C4 January 2009 Q4
4. With respect to a fixed origin \(O\) the lines \(l_1\) and \(l_2\) are given by the equations
\[l_1\colon\ \ \mathbf{r} = \begin{pmatrix}11\\2\\17\end{pmatrix} + \lambda\begin{pmatrix}-2\\1\\-4\end{pmatrix}\qquad\qquad l_2\colon\ \ \mathbf{r} = \begin{pmatrix}-5\\11\\p\end{pmatrix} + \mu\begin{pmatrix}q\\2\\2\end{pmatrix}\]
where \(\lambda\) and \(\mu\) are parameters and \(p\) and \(q\) are constants. Given that \(l_1\) and \(l_2\) are perpendicular,
Given further that \(l_1\) and \(l_2\) intersect, find
The point \(A\) lies on \(l_1\) and has position vector \(\begin{pmatrix}9\\3\\13\end{pmatrix}\). The point \(C\) lies on \(l_2\).
Given that a circle, with centre \(C\), cuts the line \(l_1\) at the points \(A\) and \(B\),
| Scheme | Marks |
|---|---|
| \(\mathbf{d}_1 = -2\mathbf{i} + \mathbf{j} - 4\mathbf{k},\quad \mathbf{d}_2 = q\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\) | |
| As \(\left\{\mathbf{d}_1 \bullet \mathbf{d}_2 = \begin{pmatrix}-2\\1\\-4\end{pmatrix} \bullet \begin{pmatrix}q\\2\\2\end{pmatrix}\right\} = \underline{(-2 \times q) + (1 \times 2) + (-4 \times 2)}\) | M1 |
| \(\mathbf{d}_1 \bullet \mathbf{d}_2 = 0 \Rightarrow -2q + 2 - 8 = 0\) \(\qquad -2q = 6 \Rightarrow \underline{q = -3}\) AG | A1 cso |
| (2) |
Notes
M1: Apply dot product calculation between two direction vectors, ie. \(\underline{(-2 \times q) + (1 \times 2) + (-4 \times 2)}\)
A1 cso: Sets \(\mathbf{d}_1 \bullet \mathbf{d}_2 = 0\) and solves to find \(\underline{q = -3}\)
4. (a) \(-2q + 2 - 8\) is sufficient for M1.
| Scheme | Marks |
|---|---|
| Lines meet where: \(\begin{pmatrix}11\\2\\17\end{pmatrix} + \lambda\begin{pmatrix}-2\\1\\-4\end{pmatrix} = \begin{pmatrix}-5\\11\\p\end{pmatrix} + \mu\begin{pmatrix}q\\2\\2\end{pmatrix}\) | |
| First two of \(\begin{aligned}\mathbf{i}&\colon\ 11 - 2\lambda = -5 + q\mu &&(1)\\ \mathbf{j}&\colon\ 2 + \lambda = 11 + 2\mu &&(2)\\ \mathbf{k}&\colon\ 17 - 4\lambda = p + 2\mu &&(3)\end{aligned}\) | M1 |
| (1) + 2(2) gives: \(15 = 17 + \mu \Rightarrow \mu = -2\) (2) gives: \(2 + \lambda = 11 - 4 \Rightarrow \lambda = 5\) | dM1 A1 A1 |
| (3) \(\Rightarrow 17 - 4(5) = p + 2(-2)\) | ddM1 |
| \(\Rightarrow p = 17 - 20 + 4 \Rightarrow \underline{p = 1}\) | A1 cso |
| (6) |
Notes
M1: Need to see equations (1) and (2). Condone one slip. (Note that \(q = -3\).)
dM1: Attempts to solve (1) and (2) to find one of either \(\lambda\) or \(\mu\)
A1: Any one of \(\underline{\lambda = 5}\) or \(\underline{\mu = -2}\) A1: Both \(\underline{\lambda = 5}\) and \(\underline{\mu = -2}\)
ddM1: Attempt to substitute their \(\lambda\) and \(\mu\) into their \(\mathbf{k}\) component to give an equation in \(p\) alone.
A1 cso: \(\underline{p = 1}\)
Aliter 4. (b) Way 2
Only apply Way 2 if candidate does not find both \(\lambda\) and \(\mu\).
| Scheme | Marks |
|---|---|
| Lines meet where: \(\begin{pmatrix}11\\2\\17\end{pmatrix} + \lambda\begin{pmatrix}-2\\1\\-4\end{pmatrix} = \begin{pmatrix}-5\\11\\p\end{pmatrix} + \mu\begin{pmatrix}q\\2\\2\end{pmatrix}\) | |
| First two of \(\begin{aligned}\mathbf{i}&\colon\ 11 - 2\lambda = -5 + q\mu &&(1)\\ \mathbf{j}&\colon\ 2 + \lambda = 11 + 2\mu &&(2)\\ \mathbf{k}&\colon\ 17 - 4\lambda = p + 2\mu &&(3)\end{aligned}\) | M1 |
| (2) gives \(\lambda = 9 + 2\mu\) | |
| (1) gives \(11 - 2(9 + 2\mu) = -5 - 3\mu\) \(\qquad 11 - 18 - 4\mu = -5 - 3\mu\) | dM1 |
| gives: \(11 - 18 + 5 = \mu \Rightarrow \mu = -2\) | A1 |
| (3) gives \(17 - 4(9 + 2\mu) = p + 2\mu\) | A1 |
| (3) \(\Rightarrow 17 - 4(9 + 2(-2)) = p + 2(-2)\) | ddM1 |
| \(\Rightarrow 17 - 20 = p - 4 \Rightarrow \underline{p = 1}\) | A1 cso |
| (6) |
M1: Need to see equations (1) and (2) (corrected from the printed mark scheme: it prints “(2) and (2)”). Condone one slip. (Note that \(q = -3\).)
dM1: Attempts to solve (1) and (2) to find one of either \(\lambda\) or \(\mu\). A1: Any one of \(\underline{\lambda = 5}\) or \(\underline{\mu = -2}\)
A1: Candidate writes down a correct equation containing \(p\) and one of either \(\lambda\) or \(\mu\) which has already been found.
ddM1: Attempt to substitute their value for \(\lambda\ (= 9 + 2\mu)\) and \(\mu\) into their \(\mathbf{k}\) component to give an equation in \(p\) alone. A1 cso: \(\underline{p = 1}\)
| Scheme | Marks |
|---|---|
| \(\mathbf{r} = \begin{pmatrix}11\\2\\17\end{pmatrix} + 5\begin{pmatrix}-2\\1\\-4\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}-5\\11\\1\end{pmatrix} - 2\begin{pmatrix}-3\\2\\2\end{pmatrix}\) | M1 |
| Intersect at \(\mathbf{r} = \underline{\begin{pmatrix}1\\7\\-3\end{pmatrix}}\) or \(\underline{(1, 7, -3)}\) | A1 |
| (2) |
Notes
M1: Substitutes their value of \(\lambda\) or \(\mu\) into the correct line \(l_1\) or \(l_2\).
A1: \(\begin{pmatrix}1\\7\\-3\end{pmatrix}\) or \(\underline{(1, 7, -3)}\)
4. (c) If no working is shown then any two out of the three coordinates can imply the first M1 mark.
| Scheme | Marks |
|---|---|
| Let \(\overrightarrow{OX} = \mathbf{i} + 7\mathbf{j} - 3\mathbf{k}\) be point of intersection | |
| \(\overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix}1\\7\\-3\end{pmatrix} - \begin{pmatrix}9\\3\\13\end{pmatrix} = \begin{pmatrix}-8\\4\\-16\end{pmatrix}\) | M1ft \(\pm\) |
| \(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OA} + 2\overrightarrow{AX}\) | |
| \(\overrightarrow{OB} = \begin{pmatrix}9\\3\\13\end{pmatrix} + 2\begin{pmatrix}-8\\4\\-16\end{pmatrix}\) | dM1ft |
| Hence, \(\overrightarrow{OB} = \underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\overrightarrow{OB} = \underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) | A1 |
| (3) | |
| (13 marks) |
Notes
M1ft \(\pm\): Finding vector \(\overrightarrow{AX}\) by finding the difference between \(\overrightarrow{OX}\) and \(\overrightarrow{OA}\). Can be ft using candidate’s \(\overrightarrow{OX}\).
dM1ft: \(\begin{pmatrix}9\\3\\13\end{pmatrix} + 2\left(\text{their } \overrightarrow{AX}\right)\)
A1: \(\underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) or \(\underline{(-7, 11, -19)}\)
Aliter 4. (d) Way 2
| Scheme | Marks |
|---|---|
| Let \(\overrightarrow{OX} = \mathbf{i} + 7\mathbf{j} - 3\mathbf{k}\) be point of intersection | |
| \(\overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix}1\\7\\-3\end{pmatrix} - \begin{pmatrix}9\\3\\13\end{pmatrix} = \begin{pmatrix}-8\\4\\-16\end{pmatrix}\) | M1ft \(\pm\) |
| \(\overrightarrow{OB} = \overrightarrow{OX} + \overrightarrow{XB} = \overrightarrow{OX} + \overrightarrow{AX}\) | |
| \(\overrightarrow{OB} = \begin{pmatrix}1\\7\\-3\end{pmatrix} + \begin{pmatrix}-8\\4\\-16\end{pmatrix}\) | dM1ft |
| Hence, \(\overrightarrow{OB} = \underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\overrightarrow{OB} = \underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) | A1 |
| (3) |
M1ft \(\pm\): Finding the difference between their \(\overrightarrow{OX}\) (can be implied) and \(\overrightarrow{OA}\). \(\overrightarrow{AX} = \pm\left(\begin{pmatrix}1\\7\\-3\end{pmatrix} - \begin{pmatrix}9\\3\\13\end{pmatrix}\right)\)
dM1ft: \(\left(\text{their } \overrightarrow{OX}\right) + \left(\text{their } \overrightarrow{AX}\right)\)
A1: \(\underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) or \(\underline{(-7, 11, -19)}\)
Aliter 4. (d) Way 3
| Scheme | Marks |
|---|---|
| At \(A\), \(\lambda = 1\). At \(X\), \(\lambda = 5\). | |
| Hence at \(B\), \(\lambda = 5 + (5 - 1) = 9\) | M1ft |
| \(\overrightarrow{OB} = \begin{pmatrix}11\\2\\17\end{pmatrix} + 9\begin{pmatrix}-2\\1\\-4\end{pmatrix}\) | dM1ft |
| Hence, \(\overrightarrow{OB} = \underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\overrightarrow{OB} = \underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) | A1 |
| (3) |
M1ft: \(\lambda_B = \left(\text{their } \lambda_X\right) + \left(\text{their } \lambda_X - \text{their } \lambda_A\right)\) or \(\lambda_B = 2\left(\text{their } \lambda_X\right) - \left(\text{their } \lambda_A\right)\)
dM1ft: Substitutes their value of \(\lambda\) into the line \(l_1\).
A1: \(\underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) or \(\underline{(-7, 11, -19)}\)
Aliter 4. (d) Way 4
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OA} = 9\mathbf{i} + 3\mathbf{j} + 13\mathbf{k}\) and the point of intersection \(\overrightarrow{OX} = \mathbf{i} + 7\mathbf{j} - 3\mathbf{k}\) | |
| \(\begin{pmatrix}9\\3\\13\end{pmatrix} \rightarrow \begin{pmatrix}\text{Minus } 8\\ \text{Plus } 4\\ \text{Minus } 16\end{pmatrix} \rightarrow \begin{pmatrix}1\\7\\-3\end{pmatrix}\) | M1ft \(\pm\) |
| \(\begin{pmatrix}1\\7\\-3\end{pmatrix} \rightarrow \begin{pmatrix}\text{Minus } 8\\ \text{Plus } 4\\ \text{Minus } 16\end{pmatrix} \rightarrow \begin{pmatrix}-7\\11\\-19\end{pmatrix}\) | dM1ft |
| Hence, \(\overrightarrow{OB} = \underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\overrightarrow{OB} = \underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) | A1 |
| (3) |
M1ft \(\pm\): Finding the difference between their \(\overrightarrow{OX}\) (can be implied) and \(\overrightarrow{OA}\). \(\left(\overrightarrow{AX} =\right) \pm\left(\begin{pmatrix}1\\7\\-3\end{pmatrix} - \begin{pmatrix}9\\3\\13\end{pmatrix}\right)\)
dM1ft: \(\left(\text{their } \overrightarrow{OX}\right) + \left(\text{their } \overrightarrow{AX}\right)\)
A1: \(\underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) or \(\underline{(-7, 11, -19)}\)
Aliter 4. (d) Way 5
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OA} = 9\mathbf{i} + 3\mathbf{j} + 13\mathbf{k}\) and \(\overrightarrow{OB} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) and the point of intersection \(\overrightarrow{OX} = \mathbf{i} + 7\mathbf{j} - 3\mathbf{k}\) | |
| As \(X\) is the midpoint of \(AB\), then | |
| \((1, 7, -3) = \left(\dfrac{9 + a}{2}, \dfrac{3 + b}{2}, \dfrac{13 + c}{2}\right)\) | M1ft |
| \(a = 2(1) - 9 = -7\) \(b = 2(7) - 3 = 11\) \(c = 2(-3) - 13 = -19\) | dM1ft |
| Hence, \(\overrightarrow{OB} = \underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\overrightarrow{OB} = \underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) | A1 |
| (3) |
M1ft: Writing down any two of these “equations” correctly. dM1ft: An attempt to find at least two of \(a\), \(b\) or \(c\).
A1: \(\underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) or \(\underline{(-7, 11, -19)}\) or \(\underline{a = -7, b = 11, c = -19}\)
Aliter 4. (d) Way 6
| Scheme | Marks |
|---|---|
| Let \(\overrightarrow{OX} = \mathbf{i} + 7\mathbf{j} - 3\mathbf{k}\) be point of intersection | |
| \(\overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix}1\\7\\-3\end{pmatrix} - \begin{pmatrix}9\\3\\13\end{pmatrix} = \begin{pmatrix}-8\\4\\-16\end{pmatrix}\) and \(\left|\overrightarrow{AX}\right| = \sqrt{64 + 16 + 256} = \sqrt{336} = 4\sqrt{21}\) | M1ft \(\pm\) |
| \(\overrightarrow{BX} = \overrightarrow{OX} - \overrightarrow{OB} = \begin{pmatrix}1\\7\\-3\end{pmatrix} - \begin{pmatrix}11 - 2\lambda\\2 + \lambda\\17 - 4\lambda\end{pmatrix} = \begin{pmatrix}-10 + 2\lambda\\5 - \lambda\\-20 + 4\lambda\end{pmatrix}\) | |
| Hence \(\left|\overrightarrow{BX}\right| = \left|\overrightarrow{AX}\right| = \sqrt{336}\) gives | |
| \(\left(-10 + 2\lambda\right)^2 + \left(5 - \lambda\right)^2 + \left(-20 + 4\lambda\right)^2 = 336\) | dM1ft |
| \(100 - 40\lambda + 4\lambda^2 + 25 - 10\lambda + \lambda^2 + 400 - 160\lambda + 16\lambda^2 = 336\) \(21\lambda^2 - 210\lambda + 525 = 336\) \(21\lambda^2 - 210\lambda + 189 = 0\) \(\lambda^2 - 10\lambda + 9 = 0\) \((\lambda - 1)(\lambda - 9) = 0\) | |
| At \(A\), \(\lambda = 1\) and at \(B\) \(\lambda = 9\), so, \(\overrightarrow{OB} = \begin{pmatrix}11 - 2(9)\\2 + 9\\17 - 4(9)\end{pmatrix}\) | |
| Hence, \(\overrightarrow{OB} = \underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\overrightarrow{OB} = \underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) | A1 |
| (3) |
M1ft \(\pm\): Finding the difference between their \(\overrightarrow{OX}\) (can be implied) and \(\overrightarrow{OA}\). \(\overrightarrow{AX} = \pm\left(\begin{pmatrix}1\\7\\-3\end{pmatrix} - \begin{pmatrix}9\\3\\13\end{pmatrix}\right)\)
Note \(\left|\overrightarrow{AX}\right| = \sqrt{336}\) would imply M1.
dM1ft: Writes distance equation of \(\left|\overrightarrow{BX}\right|^2 = 336\) where \(\overrightarrow{BX} = \overrightarrow{OX} - \overrightarrow{OB}\) and \(\overrightarrow{OB} = \begin{pmatrix}11 - 2\lambda\\2 + \lambda\\17 - 4\lambda\end{pmatrix}\)
A1: \(\underline{\begin{pmatrix}-7\\11\\-19\end{pmatrix}}\) or \(\underline{-7\mathbf{i} + 11\mathbf{j} - 19\mathbf{k}}\) or \(\underline{(-7, 11, -19)}\)