C4 June 2017 Q6
6. With respect to a fixed origin \(O\), the lines \(l_1\) and \(l_2\) are given by the equations \[l_1: \mathbf{r} = \begin{pmatrix}4\\28\\4\end{pmatrix} + \lambda\begin{pmatrix}-1\\-5\\1\end{pmatrix}, \qquad l_2: \mathbf{r} = \begin{pmatrix}5\\3\\1\end{pmatrix} + \mu\begin{pmatrix}3\\0\\-4\end{pmatrix}\] where \(\lambda\) and \(\mu\) are scalar parameters.
The lines \(l_1\) and \(l_2\) intersect at the point \(X\).
(a) Find the coordinates of the point \(X\). (3)
(b) Find the size of the acute angle between \(l_1\) and \(l_2\), giving your answer in degrees to 2 decimal places. (3)
The point \(A\) lies on \(l_1\) and has position vector \(\begin{pmatrix}2\\18\\6\end{pmatrix}\)
(c) Find the distance \(AX\), giving your answer as a surd in its simplest form. (2)
The point \(Y\) lies on \(l_2\). Given that the vector \(\overrightarrow{YA}\) is perpendicular to the line \(l_1\)
(d) find the distance \(YA\), giving your answer to one decimal place. (2)
The point \(B\) lies on \(l_1\) where \(\left|\overrightarrow{AX}\right| = 2\left|\overrightarrow{AB}\right|\).
(e) Find the two possible position vectors of \(B\). (3)
| Scheme | Marks |
|---|---|
| \(l_1: \mathbf{r} = \begin{pmatrix}4\\28\\4\end{pmatrix} + \lambda\begin{pmatrix}-1\\-5\\1\end{pmatrix},\ l_2: \mathbf{r} = \begin{pmatrix}5\\3\\1\end{pmatrix} + \mu\begin{pmatrix}3\\0\\-4\end{pmatrix};\ \overrightarrow{OA} = \begin{pmatrix}2\\18\\6\end{pmatrix}\) lies on \(l_1\) Let \(\theta_{\text{Acute}}\) be the acute angle between \(l_1\) and \(l_2\) | |
| \(\{l_1 = l_2 \Rightarrow\}\ 28 - 5\lambda = 3\ \{\Rightarrow \lambda = 5\}\) or \(4 - \lambda = 5 + 3\mu\) and \(4 + \lambda = 1 - 4\mu\ \{\Rightarrow \mu = -2\}\) \(28 - 5\lambda = 3\) or \(4 - \lambda = 5 + 3\mu\) and \(4 + \lambda = 1 - 4\mu\) or \(\lambda = 5\) or \(\mu = -2\) (Can be implied). | B1 |
| \(\left\{\overrightarrow{OX} =\right\}\ \begin{pmatrix}4\\28\\4\end{pmatrix} + 5\begin{pmatrix}-1\\-5\\1\end{pmatrix}\) or \(\begin{pmatrix}5\\3\\1\end{pmatrix} - 2\begin{pmatrix}3\\0\\-4\end{pmatrix}\) Puts \(l_1 = l_2\) and solves to find \(\lambda\) and/or \(\mu\) and substitutes their value for \(\lambda\) into \(l_1\) or their value for \(\mu\) into \(l_2\) | M1 |
| So, \(X(-1, 3, 9)\) \((-1, 3, 9)\) or \(\begin{pmatrix}-1\\3\\9\end{pmatrix}\) or \(-\mathbf{i} + 3\mathbf{j} + 9\mathbf{k}\) or condone \(\begin{matrix}-1\\3\\9\end{matrix}\) | A1 cao |
| (3) |
Notes
Note: M1 can be implied by at least two correct follow through coordinates from their \(\lambda\) or from their \(\mu\)
| Scheme | Marks |
|---|---|
| Way 1 | |
| \(\mathbf{d}_1 = \begin{pmatrix}-1\\-5\\1\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}3\\0\\-4\end{pmatrix} \Rightarrow \begin{pmatrix}-1\\-5\\1\end{pmatrix} \bullet \begin{pmatrix}3\\0\\-4\end{pmatrix}\) Realisation that the dot product is required between \(\mathbf{d}_1\) and \(\mathbf{d}_2\) or a multiple of \(\mathbf{d}_1\) and \(\mathbf{d}_2\) | M1 |
| \(\cos\theta = \dfrac{\pm\left(\begin{pmatrix}-1\\-5\\1\end{pmatrix} \bullet \begin{pmatrix}3\\0\\-4\end{pmatrix}\right)}{\sqrt{(-1)^2 + (-5)^2 + (1)^2}\,.\sqrt{(3)^2 + (0)^2 + (-4)^2}}\ \left\{= \dfrac{-7}{\sqrt{27}\,.\sqrt{25}}\right\}\) dependent on the 1st M mark. Applies dot product formula between \(\mathbf{d}_1\) and \(\mathbf{d}_2\) or a multiple of \(\mathbf{d}_1\) and \(\mathbf{d}_2\) | dM1 |
| \(\{\theta = 105.6303588\ldots \Rightarrow\}\ \theta_{\text{Acute}} = 74.36964117\ldots = 74.37\) (2 dp) awrt 74.37 seen in (b) only | A1 |
| (3) |
Notes
Note: Evaluating the dot product (i.e. \((-1)(3) + (-5)(0) + (1)(-4)\)) is not required for the M1, dM1 marks.
Note: For M1 dM1: Allow one slip in writing down their direction vectors, \(\mathbf{d}_1\) and \(\mathbf{d}_2\)
Note: Allow M1 dM1 for \(\left(\sqrt{(-1)^2 + (-5)^2 + (1)^2}\,.\sqrt{(3)^2 + (0)^2 + (-4)^2}\right)\cos\theta = \pm\begin{pmatrix}-1\\-5\\1\end{pmatrix} \bullet \begin{pmatrix}3\\0\\-4\end{pmatrix}\)
Note: \(\theta = 1.297995\ldots^{\mathrm{c}}\), (without evidence of awrt 74.37) is A0
Way 2 — Alternative Method: Vector Cross Product
Only apply this scheme if it is clear that a vector cross product method is being applied.
| Scheme | Marks |
|---|---|
| \(\mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix}-1\\-5\\1\end{pmatrix} \times \begin{pmatrix}3\\0\\-4\end{pmatrix} = \left\{\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -5 & 1 \\ 3 & 0 & -4 \end{vmatrix} = 20\mathbf{i} - \mathbf{j} + 15\mathbf{k}\right\}\) Realisation that the vector cross product is required between \(\mathbf{d}_1\) and \(\mathbf{d}_2\) or a multiple of \(\mathbf{d}_1\) and \(\mathbf{d}_2\) | M1 |
| \(\sin\theta = \dfrac{\sqrt{(20)^2 + (-1)^2 + (15)^2}}{\sqrt{(-1)^2 + (-5)^2 + (1)^2}\,.\sqrt{(3)^2 + (0)^2 + (-4)^2}}\) Applies the vector product formula between \(\mathbf{d}_1\) and \(\mathbf{d}_2\) or a multiple of \(\mathbf{d}_1\) and \(\mathbf{d}_2\) | dM1 |
| \(\sin\theta = \dfrac{\sqrt{626}}{\sqrt{27}\,.\sqrt{25}} \Rightarrow \theta = 74.36964117\ldots = 74.37\) (2 dp) awrt 74.37 seen in (b) only | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AX} = \text{"}\overrightarrow{OX}\text{"} - \overrightarrow{OA} = \begin{pmatrix}-1\\3\\9\end{pmatrix} - \begin{pmatrix}2\\18\\6\end{pmatrix} = \begin{pmatrix}-3\\-15\\3\end{pmatrix}\) or \(A_{\lambda=2},\ X_{\lambda=5} \Rightarrow AX = 3\left|\mathbf{d}_1\right|,\ \left\{\left|\mathbf{d}_1\right| = \sqrt{27}\right\}\) | |
| \(AX = \sqrt{(-3)^2 + (-15)^2 + (3)^2}\) or \(3\sqrt{27}\ \left\{= \sqrt{243}\right\} = 9\sqrt{3}\) Full method for finding \(AX\) or \(XA\) \(9\sqrt{3}\) seen in (c) only | M1 A1 cao |
| Note: You cannot recover work for part (c) in either part (d) or part (e). | |
| (2) |
Notes
M1: Finds the difference between their \(\overrightarrow{OX}\) and \(\overrightarrow{OA}\) and applies Pythagoras to the result to find \(AX\) or \(XA\)
OR applies \(\left|(\text{their } \lambda_X \text{ found in } (a)) - 2\right|.\sqrt{(-1)^2 + (-5)^2 + (1)^2}\)
Note: For M1: Allow one slip in writing down their \(\overrightarrow{OX}\) and \(\overrightarrow{OA}\)
Note: Allow M1A1 for \(\begin{pmatrix}3\\15\\3\end{pmatrix}\) leading to \(AX = \sqrt{(3)^2 + (15)^2 + (3)^2} = \sqrt{243} = 9\sqrt{3}\)
| Scheme | Marks |
|---|---|
| Way 1 | |
| \(\dfrac{YA}{\text{"}9\sqrt{3}\text{"}} = \tan(\text{"}74.36964\ldots\text{"})\) \(\dfrac{YA}{\text{their } \left|\overrightarrow{AX}\right|} = \tan\theta\) or \(YA = \left(\text{their } \left|\overrightarrow{AX}\right|\right)\tan\theta\), where \(\theta\) is their acute or obtuse angle between \(l_1\) and \(l_2\) | M1 |
| \(YA = 55.71758\ldots = 55.7\) (1 dp) anything that rounds to 55.7 | A1 |
| (2) |
Notes
Way 2
| Scheme | Marks |
|---|---|
| \(\dfrac{\text{"}9\sqrt{3}\text{"}}{YA} = \tan(90 - \text{"}74.36964\ldots\text{"})\) \(\dfrac{\text{their } \left|\overrightarrow{AX}\right|}{YA} = \tan(90 - \theta)\) or \(AY = \dfrac{\text{their } \left|\overrightarrow{AX}\right|}{\tan(90 - \theta)}\), where \(\theta\) is the acute or obtuse angle between \(l_1\) and \(l_2\) | M1 |
| \(YA = 55.71758\ldots = 55.7\) (1 dp) anything that rounds to 55.7 | A1 |
| (2) |
Way 3
| Scheme | Marks |
|---|---|
| \(\dfrac{YA}{\sin(\text{"}74.36964\ldots\text{"})} = \dfrac{\text{"}9\sqrt{3}\text{"}}{\sin(90 - \text{"}74.36964\ldots\text{"})}\) \(\dfrac{YA}{\sin\theta} = \dfrac{\text{their } \left|\overrightarrow{AX}\right|}{\sin(90 - \theta)}\) o.e., where \(\theta\) is the acute or obtuse angle between \(l_1\) and \(l_2\) | M1 |
| \(YA = \dfrac{9\sqrt{3}\sin(74.36964\ldots)}{\sin(15.63036\ldots)} = 55.71758\ldots = 55.7\) (1 dp) anything that rounds to 55.7 | A1 |
| (2) |
Way 4
| Scheme | Marks |
|---|---|
| \(\mathbf{d}_1 = \begin{pmatrix}-1\\-5\\1\end{pmatrix},\ \overrightarrow{OY} = \begin{pmatrix}5\\3\\1\end{pmatrix} + \mu\begin{pmatrix}3\\0\\-4\end{pmatrix} = \begin{pmatrix}5 + 3\mu\\3\\1 - 4\mu\end{pmatrix}\) | |
| \(\overrightarrow{YA} = \begin{pmatrix}2\\18\\6\end{pmatrix} - \begin{pmatrix}5 + 3\mu\\3\\1 - 4\mu\end{pmatrix} = \begin{pmatrix}-3 - 3\mu\\15\\5 + 4\mu\end{pmatrix}\) \(\overrightarrow{YA} \bullet \mathbf{d}_1 = 0 \Rightarrow \begin{pmatrix}-3 - 3\mu\\15\\5 + 4\mu\end{pmatrix} \bullet \begin{pmatrix}-1\\-5\\1\end{pmatrix} = 0\) \(\Rightarrow 3 + 3\mu - 75 + 5 + 4\mu = 0 \Rightarrow \mu = \dfrac{67}{7}\) \(YA^2 = \left(-3 - 3\left(\dfrac{67}{7}\right)\right)^2 + (15)^2 + \left(5 + 4\left(\dfrac{67}{7}\right)\right)^2\) (Allow a sign slip in copying \(\mathbf{d}_1\)) Applies \(\overrightarrow{YA} \bullet \mathbf{d}_1 = 0\) or \(\overrightarrow{AY} \bullet \mathbf{d}_1 = 0\) or \(\overrightarrow{YA} \bullet (K\mathbf{d}_1) = 0\) or \(\overrightarrow{AY} \bullet (K\mathbf{d}_1) = 0\) to find \(\mu\) and applies Pythagoras to find a numerical expression for \(AY^2\) or for the distance \(AY\) | M1 |
| So, \(YA = \sqrt{\left(-\dfrac{222}{7}\right)^2 + (15)^2 + \left(\dfrac{303}{7}\right)^2}\) \(= 55.71758\ldots = 55.7\) (1 dp) anything that rounds to 55.7 | A1 |
| Note: \(\overrightarrow{OY} = \dfrac{236}{7}\mathbf{i} + 3\mathbf{j} - \dfrac{261}{7}\mathbf{k},\ \overrightarrow{YA} = -\dfrac{222}{7}\mathbf{i} + 15\mathbf{j} + \dfrac{303}{7}\mathbf{k}\) | |
| (2) |
(corrected from the printed mark scheme: the last vector in this note is printed as \(\overrightarrow{AY}\); the vector \(-\frac{222}{7}\mathbf{i} + 15\mathbf{j} + \frac{303}{7}\mathbf{k}\) is \(\overrightarrow{YA}\), as found above.)
| Scheme | Marks |
|---|---|
| Way 1 | |
| \(\{A_{\lambda=2},\ X_{\lambda=5} \Rightarrow \text{So } AX = 2AB \Rightarrow \text{So at } B,\ \lambda = 3.5 \text{ or } \lambda = 0.5)\}\) | |
| \(\overrightarrow{OB} = \begin{pmatrix}4\\28\\4\end{pmatrix} + 3.5\begin{pmatrix}-1\\-5\\1\end{pmatrix};\ = \begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}\) \(\overrightarrow{OB} = \begin{pmatrix}4\\28\\4\end{pmatrix} + 0.5\begin{pmatrix}-1\\-5\\1\end{pmatrix};\ = \begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}\) Substitutes either \(\lambda = \dfrac{(\text{their } \lambda_X \text{ found in } (a)) + 2}{2}\) or \(\lambda_\beta = 3 - \dfrac{(\text{their } \lambda_X \text{ found in } (a))}{2}\) into \(l_1\) At least one position vector is correct (Also allow coordinates). Both position vectors are correct (Also allow coordinates). | M1; A1 A1 |
| (3) | |
| (13 marks) |
Notes
Note: Imply M1 for no working leading to any two components of one of the \(\overrightarrow{OB}\) which are correct.
Way 2
| Scheme | Marks |
|---|---|
| \(\left\{AX = 2AB \Rightarrow AB = \tfrac{1}{2}AX.\ \text{So, } \overrightarrow{OB} = \overrightarrow{OA} \pm \overrightarrow{AB} \Rightarrow \overrightarrow{OB} = \overrightarrow{OA} \pm \tfrac{1}{2}\overrightarrow{AX}\right\}\) | |
| \(\overrightarrow{OB} = \begin{pmatrix}2\\18\\6\end{pmatrix} + 0.5\begin{pmatrix}-3\\-15\\3\end{pmatrix};\ = \begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}\) \(\overrightarrow{OB} = \begin{pmatrix}2\\18\\6\end{pmatrix} - 0.5\begin{pmatrix}-3\\-15\\3\end{pmatrix};\ = \begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}\) Applies either \(\overrightarrow{OA} + 0.5\overrightarrow{AX}\) or \(\overrightarrow{OA} - 0.5\overrightarrow{AX}\) where \((\text{their } \overrightarrow{AX}) = \pm\big[(\text{their } \overrightarrow{OX}) - \overrightarrow{OA}\big]\) At least one position vector is correct (Also allow coordinates). Both position vectors are correct (Also allow coordinates). | M1; A1 A1 |
| (3) |
Way 3
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix}4 - \lambda\\28 - 5\lambda\\4 + \lambda\end{pmatrix} - \begin{pmatrix}2\\18\\6\end{pmatrix} = \begin{pmatrix}2 - \lambda\\10 - 5\lambda\\-2 + \lambda\end{pmatrix} = \begin{pmatrix}1(2 - \lambda)\\5(2 - \lambda)\\-1(2 - \lambda)\end{pmatrix};\ \overrightarrow{AX} = \begin{pmatrix}-3\\-15\\3\end{pmatrix}\) \(AX^2 = 243 \Rightarrow AB^2 = 27(2 - \lambda)^2\) \(AX = 2AB \Rightarrow AX^2 = 4AB^2 \Rightarrow 243 = 4(27)(2 - \lambda)^2 \Rightarrow (2 - \lambda)^2 = \dfrac{9}{4}\) or \(27\lambda^2 - 108\lambda + \dfrac{189}{4} = 0\) or \(108\lambda^2 - 432\lambda + 189 = 0\) or \(4\lambda^2 - 16\lambda + 7 = 0 \Rightarrow \lambda = 3.5\) or \(\lambda = 0.5\) | |
| \(\overrightarrow{OB} = \begin{pmatrix}4\\28\\4\end{pmatrix} + 3.5\begin{pmatrix}-1\\-5\\1\end{pmatrix};\ = \begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}\) \(\overrightarrow{OB} = \begin{pmatrix}4\\28\\4\end{pmatrix} + 0.5\begin{pmatrix}-1\\-5\\1\end{pmatrix};\ = \begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}\) Full method of solving for \(\lambda\) the equation \(AX^2 = 4AB^2\) using (their \(\overrightarrow{AX}\)) and \(\overrightarrow{AB}\) and substitutes at least one of their values for \(\lambda\) into \(l_1\) At least one position vector is correct (Also allow coordinates). Both position vectors are correct (Also allow coordinates). | M1; A1 A1 |
| (3) |
Note: \(AX = 2AB \Rightarrow \overrightarrow{AX} = \pm 2\overrightarrow{AB}\). Hence, \(\lambda = 3.5\) or \(\lambda = 0.5\) can be found from solving either \(x: -3 = \pm 2(2 - \lambda)\) or \(y: -15 = \pm 2(10 - 5\lambda)\) or \(z: -3 = \pm 2(-2 + \lambda)\)
Way 4
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OB} = \begin{pmatrix}-1\\3\\9\end{pmatrix} + 0.5\begin{pmatrix}3\\15\\-3\end{pmatrix};\ = \begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}\) \(\overrightarrow{OB} = \begin{pmatrix}-1\\3\\9\end{pmatrix} + 1.5\begin{pmatrix}3\\15\\-3\end{pmatrix};\ = \begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}\) Applies either (their \(\overrightarrow{OX}\)) + \(0.5\overrightarrow{XA}\) or (their \(\overrightarrow{OX}\)) + \(1.5\overrightarrow{XA}\) where (their \(\overrightarrow{XA}\)) = \(\overrightarrow{OA}\) − (their \(\overrightarrow{OX}\)) At least one position vector is correct (Also allow coordinates). Both position vectors are correct (Also allow coordinates). | M1; A1 A1 |
| (3) |
Way 5
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OB} = 0.5\left(\begin{pmatrix}-1\\3\\9\end{pmatrix} + \begin{pmatrix}2\\18\\6\end{pmatrix}\right);\ = \begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}\) \(\overrightarrow{OB} = \begin{pmatrix}2\\18\\6\end{pmatrix} - 0.5\begin{pmatrix}-3\\-15\\3\end{pmatrix};\ = \begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}\) Applies \(\dfrac{1}{2}\big[(\text{their } \overrightarrow{OX}) + \overrightarrow{OA}\big]\) At least one position vector is correct (Also allow coordinates). Both position vectors are correct (Also allow coordinates). | M1; A1 A1 |
| (3) |
Way 6
| Scheme | Marks |
|---|---|
| \(\left\{\left|\overrightarrow{AX}\right| = 9\sqrt{3},\ |d_1| = 3\sqrt{3} \Rightarrow K = \dfrac{9\sqrt{3}}{3\sqrt{3}} = 3 \Rightarrow \overrightarrow{AX} = 3\mathbf{d}_1;\ \text{So, } \overrightarrow{OB} = \overrightarrow{OA} \pm \tfrac{1}{2}\overrightarrow{AX} = \overrightarrow{OA} \pm \tfrac{1}{2}(3\mathbf{d}_1)\right\}\) | |
| \(\overrightarrow{OB} = \begin{pmatrix}2\\18\\6\end{pmatrix} + 0.5\left(3\begin{pmatrix}-1\\-5\\1\end{pmatrix}\right);\ = \begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}\) \(\overrightarrow{OB} = \begin{pmatrix}2\\18\\6\end{pmatrix} - 0.5\left(3\begin{pmatrix}-1\\-5\\1\end{pmatrix}\right);\ = \begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}\) Applies either \(\overrightarrow{OA} + 0.5(K\mathbf{d}_1)\) or \(\overrightarrow{OA} - 0.5(K\mathbf{d}_1)\), where \(K = \dfrac{\text{their } \left|\overrightarrow{AX}\right|}{3\sqrt{3}}\) At least one position vector is correct (Also allow coordinates). Both position vectors are correct (Also allow coordinates). | M1; A1 A1 |
| (3) |