C4 January 2013 Q7
7. With respect to a fixed origin \(O\), the lines \(l_1\) and \(l_2\) are given by the equations\[l_1:\ \mathbf{r} = (9\mathbf{i} + 13\mathbf{j} - 3\mathbf{k}) + \lambda(\mathbf{i} + 4\mathbf{j} - 2\mathbf{k})\]\[l_2:\ \mathbf{r} = (2\mathbf{i} - \mathbf{j} + \mathbf{k}) + \mu(2\mathbf{i} + \mathbf{j} + \mathbf{k})\]where \(\lambda\) and \(\mu\) are scalar parameters.
Given that the point \(A\) has position vector \(4\mathbf{i} + 16\mathbf{j} - 3\mathbf{k}\) and that the point \(P\) lies on \(l_1\) such that \(AP\) is perpendicular to \(l_1\),
| Scheme | Marks |
|---|---|
| \(\mathbf{i}:\quad 9 + \lambda = 2 + 2\mu \quad \textbf{(1)}\) \(\mathbf{j}:\quad 13 + 4\lambda = -1 + \mu \quad \textbf{(2)}\) \(\mathbf{k}:\quad -3 - 2\lambda = 1 + \mu \quad \textbf{(3)}\) Any two equations. (Allow one slip). | M1 |
| Eg: \(\textbf{(2)} - \textbf{(3)}:\ 16 + 6\lambda = -2\) or \(\textbf{(2)} - 4\textbf{(1)}:\ -23 = -9 - 7\mu\) An attempt to eliminate one of the parameters. | dM1 |
| Leading to \(\lambda = -3\) or \(\mu = 2\) Either \(\lambda = -3\) or \(\mu = 2\) | A1 |
| \(l_1:\ \mathbf{r} = \begin{pmatrix} 9 \\ 13 \\ -3 \end{pmatrix} - 3\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 6 \\ 1 \\ 3 \end{pmatrix}\) or \(l_2:\ \mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} + 2\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 1 \\ 3 \end{pmatrix}\) See notes | ddM1 A1 |
| (5) |
Notes
M1: Writes down any two equations. Allow one slip.
dM1: Attempts to eliminate either \(\lambda\) or \(\mu\) to form an equation in one parameter only.
A1: For either \(\lambda = -3\) or \(\mu = 2\). Note: candidates only need to find one of the parameters.
ddM1: For either substituting their value of \(\lambda\) into \(l_1\) or their \(\mu\) into \(l_2\).
2nd A1: For either \(\begin{pmatrix} 6 \\ 1 \\ 3 \end{pmatrix}\) or \(6\mathbf{i} + \mathbf{j} + 3\mathbf{k}\) or \(\begin{pmatrix} 6 & 1 & 3 \end{pmatrix}\).
Note: Each of the method marks in this part are dependent upon the previous method marks.
| Scheme | Marks |
|---|---|
| \(\mathbf{d}_1 = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \Rightarrow \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} \bullet \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\) Realisation that the dot product is required between \(\pm A\mathbf{d}_1\) and \(\pm B\mathbf{d}_2\). | M1 |
| \(\cos\theta = \pm\left(\dfrac{2 + 4 - 2}{\sqrt{(1)^2 + (4)^2 + (-2)^2}.\sqrt{(2)^2 + (1)^2 + (1)^2}}\right)\) Correct equation. | A1 |
| \(\cos\theta = \dfrac{4}{\sqrt{21}.\sqrt{6}} \Rightarrow \theta = 69.1238974\ldots = 69.1\ \text{(1 dp)}\) awrt 69.1 | A1 |
| (3) |
Notes
M1: Realisation that the dot product is required between \(\pm A\mathbf{d}_1\) and \(\pm B\mathbf{d}_2\). Allow one slip in \(\mathbf{d}_1 = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\).
A1: Correct application of the dot product formula \(\mathbf{d}_1 \bullet \mathbf{d}_2 = \pm|\mathbf{d}_1||\mathbf{d}_2|\cos\theta\) or \(\cos\theta = \pm\left(\dfrac{\mathbf{d}_1 \bullet \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}\right)\)
The dot product must be correctly applied and the square roots although they can be un-simplified must be correctly applied.
A1: awrt 69.1. This can be also be achieved by \(180 - 110.876 = \) awrt 69.1. \(\theta = 1.2064\ldots^{c}\) is A0.
Common response: \(\cos\theta = \left(\dfrac{-12 - 24 + 12}{\sqrt{(-3)^2 + (-12)^2 + (6)^2}.\sqrt{(4)^2 + (2)^2 + (2)^2}}\right) = \dfrac{-24}{\sqrt{189}.\sqrt{24}}\) is M1A1...
Alternative Method: Vector Cross Product
| Scheme | Marks |
|---|---|
| Only apply this scheme if it is clear that a candidate is applying a vector cross product method. | |
| \(\mathbf{d}_1 \times \mathbf{d}_2 = \underline{\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}} = \left\{\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 4 & -2 \\ 2 & 1 & 1 \end{vmatrix} = 6\mathbf{i} - 5\mathbf{j} - 7\mathbf{k}\right\}\) M1: Realisation that the vector cross product is required between \(\pm A\mathbf{d}_1\) and \(\pm B\mathbf{d}_2\). Allow one slip in \(\mathbf{d}_1 = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\). | M1 |
| \(\sin\theta = \dfrac{\sqrt{(6)^2 + (5)^2 + (-7)^2}}{\sqrt{(1)^2 + (4)^2 + (-2)^2}.\sqrt{(2)^2 + (1)^2 + (1)^2}}\) A1: Correct applied equation. | A1 |
| \(\sin\theta = \dfrac{\sqrt{110}}{\sqrt{21}.\sqrt{6}} \Rightarrow \theta = 69.1238974\ldots = 69.1\ \text{(1 dp)}\) | A1: awrt 69.1 |
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OA} = \begin{pmatrix} 4 \\ 16 \\ -3 \end{pmatrix},\quad \overrightarrow{OP} = \begin{pmatrix} 9 \\ 13 \\ -3 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 9 + \lambda \\ 13 + 4\lambda \\ -3 - 2\lambda \end{pmatrix}\) | |
| \(\overrightarrow{AP} = \begin{pmatrix} 9 + \lambda \\ 13 + 4\lambda \\ -3 - 2\lambda \end{pmatrix} - \begin{pmatrix} 4 \\ 16 \\ -3 \end{pmatrix} = \begin{pmatrix} \lambda + 5 \\ 4\lambda - 3 \\ -2\lambda \end{pmatrix}\) | M1 A1 |
| \(\overrightarrow{AP} \bullet \mathbf{d}_1 = 0 \Rightarrow \begin{pmatrix} \lambda + 5 \\ 4\lambda - 3 \\ -2\lambda \end{pmatrix} \bullet \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = \lambda + 5 + 16\lambda - 12 + 4\lambda = 0\) | dM1 |
| leading to \(\{21\lambda - 7 = 0 \Rightarrow\}\ \lambda = \dfrac{1}{3}\) \(\lambda = \dfrac{1}{3}\) | A1 |
| Position vector \(\overrightarrow{OP} = \begin{pmatrix} 9 \\ 13 \\ -3 \end{pmatrix} + \frac{1}{3}\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 9\frac{1}{3} \\ 14\frac{1}{3} \\ -3\frac{2}{3} \end{pmatrix}\text{ or }\begin{pmatrix} \frac{28}{3} \\ \frac{43}{3} \\ -\frac{11}{3} \end{pmatrix}\) | ddM1 A1 |
| (6) | |
| (14 marks) |
Notes
M1: Attempts to find \(\overrightarrow{AP}\) in terms of the parameter by subtracting the components of \(\overrightarrow{OP}\) from \(l_1\) and \(\overrightarrow{OA}\). Ignore the direction of subtraction and ignore any confusion between \(\overrightarrow{OP}\) and \(\overrightarrow{PO}\) or between \(\overrightarrow{OA}\) and \(\overrightarrow{AO}\). The correct subtraction of two components is enough to establish that subtraction is intended. The coordinates or position vector of \(P\) must be given in terms of a parameter. Taking \(P: (x, y, z)\) gains no marks although this can be recovered later. See Additional Solutions.
A1: (M1 on epen) A correct expression for \(\overrightarrow{AP}\). Again accept the reverse direction.
dM1: Depends on the previous M. Taking the scalar product of their expression for \(\overrightarrow{AP}\) with \(\mathbf{d}_1\) or a multiple of \(\mathbf{d}_1\) and equating to 0 and obtaining an equation for \(\lambda\). The equation must derive from an expression of the form \(x_1x_2 + y_1y_2 + z_1z_2 = 0\). Differentiation can be used. See Additional Solutions.
A1: Solving to find \(\lambda = \tfrac{1}{3}\).
ddM1: Depends on both previous Ms. Substitutes their value of the parameter into their expression for \(\overrightarrow{OP}\). Substituting into \(\overrightarrow{AP}\) is a common error which loses the mark.
Note: Needs 2 correct co-ordinates if \(\lambda = \tfrac{1}{3}\) found and then \(P\) stated without method to gain ddM1.
A1: \(9\dfrac{1}{3}\mathbf{i} + 14\dfrac{1}{3}\mathbf{j} - 3\dfrac{2}{3}\mathbf{k}\). Accept vector notation or coordinates. Must be exact.
Additional Solution 1:
| Scheme | Marks |
|---|---|
| Taking \(\overrightarrow{OP} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\), in itself, can gain no marks but this may be converted to a parameter at a later stage in the solution and, at that stage, any relevant marks can be awarded. | |
| For example, \(\overrightarrow{AP} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} 4 \\ 16 \\ -3 \end{pmatrix} = \begin{pmatrix} x - 4 \\ y - 16 \\ z + 3 \end{pmatrix}\) leading to: \(\begin{pmatrix} x - 4 \\ y - 16 \\ z + 3 \end{pmatrix} \bullet \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = x - 4 + 4y - 64 - 2z - 6 = 0\) | No marks gained at this stage. |
| Using, \(\overrightarrow{OP} = \begin{pmatrix} 9 \\ 13 \\ -3 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 9 + \lambda \\ 13 + 4\lambda \\ -3 - 2\lambda \end{pmatrix}\) on \(x + 4y - 2z = 74\) which gives: \(9 + \lambda + 4(13 + 4\lambda) - 2(-3 - 2\lambda) = 74\) | M1A1 dM1 |
| \(\Rightarrow 21\lambda + 67 = 74 \Rightarrow \underline{\lambda = \dfrac{1}{3}}\) | A1 |
| Position vector \(\overrightarrow{OP} = \begin{pmatrix} 9 \\ 13 \\ -3 \end{pmatrix} + \frac{1}{3}\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 9\frac{1}{3} \\ 14\frac{1}{3} \\ -3\frac{2}{3} \end{pmatrix}\text{ or }\begin{pmatrix} \frac{28}{3} \\ \frac{43}{3} \\ -\frac{11}{3} \end{pmatrix}\) | ddM1 A1 |
At the \(9 + \lambda + 4(13 + 4\lambda) - 2(-3 - 2\lambda) = 74\) stage award M1A1 and dM1 (which is implied by an equation). A1: Solving to find \(\lambda = \dfrac{1}{3}\).
Additional Solution 2: Using Differentiation
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AP} = \begin{pmatrix} 9 + \lambda \\ 13 + 4\lambda \\ -3 - 2\lambda \end{pmatrix} - \begin{pmatrix} 4 \\ 16 \\ -3 \end{pmatrix} = \begin{pmatrix} \lambda + 5 \\ 4\lambda - 3 \\ -2\lambda \end{pmatrix}\) M1A1: As main scheme | M1A1 |
| \(AP^2 = (\lambda + 5)^2 + (4\lambda - 3)^2 + (-2\lambda)^2 = \left\{21\lambda^2 - 14\lambda + 34\right\}\) \(\dfrac{\mathrm{d}}{\mathrm{d}\lambda}\left(AP^2\right) = 42\lambda - 14 = 0\) | M1 |
| leading to \(\lambda = \dfrac{1}{3}\) A1: Solving to find \(\lambda = \dfrac{1}{3}\). | A1 |
... then apply the main scheme.