C4 June 2015 Q4
4. With respect to a fixed origin \(O\), the lines \(l_1\) and \(l_2\) are given by the equations \[l_1: \mathbf{r} = \begin{pmatrix}5\\-3\\p\end{pmatrix} + \lambda\begin{pmatrix}0\\1\\-3\end{pmatrix}, \qquad l_2: \mathbf{r} = \begin{pmatrix}8\\5\\-2\end{pmatrix} + \mu\begin{pmatrix}3\\4\\-5\end{pmatrix}\] where \(\lambda\) and \(\mu\) are scalar parameters and \(p\) is a constant.
The lines \(l_1\) and \(l_2\) intersect at the point \(A\).
The point \(B\) lies on \(l_2\) where \(\mu = 1\)
| Scheme | Marks |
|---|---|
| \(l_1: \mathbf{r} = \begin{pmatrix}5\\-3\\p\end{pmatrix} + \lambda\begin{pmatrix}0\\1\\-3\end{pmatrix},\ l_2: \mathbf{r} = \begin{pmatrix}8\\5\\-2\end{pmatrix} + \mu\begin{pmatrix}3\\4\\-5\end{pmatrix}\). Let \(\theta\) = acute angle between \(l_1\) and \(l_2\). Note: You can mark parts (a) and (b) together. | |
| \(\{l_1 = l_2 \Rightarrow \mathbf{i}:\}\ 5 = 8 + 3\mu \Rightarrow \mu = -1\) Finds \(\mu\) and substitutes their \(\mu\) into \(l_2\) | M1 |
| So, \(\left\{\overrightarrow{OA}\right\} = \begin{pmatrix}8\\5\\-2\end{pmatrix} - 1\begin{pmatrix}3\\4\\-5\end{pmatrix} = \begin{pmatrix}5\\1\\3\end{pmatrix}\) \(5\mathbf{i} + \mathbf{j} + 3\mathbf{k}\) or \(\begin{pmatrix}5\\1\\3\end{pmatrix}\) or \((5, 1, 3)\) | A1 |
| (2) |
Notes
M1: Finds \(\mu\) and substitutes their \(\mu\) into \(l_2\)
A1: Point of intersection of \(5\mathbf{i} + \mathbf{j} + 3\mathbf{k}\). Allow \(\begin{pmatrix}5\\1\\3\end{pmatrix}\) or \((5, 1, 3)\).
Note: You cannot recover the answer for part (a) in part (c) or part (d).
| Scheme | Marks |
|---|---|
| \(\{\mathbf{j}: -3 + \lambda = 5 + 4\mu \Rightarrow\}\ -3 + \lambda = 5 + 4(-1) \Rightarrow \lambda = 4\) Equates \(\mathbf{j}\) components, substitutes their \(\mu\) and solves to give \(\lambda = \ldots\) | M1 |
| \(\mathbf{k}: p - 3\lambda = -2 - 5\mu \Rightarrow\) \(p - 3(4) = -2 - 5(-1) \Rightarrow \underline{p = 15}\) or \(\mathbf{k}: p - 3\lambda = 3 \Rightarrow\) \(p - 3(4) = 3 \Rightarrow \underline{p = 15}\) Equates \(\mathbf{k}\) components, substitutes their \(\lambda\) and their \(\mu\) and solves to give \(p = \ldots\) or equates \(\mathbf{k}\) components to give their "\(p - 3\lambda\) = the \(\mathbf{k}\) value of \(A\) found in part (a)", substitutes their \(\lambda\) and solves to give \(p = \ldots\) \(p = 15\) | M1 A1 |
| (3) |
Notes
M1: Equates \(\mathbf{j}\) components, substitutes their \(\mu\) and solves to give \(\lambda = \ldots\)
M1: Equates \(\mathbf{k}\) components, substitutes their \(\lambda\) and their \(\mu\) and solves to give \(p = \ldots\)
or equates \(\mathbf{k}\) components to give their "\(p - 3\lambda\) = the \(\mathbf{k}\) value of \(A\)” found in part (b).
A1: \(p = 15\)
Alternative method for part (b)
| Scheme | Marks |
|---|---|
| \(\left\{\begin{aligned} 3 \times \mathbf{j}&: -9 + 3\lambda = 15 + 12\mu \\ \mathbf{k}&: p - 3\lambda = -2 + 5\mu \end{aligned}\right\}\ p - 9 = 13 + 7\mu\) Eliminates \(\lambda\) to write down an equation in \(p\) and \(\mu\) | M1 |
| \(p - 9 = 13 + 7(-1) \Rightarrow \underline{p = 15}\) Substitutes their \(\mu\) and solves to give \(p = \ldots\) \(p = 15\) | M1 A1 |
| Scheme | Marks |
|---|---|
| \(\mathbf{d}_1 = \begin{pmatrix}0\\1\\-3\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}3\\4\\-5\end{pmatrix} \Rightarrow \begin{pmatrix}0\\1\\-3\end{pmatrix} \bullet \begin{pmatrix}3\\4\\-5\end{pmatrix}\) Realisation that the dot product is required between \(\pm A\mathbf{d}_1\) and \(\pm B\mathbf{d}_2\). | M1 |
| \(\cos\theta = \pm K\left(\dfrac{0(3) + (1)(4) + (-3)(-5)}{\sqrt{(0)^2 + (1)^2 + (-3)^2}\,.\sqrt{(3)^2 + (4)^2 + (-5)^2}}\right)\) An attempt to apply the dot product formula between \(\pm A\mathbf{d}_1\) and \(\pm B\mathbf{d}_2\). | dM1 (A1 on ePEN) |
| \(\cos\theta = \dfrac{19}{\sqrt{10}\,.\sqrt{50}} \Rightarrow \theta = 31.8203116\ldots = 31.82\) (2 dp) anything that rounds to 31.82 | A1 |
| (3) |
Notes
NOTE: Part (c) appears as M1A1A1 on ePEN, but now is marked as M1M1A1.
M1: Realisation that the dot product is required between \(\pm A\mathbf{d}_1\) and \(\pm B\mathbf{d}_2\).
Note: Allow one slip in candidates copying down their direction vectors, \(\mathbf{d}_1\) and \(\mathbf{d}_2\).
dM1: dependent on the FIRST method mark being awarded.
An attempt to apply the dot product formula between \(\pm A\mathbf{d}_1\) and \(\pm B\mathbf{d}_2\).
A1: anything that rounds to 31.82. This can also be achieved by \(180 - 148.1796\ldots\) = awrt 31.82
Note: \(\theta = 0.5553\ldots^{\mathrm{c}}\) is A0.
Note: M1A1 for \(\cos\theta = \left(\dfrac{0 - 16 - 60}{\sqrt{(0)^2 + (4)^2 + (-12)^2}\,.\sqrt{(-3)^2 + (-4)^2 + (5)^2}}\right) = \dfrac{-76}{\sqrt{160}\,.\sqrt{50}}\)
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 |
|---|---|
| \(\mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix}0\\1\\-3\end{pmatrix} \times \begin{pmatrix}3\\4\\-5\end{pmatrix} = \left\{\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & -3 \\ 3 & 4 & -5 \end{vmatrix}\right\} = 7\mathbf{i} - 9\mathbf{j} - 3\mathbf{k}\) Realisation that the vector cross product is required between \(\pm A\mathbf{d}_1\) and \(\pm B\mathbf{d}_2\). | M1 |
| \(\sin\theta = \dfrac{\sqrt{(7)^2 + (-9)^2 + (3)^2}}{\sqrt{(0)^2 + (1)^2 + (-3)^2}\,.\sqrt{(3)^2 + (4)^2 + (-5)^2}}\) An attempt to apply the vector cross product formula | dM1 (A1 on ePEN) |
| \(\sin\theta = \dfrac{\sqrt{139}}{\sqrt{10}\,.\sqrt{50}} \Rightarrow \theta = 31.8203116\ldots = 31.82\) (2 dp) anything that rounds to 31.82 | A1 |
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OB} = \begin{pmatrix}11\\9\\-7\end{pmatrix};\ \overrightarrow{AB} = \begin{pmatrix}11\\9\\-7\end{pmatrix} - \begin{pmatrix}5\\1\\3\end{pmatrix} = \begin{pmatrix}6\\8\\-10\end{pmatrix}\) or \(\overrightarrow{AB} = 2\begin{pmatrix}3\\4\\-5\end{pmatrix} = \begin{pmatrix}6\\8\\-10\end{pmatrix}\) \(\left|\overrightarrow{AB}\right| = \sqrt{6^2 + 8^2 + (-10)^2}\ \left\{= 10\sqrt{2}\right\}\) See notes | M1 |
| \(\dfrac{d}{10\sqrt{2}} = \sin\theta\) Writes down a correct trigonometric equation involving the shortest distance, \(d\). Eg: \(\dfrac{d}{\text{their } AB} = \sin\theta\), oe. | dM1 |
| \(\left\{d = 10\sqrt{2}\sin 31.82\ldots \Rightarrow\right\} d = 7.456540753\ldots = 7.46\) (3sf) anything that rounds to 7.46 | A1 |
| (3) | |
| (11 marks) |
Notes
M1: Full method for finding \(B\) and for finding the magnitude of \(\overrightarrow{AB}\) or the magnitude of \(\overrightarrow{BA}\).
dM1: dependent on the first method mark being awarded.
Writes down correct trigonometric equation involving the shortest distance, \(d\).
Eg: \(\dfrac{d}{\text{their } AB} = \sin\theta\) or \(\dfrac{d}{\text{their } AB} = \cos(90 - \theta)\), o.e., where “their \(AB\)” is a value.
and \(\theta\) = "their \(\theta\)" or stated as \(\theta\)
A1: anything that rounds to 7.46
Alternative Methods for part (d)
Let \(X\) be the foot of the perpendicular from \(B\) onto \(l_1\)
\(\mathbf{d}_1 = \begin{pmatrix}0\\1\\-3\end{pmatrix},\ \overrightarrow{OX} = \begin{pmatrix}5\\-3\\15\end{pmatrix} + \lambda\begin{pmatrix}0\\1\\-3\end{pmatrix} = \begin{pmatrix}5\\-3 + \lambda\\15 - 3\lambda\end{pmatrix}\)
\(\overrightarrow{BX} = \begin{pmatrix}5\\-3 + \lambda\\15 - 3\lambda\end{pmatrix} - \begin{pmatrix}11\\9\\-7\end{pmatrix} = \begin{pmatrix}-6\\-12 + \lambda\\22 - 3\lambda\end{pmatrix}\)
Method 1
| Scheme | Marks |
|---|---|
| \(\overrightarrow{BX} \bullet \mathbf{d}_1 = 0 \Rightarrow \begin{pmatrix}-6\\-12 + \lambda\\22 - 3\lambda\end{pmatrix} \bullet \begin{pmatrix}0\\1\\-3\end{pmatrix} = -12 + \lambda - 66 + 9\lambda = 0\) leading to \(10\lambda - 78 = 0 \Rightarrow \lambda = \dfrac{39}{5}\) (Allow a sign slip in copying \(\mathbf{d}_1\)) Applies \(\overrightarrow{BX} \bullet \mathbf{d}_1 = 0\) and solves the resulting equation to find a value for \(\lambda\). | M1 |
| \(\overrightarrow{BX} = \begin{pmatrix}-6\\-12 + \dfrac{39}{5}\\22 - 3\left(\dfrac{39}{5}\right)\end{pmatrix} = \begin{pmatrix}-6\\-\dfrac{21}{5}\\-\dfrac{7}{5}\end{pmatrix}\) Substitutes their value of \(\lambda\) into their \(\overrightarrow{BX}\). Note: This mark is dependent upon the previous M1 mark. | dM1 |
| \(d = BX = \sqrt{(-6)^2 + \left(-\dfrac{21}{5}\right)^2 + \left(-\dfrac{7}{5}\right)^2} = 7.456540753\ldots\) awrt 7.46 | A1 |
Method 2
| Scheme | Marks |
|---|---|
| Let \(\beta = \left|\overrightarrow{BX}\right|^2 = 36 + 144 - 24\lambda + \lambda^2 + 484 - 132\lambda + 9\lambda^2\) \(= 10\lambda^2 - 156\lambda + 664\) So \(\dfrac{\mathrm{d}\beta}{\mathrm{d}\lambda} = 20\lambda - 156 = 0 \Rightarrow \lambda = \dfrac{39}{5}\) Finds \(\beta = \left|\overrightarrow{BX}\right|^2\) in terms of \(\lambda\), finds \(\dfrac{\mathrm{d}\beta}{\mathrm{d}\lambda}\) and sets this result equal to 0 and finds a value for \(\lambda\). | M1 |
| \(\left|\overrightarrow{BX}\right|^2 = 10\left(\dfrac{39}{5}\right)^2 - 156\left(\dfrac{39}{5}\right) + 664 = \dfrac{278}{5}\) Substitutes their value of \(\lambda\) into their \(\left|\overrightarrow{BX}\right|^2\). Note: This mark is dependent upon the previous M1 mark. | dM1 |
| \(d = BX = \sqrt{\dfrac{278}{5}} = 7.456540753\ldots\) awrt 7.46 | A1 |