}, \qquad \text{where } \mu \text{ is a scalar parameter.}\]
| Scheme | Marks |
|---|---|
| \(l_1: \mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + \lambda\begin{pmatrix}0\\2\\-1\end{pmatrix},\quad l_2: \mathbf{r} = \begin{pmatrix}7\\0\\7\end{pmatrix} + \mu\begin{pmatrix}3\\-5\\4\end{pmatrix},\quad \overrightarrow{OA} = \begin{pmatrix}1\\2\\3\end{pmatrix},\ \overrightarrow{OB} = \begin{pmatrix}4\\p\\3\end{pmatrix}\) \(A\) lies on \(l_1\) and \(B\) lies on \(l_2\) | |
| \(\{B \text{ lies on } l_2 \Rightarrow \mu = -1 \Rightarrow\}\quad p = 5\) \(p = 5\) | B1 |
| (1) |
Notes
B1: \(p = 5\) (Ignore working.)
| Scheme | Marks |
|---|---|
| \(\{l_1 = l_2 \Rightarrow\}\ \left\{\begin{aligned} \mathbf{i}&:\quad 1 = 7 + 3\mu\\ \mathbf{j}&:\quad 2 + 2\lambda = -5\mu\\ \mathbf{k}&:\quad 3 - \lambda = 7 + 4\mu\end{aligned}\right\}\) | |
| e.g. \(\mathbf{i}\): \(7 + 3\mu = 1\) Writes down an equation involving only one parameter. | M1 |
| So, \(\mu = -2\) \(\mu = -2\) | A1 |
| Point of intersection is \(\overrightarrow{OC} = \mathbf{i} + 10\mathbf{j} - \mathbf{k}\) \(\mathbf{i} + 10\mathbf{j} - \mathbf{k}\) | B1 |
Finds \(\lambda = 4\) and either
| B1 |
| (4) |
Alternative Method: Solving \(\mathbf{j}\) and \(\mathbf{k}\) simultaneously gives
| Scheme | Marks |
|---|---|
| \(8 = 14 + 3\mu\) or \(23 + 3\lambda = 35\) Writes down an equation involving only one parameter. | M1 |
| So, \(\mu = -2\) or \(\lambda = 4\) Either \(\mu = -2\) or \(\lambda = 4\) | A1 |
| Point of intersection is \(\overrightarrow{OC} = \mathbf{i} + 10\mathbf{j} - \mathbf{k}\) \(\mathbf{i} + 10\mathbf{j} - \mathbf{k}\) | B1 |
Finds \(\lambda = 4\) and either
| B1 |
| (4) |
Notes
Method 1
M1: Writes down an equation involving only one parameter.
This equation will usually be \(7 + 3\mu = 1\) which is found from equating the \(\mathbf{i}\) components of \(l_1\) and \(l_2\).
A1: Finds \(\mu = -2\)
B1: Point of intersection of \(\mathbf{i} + 10\mathbf{j} - \mathbf{k}\). Allow \((1, 10, -1)\) or \(\begin{pmatrix}1\\10\\-1\end{pmatrix}\).
B1: Finds \(\lambda = 4\) and either
- checks \(\lambda = 4\) and \(\mu = -2\) is true for the third component.
- substitutes \(\mu = -2\) into \(l_1\) to give \(\mathbf{i} + 10\mathbf{j} - \mathbf{k}\) and substitutes \(\lambda = 4\) into \(l_2\) to give \(\mathbf{i} + 10\mathbf{j} - \mathbf{k}\)
Alternative Method
M1: Writes down an equation involving only one parameter.
Solving the \(\mathbf{j}\) and \(\mathbf{k}\) components simultaneously will usually give either \(8 = 14 + 3\mu\) or \(23 + 3\lambda = 35\)
A1: Finds either \(\mu = -2\) or \(\lambda = 4\)
B1: Point of intersection of \(\mathbf{i} + 10\mathbf{j} - \mathbf{k}\). Allow \((1, 10, -1)\) or \(\begin{pmatrix}1\\10\\-1\end{pmatrix}\).
B1: Finds \(\lambda = 4\) and either
- checks \(\mu = -2\) is true for the \(\mathbf{i}\) component.
- substitutes \(\mu = -2\) into \(l_1\) to give \(\mathbf{i} + 10\mathbf{j} - \mathbf{k}\) and substitutes \(\lambda = 4\) into \(l_2\) to give \(\mathbf{i} + 10\mathbf{j} - \mathbf{k}\)
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AC} = \begin{pmatrix}1\\10\\-1\end{pmatrix} - \begin{pmatrix}1\\2\\3\end{pmatrix} = \begin{pmatrix}0\\8\\-4\end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix}1\\10\\-1\end{pmatrix} - \begin{pmatrix}4\\5\\3\end{pmatrix} = \begin{pmatrix}-3\\5\\-4\end{pmatrix}\) An attempt to find both the vectors \(\left(\overrightarrow{AC} \text{ or } \overrightarrow{CA}\right)\) and \(\left(\overrightarrow{BC} \text{ or } \overrightarrow{CB}\right)\). | M1 |
| \(\cos ACB = \dfrac{\overrightarrow{AC} \bullet \overrightarrow{BC}}{\left|\overrightarrow{AC}\right| . \left|\overrightarrow{BC}\right|} = \dfrac{\pm\left(\begin{pmatrix}0\\8\\-4\end{pmatrix} \bullet \begin{pmatrix}-3\\5\\-4\end{pmatrix}\right)}{\sqrt{(0)^2 + (8)^2 + (-4)^2} . \sqrt{(-3)^2 + (5)^2 + (-4)^2}}\) Applies dot product formula between their \(\left(\overrightarrow{AC} \text{ or } \overrightarrow{CA}\right)\) and their \(\left(\overrightarrow{BC} \text{ or } \overrightarrow{CB}\right)\). | M1 |
| \(\left\{\cos ACB = \dfrac{0 + 40 + 16}{\sqrt{80} . \sqrt{50}} = \dfrac{56}{\sqrt{4000}} \Rightarrow\right\}\ ACB = 27.69446\ldots = 27.7\) (3 sf) Anything that rounds to 27.7 | A1 |
| (3) |
Alternative Method 1: Using the direction vectors of Line 1 and Line 2
| Scheme | Marks |
|---|---|
| \(\mathbf{d}_1 = \begin{pmatrix}0\\2\\-1\end{pmatrix},\quad \mathbf{d}_2 = \begin{pmatrix}3\\-5\\4\end{pmatrix}\) | |
| \(\cos\theta = \dfrac{\mathbf{d}_1 \bullet \mathbf{d}_2}{\left|\mathbf{d}_1\right| . \left|\mathbf{d}_2\right|} = \dfrac{\begin{pmatrix}0\\2\\-1\end{pmatrix} \bullet \begin{pmatrix}3\\-5\\4\end{pmatrix}}{\sqrt{(0)^2 + (2)^2 + (-1)^2} . \sqrt{(3)^2 + (-5)^2 + (4)^2}}\) (corrected from the printed mark scheme: \(\mathbf{d}_1 \bullet \mathbf{d}_1\) in the numerator) Applies dot product formula between their \(\mathbf{d}_1\) and \(\mathbf{d}_2\) | M2 |
| \(\left\{\cos\theta = \dfrac{0 - 10 - 4}{\sqrt{5} . \sqrt{50}} = \dfrac{-7\sqrt{10}}{25} \Rightarrow\right\}\ \theta = 152.3054385\ldots\) | |
| Angle \(ACB = 180 - 152.3054385\ldots = 27.69446145\ldots = 27.7\) (3 sf) Anything that rounds to 27.7 | A1 |
| (3) |
Alternative Method 2: The Cosine Rule
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AC} = \begin{pmatrix}1\\10\\-1\end{pmatrix} - \begin{pmatrix}1\\2\\3\end{pmatrix} = \begin{pmatrix}0\\8\\-4\end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix}1\\10\\-1\end{pmatrix} - \begin{pmatrix}4\\5\\3\end{pmatrix} = \begin{pmatrix}-3\\5\\-4\end{pmatrix}\) An attempt to find both the vectors \(\left(\overrightarrow{AC} \text{ or } \overrightarrow{CA}\right)\) and \(\left(\overrightarrow{BC} \text{ or } \overrightarrow{CB}\right)\). | M1 |
| Also \(\overrightarrow{AB} = \begin{pmatrix}4\\5\\3\end{pmatrix} - \begin{pmatrix}1\\2\\3\end{pmatrix} = \begin{pmatrix}3\\3\\0\end{pmatrix}\) | |
| Note \(\left|\overrightarrow{AC}\right| = \sqrt{80},\ \left|\overrightarrow{BC}\right| = \sqrt{50}\) and \(\left|\overrightarrow{AB}\right| = \sqrt{18}\) | |
| \(\left(\sqrt{18}\right)^2 = \left(\sqrt{80}\right)^2 + \left(\sqrt{50}\right)^2 - 2\left(\sqrt{80}\right)\left(\sqrt{50}\right)\cos\theta\) Applies the cosine rule the correct way round. | M1 oe |
| \(\left\{\cos\theta = \dfrac{7\sqrt{10}}{25}\right\} \Rightarrow \theta = 27.69446145\ldots = 27.7\) (3 sf) Anything that rounds to 27.7 | A1 |
| (3) |
Alternative Method 3: Vector Cross Product
Only apply this scheme if it is clear that a candidate is applying a vector cross product method.
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AC} = \begin{pmatrix}1\\10\\-1\end{pmatrix} - \begin{pmatrix}1\\2\\3\end{pmatrix} = \begin{pmatrix}0\\8\\-4\end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix}1\\10\\-1\end{pmatrix} - \begin{pmatrix}4\\5\\3\end{pmatrix} = \begin{pmatrix}-3\\5\\-4\end{pmatrix}\) An attempt to find both the vectors \(\left(\overrightarrow{AC} \text{ or } \overrightarrow{CA}\right)\) and \(\left(\overrightarrow{BC} \text{ or } \overrightarrow{CB}\right)\). | M1 |
| \(\overrightarrow{AC} \times \overrightarrow{BC} = \begin{pmatrix}0\\8\\-4\end{pmatrix} \times \begin{pmatrix}-3\\5\\-4\end{pmatrix} = \left\{\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0 & 8 & -4\\ -3 & 5 & -4\end{vmatrix} = -12\mathbf{i} + 12\mathbf{j} + 24\mathbf{k}\right\}\) (corrected from the printed mark scheme: \(24\mathbf{i} + 12\mathbf{j} + 24\mathbf{k}\)) \(\sin ACB = \dfrac{\sqrt{(-12)^2 + (12)^2 + (24)^2}}{\sqrt{(0)^2 + (8)^2 + (-4)^2} . \sqrt{(-3)^2 + (5)^2 + (-4)^2}}\) (corrected from the printed mark scheme: \((24)^2 + (12)^2 + (12)^2\)) Full method for applying the vector cross product formula between their \(\left(\overrightarrow{AC} \text{ or } \overrightarrow{CA}\right)\) and their \(\left(\overrightarrow{BC} \text{ or } \overrightarrow{CB}\right)\). | M1 |
| \(\left\{\sin ACB = \dfrac{\sqrt{864}}{\sqrt{80} . \sqrt{50}} = \dfrac{3\sqrt{15}}{25} \Rightarrow\right\}\ \theta = 27.69446145\ldots = 27.7\) (3 sf) Anything that rounds to 27.7 | A1 |
| (3) |
Notes
M1: An attempt to find both the vectors \(\left(\overrightarrow{AC} \text{ or } \overrightarrow{CA}\right)\) and \(\left(\overrightarrow{BC} \text{ or } \overrightarrow{CB}\right)\) by subtracting.
M1: Applies dot product formula between their \(\left(\overrightarrow{AC} \text{ or } \overrightarrow{CA}\right)\) and their \(\left(\overrightarrow{BC} \text{ or } \overrightarrow{CB}\right)\).
A1: anything that rounds to 27.7
Note: An answer of 0.48336… in radians without the correct answer in degrees is A0.
Note: Some candidates will apply the dot product formula between vectors which are the wrong way round and achieve 152.3054385…°. If they give the acute equivalent of awrt 27.7 then award A1.
| Scheme | Marks |
|---|---|
| \(\text{Area } ACB = \dfrac{1}{2}\left(\sqrt{80}\right)\left(\sqrt{50}\right)\sin 27.69446\ldots^\circ = 14.696888\ldots\) See notes Anything that rounds to 14.7 | M1 A1 |
| (2) | |
| (10 marks) |
Notes
M1: \(\dfrac{1}{2}(\text{their length } AC)(\text{their length } BC)\sin(\text{their } 27.7^\circ \text{ from part (c)})\)
A1: anything that rounds to 14.7. Also allow \(6\sqrt{6}\).
Note: \(\text{Area } ACB = \dfrac{1}{2}\left(\sqrt{80}\right)\left(\sqrt{50}\right)\sin(152.3054385\ldots^\circ) = \text{awrt } 14.7\) is M1A1.