C4 June 2016 Q8
8. With respect to a fixed origin \(O\), the line \(l_1\) is given by the equation \[\mathbf{r} = \begin{pmatrix}8\\1\\-3\end{pmatrix} + \mu\begin{pmatrix}-5\\4\\3\end{pmatrix}\] where \(\mu\) is a scalar parameter.
The point \(A\) lies on \(l_1\) where \(\mu = 1\)
The point \(P\) has position vector \(\begin{pmatrix}1\\5\\2\end{pmatrix}\).
The line \(l_2\) passes through the point \(P\) and is parallel to the line \(l_1\)
Give your answer in the form \(k\sqrt{2}\), where \(k\) is a constant to be determined. (2)
The acute angle between \(AP\) and \(l_2\) is \(\theta\).
A point \(E\) lies on the line \(l_2\)
Given that \(AP = PE\),
| Scheme | Marks |
|---|---|
| \(l_1: \mathbf{r} = \begin{pmatrix}8\\1\\-3\end{pmatrix} + \mu\begin{pmatrix}-5\\4\\3\end{pmatrix}\) So \(\mathbf{d}_1 = \begin{pmatrix}-5\\4\\3\end{pmatrix}\). \(\overline{OA}\) occurs when \(\mu = 1\). \(\overrightarrow{OP} = \begin{pmatrix}1\\5\\2\end{pmatrix}\) | |
| \(A(3, 5, 0)\) \((3, 5, 0)\) | B1 |
| (1) |
Notes
B1: Allow \(A(3, 5, 0)\) or \(3\mathbf{i} + 5\mathbf{j}\) or \(3\mathbf{i} + 5\mathbf{j} + 0\mathbf{k}\) or \(\begin{pmatrix}3\\5\\0\end{pmatrix}\) or benefit of the doubt \(\begin{matrix}3\\5\\0\end{matrix}\)
| Scheme | Marks |
|---|---|
| \(\{l_2:\}\ \mathbf{r} = \begin{pmatrix}1\\5\\2\end{pmatrix} + \lambda\begin{pmatrix}-5\\4\\3\end{pmatrix}\) \(\mathbf{a} + \lambda\mathbf{d}\) or \(\mathbf{a} + \mu\mathbf{d}\), \(\mathbf{a} + t\mathbf{d}\), \(\mathbf{a} \neq 0\), \(\mathbf{d} \neq 0\) with either \(\mathbf{a} = \mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\) or \(\mathbf{d} = -5\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\), or a multiple of \(-5\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\) Correct vector equation using \(\mathbf{r} =\) or \(l =\) or \(l_2 =\) | M1 A1 |
| \(\mathbf{d}_2\) is the direction vector of \(l_2\) Do not allow \(l_2:\) or \(l_2 \to\) or \(l_1 =\) for the A1 mark. | |
| (2) |
Notes
A1: Correct vector equation using \(\mathbf{r} =\) or \(l =\) or \(l_2 =\) or Line 2 =
i.e. Writing \(\mathbf{r} = \begin{pmatrix}1\\5\\2\end{pmatrix} + \lambda\begin{pmatrix}-5\\4\\3\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}1\\5\\2\end{pmatrix} + \lambda\mathbf{d}\), where \(\mathbf{d}\) is a multiple of \(\begin{pmatrix}-5\\4\\3\end{pmatrix}\).
Note: Allow the use of parameters \(\mu\) or \(t\) instead of \(\lambda\).
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = \begin{pmatrix}1\\5\\2\end{pmatrix} - \begin{pmatrix}3\\5\\0\end{pmatrix} = \begin{pmatrix}-2\\0\\2\end{pmatrix}\) | |
| \(AP = \sqrt{(-2)^2 + (0)^2 + (2)^2} = \sqrt{8} = 2\sqrt{2}\) Full method for finding \(AP\) \(2\sqrt{2}\) | M1 A1 |
| (2) |
Notes
M1: Finds the difference between \(\overrightarrow{OP}\) and their \(\overrightarrow{OA}\) and applies Pythagoras to the result to find \(AP\)
Note: Allow M1A1 for \(\begin{pmatrix}2\\0\\2\end{pmatrix}\) leading to \(AP = \sqrt{(2)^2 + (0)^2 + (2)^2} = \sqrt{8} = 2\sqrt{2}\).
| Scheme | Marks |
|---|---|
| So \(\overrightarrow{AP} = \begin{pmatrix}-2\\0\\2\end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix}-5\\4\\3\end{pmatrix} \Rightarrow \begin{pmatrix}-2\\0\\2\end{pmatrix} \bullet \begin{pmatrix}-5\\4\\3\end{pmatrix}\) Realisation that the dot product is required between \(\left(\overrightarrow{AP} \text{ or } \overrightarrow{PA}\right)\) and \(\pm K\mathbf{d}_2\) or \(\pm K\mathbf{d}_1\) | M1 |
| \(\{\cos\theta =\}\ \dfrac{\overrightarrow{AP} \bullet \mathbf{d}_2}{\left|\overrightarrow{AP}\right|\left|\mathbf{d}_2\right|} = \dfrac{\pm\left(\begin{pmatrix}-2\\0\\2\end{pmatrix} \bullet \begin{pmatrix}-5\\4\\3\end{pmatrix}\right)}{\sqrt{(-2)^2 + (0)^2 + (2)^2}\,.\sqrt{(-5)^2 + (4)^2 + (3)^2}}\) dependent on the previous M mark. Applies dot product formula between their \(\left(\overrightarrow{AP} \text{ or } \overrightarrow{PA}\right)\) and \(\pm K\mathbf{d}_2\) or \(\pm K\mathbf{d}_1\) | dM1 |
| \(\{\cos\theta\} = \dfrac{\pm(10 + 0 + 6)}{\sqrt{8}\,.\sqrt{50}} = \underline{\dfrac{4}{5}}\) \(\{\cos\theta\} = \dfrac{4}{5}\) or 0.8 or \(\dfrac{8}{10}\) or \(\dfrac{16}{20}\) | A1 cso |
| (3) |
Notes
Note: For both the M1 and dM1 marks \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) must be the vector used in part (c) or the difference \(\overrightarrow{OP}\) and their \(\overrightarrow{OA}\) from part (a).
Note: Applying the dot product formula correctly without \(\cos\theta\) as the subject is fine for M1dM1
Note: Evaluating the dot product (i.e. \((-2)(-5) + (0)(4) + (2)(3)\)) is not required for M1 and dM1 marks.
Note: In part (d) allow one slip in writing \(\overrightarrow{AP}\) and \(\mathbf{d}_2\)
Note: \(\cos\theta = \dfrac{-10 + 0 - 6}{\sqrt{8}\,.\sqrt{50}} = -\dfrac{4}{5}\) followed by \(\cos\theta = \dfrac{4}{5}\) is fine for A1 cso
Note: Give M1dM1A1 for \(\{\cos\theta =\} = \dfrac{\begin{pmatrix}-2\\0\\2\end{pmatrix} \bullet \begin{pmatrix}-10\\8\\6\end{pmatrix}}{\sqrt{8}\,.10\sqrt{2}} = \dfrac{20 + 12}{40} = \dfrac{4}{5}\)
Note: Allow final A1 (ignore subsequent working) for \(\cos\theta = 0.8\) followed by \(36.869\ldots^\circ\)
Alternative Method: Vector Cross Product
Only apply this scheme if it is clear that a candidate is applying a vector cross product method.
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AP} \times \mathbf{d}_2 = \begin{pmatrix}-2\\0\\2\end{pmatrix} \times \begin{pmatrix}-5\\4\\3\end{pmatrix} = \left\{\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 0 & 2 \\ -5 & 4 & 3 \end{vmatrix} = -8\mathbf{i} - 4\mathbf{j} - 8\mathbf{k}\right\}\) Realisation that the vector cross product is required between their \(\left(\overrightarrow{AP} \text{ or } \overrightarrow{PA}\right)\) and \(\pm K\mathbf{d}_2\) or \(\pm K\mathbf{d}_1\) | M1 |
| \(\sin\theta = \dfrac{\sqrt{(-8)^2 + (-4)^2 + (-8)^2}}{\sqrt{(-2)^2 + (0)^2 + (2)^2}\,.\sqrt{(-5)^2 + (4)^2 + (3)^2}}\) Applies the vector product formula between their \(\left(\overrightarrow{AP} \text{ or } \overrightarrow{PA}\right)\) and \(\pm K\mathbf{d}_2\) or \(\pm K\mathbf{d}_1\) | dM1 |
| \(\sin\theta = \dfrac{12}{\sqrt{8}\,.\sqrt{50}} = \dfrac{3}{5} \Rightarrow \underline{\cos\theta = \dfrac{4}{5}}\) \(\cos\theta = \dfrac{4}{5}\) or 0.8 or \(\dfrac{8}{10}\) or \(\dfrac{16}{20}\) | A1 |
| Scheme | Marks |
|---|---|
| \(\{\text{Area } APE =\}\ \dfrac{1}{2}(\text{their } 2\sqrt{2})^2\sin\theta\) \(\dfrac{1}{2}(\text{their } 2\sqrt{2})^2\sin\theta\) or \(\dfrac{1}{2}(\text{their } 2\sqrt{2})^2\sin(\text{their } \theta)\) | M1 |
| \(= 2.4\) 2.4 or \(\dfrac{12}{5}\) or \(\dfrac{24}{10}\) or awrt 2.40 | A1 |
| (2) |
Notes
Note: Allow M1;A1 for \(\dfrac{1}{2}(2\sqrt{2})^2\sin(36.869\ldots^\circ)\) or \(\dfrac{1}{2}(2\sqrt{2})^2\sin(180^\circ - 36.869\ldots^\circ)\); = awrt 2.40
Note: Candidates must use their \(\theta\) from part (d) or apply a correct method of finding their \(\sin\theta = \dfrac{3}{5}\) from their \(\cos\theta = \dfrac{4}{5}\)
| Scheme | Marks |
|---|---|
| \(\overrightarrow{PE} = (-5\lambda)\mathbf{i} + (4\lambda)\mathbf{j} + (3\lambda)\mathbf{k}\) and \(PE\) = their \(2\sqrt{2}\) from part (c) | |
| \(\left\{PE^2 =\right\}\ (-5\lambda)^2 + (4\lambda)^2 + (3\lambda)^2 = (\text{their } 2\sqrt{2})^2\) This mark can be implied. | M1 |
| \(\left\{\Rightarrow 50\lambda^2 = 8 \Rightarrow \lambda^2 = \dfrac{4}{25} \Rightarrow\right\}\ \lambda = \pm\dfrac{2}{5}\) Either \(\lambda = \dfrac{2}{5}\) or \(\lambda = -\dfrac{2}{5}\) | A1 |
| \(l_2: \mathbf{r} = \begin{pmatrix}1\\5\\2\end{pmatrix} \pm \dfrac{2}{5}\begin{pmatrix}-5\\4\\3\end{pmatrix}\) dependent on the previous M mark Substitutes at least one of their values of \(\lambda\) into \(l_2\). | dM1 |
| \(\left\{\overrightarrow{OE}\right\} = \begin{pmatrix}3\\\frac{17}{5}\\\frac{4}{5}\end{pmatrix} \text{ or } \begin{pmatrix}3\\3.4\\0.8\end{pmatrix},\ \left\{\overrightarrow{OE}\right\} = \begin{pmatrix}-1\\\frac{33}{5}\\\frac{16}{5}\end{pmatrix} \text{ or } \begin{pmatrix}-1\\6.6\\3.2\end{pmatrix}\) At least one set of coordinates are correct. Both sets of coordinates are correct. | A1 A1 |
| (5) | |
| (15 marks) |
Notes
Note: Allow the first M1A1 for deducing \(\lambda = \dfrac{2}{5}\) or \(\lambda = -\dfrac{2}{5}\) from no incorrect working
SC: Allow special case 1st M1 for \(\lambda = 2.5\) from comparing lengths or from no working
Note: Give 1st M1 for \(\sqrt{(-5\lambda)^2 + (4\lambda)^2 + (3\lambda)^2} = (\text{their } 2\sqrt{2})\)
Note: Give 1st M0 for \((-5\lambda)^2 + (4\lambda)^2 + (3\lambda)^2 = (\text{their } 2\sqrt{2})\) or equivalent
Note: Give 1st M1 for \(\lambda = \dfrac{\text{their } AP = \text{"}2\sqrt{2}\text{"}}{\sqrt{(-5)^2 + (4)^2 + (3)^2}}\) and 1st A1 for \(\lambda = \dfrac{2\sqrt{2}}{5\sqrt{2}}\)
Note: So \(\left\{\hat{\mathbf{d}}_1 = \dfrac{1}{5\sqrt{2}}\begin{pmatrix}-5\\4\\3\end{pmatrix} \Rightarrow\right\}\) "vector" \(= \dfrac{2\sqrt{2}}{5\sqrt{2}}\begin{pmatrix}-5\\4\\3\end{pmatrix}\) is M1A1
Note: The 2nd dM1 in part (f) can be implied for at least 2 (out of 6) correct \(x\), \(y\), \(z\) ordinates from their values of \(\lambda\).
Note: Giving their “coordinates” as a column vector or position vector is fine for the final A1A1.
CAREFUL: Putting \(l_2\) equal to \(A\) gives \(\begin{pmatrix}1\\5\\2\end{pmatrix} + \lambda\begin{pmatrix}-5\\4\\3\end{pmatrix} = \begin{pmatrix}3\\5\\0\end{pmatrix} \to \begin{pmatrix}\lambda = \frac{2}{5}\\ \lambda = 0\\ \lambda = -\frac{2}{3}\end{pmatrix}\)
Give M0 dM0 for finding and using \(\lambda = \dfrac{2}{5}\) from this incorrect method.
CAREFUL: Putting \(\lambda\mathbf{d}_2 = \overrightarrow{AP}\) gives \(\lambda\begin{pmatrix}-5\\4\\3\end{pmatrix} = \begin{pmatrix}2\\0\\-2\end{pmatrix} \to \begin{pmatrix}\lambda = -\frac{2}{5}\\ \lambda = 0\\ \lambda = -\frac{2}{3}\end{pmatrix}\)
Give M0 dM0 for finding and using \(\lambda = -\dfrac{2}{5}\) from this incorrect method.
General: You can follow through the part (c) answer of their \(AP = 2\sqrt{2}\) for (d) M1dM1, (e) M1, (f) M1dM1
General: You can follow through their \(\mathbf{d}_2\) in part (b) for (d) M1dM1, (f) M1dM1.