C4 June 2007 Q5
5.
The line \(l_1\) has equation \(\mathbf{r} = \begin{pmatrix}1\\0\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\0\end{pmatrix}\).
The line \(l_2\) has equation \(\mathbf{r} = \begin{pmatrix}1\\3\\6\end{pmatrix} + \mu\begin{pmatrix}2\\1\\-1\end{pmatrix}\).
The point \(A\) is on \(l_1\) where \(\lambda = 1\), and the point \(B\) is on \(l_2\) where \(\mu = 2\).
| Scheme | Marks |
|---|---|
| If \(l_1\) and \(l_2\) intersect then: | |
| \(\begin{pmatrix}1\\0\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\0\end{pmatrix} = \begin{pmatrix}1\\3\\6\end{pmatrix} + \mu\begin{pmatrix}2\\1\\-1\end{pmatrix}\) | |
| Any two of \(\begin{aligned}\mathbf{i}:&\quad 1 + \lambda = 1 + 2\mu &&(1)\\ \mathbf{j}:&\quad \lambda = 3 + \mu &&(2)\\ \mathbf{k}:&\quad -1 = 6 - \mu &&(3)\end{aligned}\) | M1 |
| (1) & (2) yields \(\lambda = 6,\ \mu = 3\) (1) & (3) yields \(\lambda = 14,\ \mu = 7\) (2) & (3) yields \(\lambda = 10,\ \mu = 7\) | A1 A1 |
| Either checking eqn (3), \(-1 \ne 3\) checking eqn (2), \(14 \ne 10\) checking eqn (1), \(11 \ne 15\) or for example: checking eqn (3), LHS \(= -1\), RHS \(= 3\) \(\Rightarrow\) Lines \(l_1\) and \(l_2\) do not intersect | B1ft |
| (4) |
Notes
M1: Writes down any two of these equations correctly.
Solves two of the above equations to find … A1: either one of \(\lambda\) or \(\mu\) correct A1: both \(\lambda\) and \(\mu\) correct
B1ft: Complete method of putting their values of \(\lambda\) and \(\mu\) into a third equation to show a contradiction. The “LHS \(= -1\), RHS \(= 3\)” type of explanation is also allowed for B1ft.
Aliter 5. (a) Way 2
| Scheme | Marks |
|---|---|
| \(\mathbf{k}:\ -1 = 6 - \mu \Rightarrow \mu = 7\) | |
| \(\mathbf{i}:\ 1 + \lambda = 1 + 2\mu \Rightarrow 1 + \lambda = 1 + 2(7)\) \(\mathbf{j}:\ \lambda = 3 + \mu \Rightarrow \lambda = 3 + (7)\) | M1 |
| \(\mathbf{i}:\ \lambda = 14\) \(\mathbf{j}:\ \lambda = 10\) | A1 A1 |
| Either: These equations are then inconsistent Or: \(14 \ne 10\) Or: Lines \(l_1\) and \(l_2\) do not intersect | B1ft |
| (4) |
M1: Uses the k component to find \(\mu\) and substitutes their value of \(\mu\) into either one of the i or j component. A1: either one of the \(\lambda\)’s correct. A1: both of the \(\lambda\)’s correct. B1ft: Complete method giving rise to any one of these three explanations.
Aliter 5. (a) Way 3
| Scheme | Marks |
|---|---|
| If \(l_1\) and \(l_2\) intersect then: | |
| \(\begin{pmatrix}1\\0\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\0\end{pmatrix} = \begin{pmatrix}1\\3\\6\end{pmatrix} + \mu\begin{pmatrix}2\\1\\-1\end{pmatrix}\) | |
| Any two of \(\begin{aligned}\mathbf{i}:&\quad 1 + \lambda = 1 + 2\mu &&(1)\\ \mathbf{j}:&\quad \lambda = 3 + \mu &&(2)\\ \mathbf{k}:&\quad -1 = 6 - \mu &&(3)\end{aligned}\) | M1 |
| (1) & (2) yields \(\mu = 3\) (3) yields \(\mu = 7\) | A1 A1 |
| Either: These equations are then inconsistent Or: \(3 \ne 7\) Or: Lines \(l_1\) and \(l_2\) do not intersect | B1ft |
| (4) |
M1: Writes down any two of these equations. A1: either one of the \(\mu\)’s correct. A1: both of the \(\mu\)’s correct. B1ft: Complete method giving rise to any one of these three explanations.
Aliter 5. (a) Way 4
| Scheme | Marks |
|---|---|
| Any two of \(\begin{aligned}\mathbf{i}:&\quad 1 + \lambda = 1 + 2\mu &&(1)\\ \mathbf{j}:&\quad \lambda = 3 + \mu &&(2)\\ \mathbf{k}:&\quad -1 = 6 - \mu &&(3)\end{aligned}\) | M1 |
| (1) & (2) yields \(\mu = 3\) | A1 |
| (3) RHS \(= 6 - 3 = 3\) | A1 |
| (3) yields \(-1 \ne 3\) | B1ft |
| (4) |
M1: Writes down any two of these equations. A1: \(\mu = 3\). A1: RHS of (3) \(= 3\). B1ft: Complete method giving rise to this explanation.
| Scheme | Marks |
|---|---|
| \(\lambda = 1 \Rightarrow \overrightarrow{OA} = \begin{pmatrix}2\\1\\-1\end{pmatrix}\) & \(\mu = 2 \Rightarrow \overrightarrow{OB} = \begin{pmatrix}5\\5\\4\end{pmatrix}\) | B1 |
| \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \underline{\begin{pmatrix}5\\5\\4\end{pmatrix} - \begin{pmatrix}2\\1\\-1\end{pmatrix}} = \begin{pmatrix}3\\4\\5\end{pmatrix}\) or \(\overrightarrow{BA} = \underline{\begin{pmatrix}-3\\-4\\-5\end{pmatrix}}\) | M1ft |
| \(\overrightarrow{AB} = 3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\), \(\mathbf{d}_1 = \mathbf{i} + \mathbf{j} + 0\mathbf{k}\) & \(\theta\) is angle | M1 |
| \(\cos\theta = \dfrac{\overrightarrow{AB} \bullet \mathbf{d}_1}{\left|\overrightarrow{AB}\right|.\left|\mathbf{d}_1\right|} = \pm\left(\dfrac{3 + 4 + 0}{\sqrt{50}.\sqrt{2}}\right)\) | M1ft A1 |
| \(\cos\theta = \underline{\tfrac{7}{10}}\) | A1 cao |
| (6) | |
| (10 marks) |
Notes
B1: Only one of either \(\overrightarrow{OA} = \begin{pmatrix}2\\1\\-1\end{pmatrix}\) or \(\overrightarrow{OB} = \begin{pmatrix}5\\5\\4\end{pmatrix}\) or \(A(2, 1, -1)\) or \(B(5, 5, 4)\). (can be implied)
M1ft: Finding the difference between their \(\overrightarrow{OB}\) and \(\overrightarrow{OA}\). (can be implied)
M1: Applying the dot product formula between “allowable” vectors. See notes below.
M1ft: Applies dot product formula between \(\mathbf{d}_1\) and their \(\pm\overrightarrow{AB}\). A1: Correct expression.
A1 cao: \(\underline{\tfrac{7}{10}}\) or \(\underline{0.7}\) or \(\tfrac{7}{\sqrt{100}}\) but not \(\tfrac{7}{\sqrt{50}\sqrt{2}}\)
Candidates can score this mark if there is a complete method for finding the dot product between their vectors in the following cases:
Case 1: their ft \(\pm\overrightarrow{AB} = \pm(3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k})\) and \(\mathbf{d}_1 = \mathbf{i} + \mathbf{j} + 0\mathbf{k}\) \(\Rightarrow \cos\theta = \pm\left(\dfrac{3 + 4 + 0}{\sqrt{50}.\sqrt{2}}\right)\)
Case 2: \(\mathbf{d}_1 = \mathbf{i} + \mathbf{j} + 0\mathbf{k}\) and \(\mathbf{d}_2 = 2\mathbf{i} + \mathbf{j} - 1\mathbf{k}\) \(\Rightarrow \cos\theta = \dfrac{2 + 1 + 0}{\sqrt{2}.\sqrt{6}}\)
Case 3: \(\mathbf{d}_1 = \mathbf{i} + \mathbf{j} + 0\mathbf{k}\) and \(\mathbf{d}_2 = 2(2\mathbf{i} + \mathbf{j} - 1\mathbf{k})\) \(\Rightarrow \cos\theta = \dfrac{4 + 2 + 0}{\sqrt{2}.\sqrt{24}}\)
Case 4: their ft \(\pm\overrightarrow{AB} = \pm(3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k})\) and \(\mathbf{d}_2 = 2\mathbf{i} + \mathbf{j} - \mathbf{k}\) \(\Rightarrow \cos\theta = \pm\left(\dfrac{6 + 4 - 5}{\sqrt{50}.\sqrt{6}}\right)\)
Case 5: their ft \(\overrightarrow{OA} = 2\mathbf{i} + 1\mathbf{j} - 1\mathbf{k}\) and their ft \(\overrightarrow{OB} = 5\mathbf{i} + 5\mathbf{j} + 4\mathbf{k}\) \(\Rightarrow \cos\theta = \pm\left(\dfrac{10 + 5 - 4}{\sqrt{6}.\sqrt{66}}\right)\)
Note: If candidate use cases 2, 3, 4 and 5 they cannot gain the final three marks for this part.
Note: Candidate can only gain some/all of the final three marks if they use case 1.
Examples of awarding marks M1M1A1 in 5.(b)
| Example | Marks |
|---|---|
| \(\sqrt{50}.\sqrt{2}\cos\theta = \pm(3 + 4 + 0)\) | M1M1A1 (Case 1) |
| \(\sqrt{2}.\sqrt{6}\cos\theta = 3\) | M1M0A0 (Case 2) |
| \(\sqrt{2}.\sqrt{24}\cos\theta = 4 + 2\) | M1M0A0 (Case 3) |