C4 June 2010 Q7
7. The line \(l_1\) has equation \(\mathbf{r} = \begin{pmatrix}2\\3\\-4\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\1\end{pmatrix}\), where \(\lambda\) is a scalar parameter.
The line \(l_2\) has equation \(\mathbf{r} = \begin{pmatrix}0\\9\\-3\end{pmatrix} + \mu\begin{pmatrix}5\\0\\2\end{pmatrix}\), where \(\mu\) is a scalar parameter.
Given that \(l_1\) and \(l_2\) meet at the point \(C\), find
(a) the coordinates of \(C\). (3)
The point \(A\) is the point on \(l_1\) where \(\lambda = 0\) and the point \(B\) is the point on \(l_2\) where \(\mu = -1\).
(b) Find the size of the angle \(ACB\). Give your answer in degrees to 2 decimal places. (4)
(c) Hence, or otherwise, find the area of the triangle \(ABC\). (5)
| Scheme | Marks |
|---|---|
| \(\mathbf{j}\) components \(3 + 2\lambda = 9 \Rightarrow \lambda = 3\) \((\mu = 1)\) | M1 A1 |
| Leading to \(C{:}\ (5, 9, -1)\) accept vector forms | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Choosing correct directions or finding \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\) | M1 |
| \(\begin{pmatrix}1\\2\\1\end{pmatrix}.\begin{pmatrix}5\\0\\2\end{pmatrix} = 5 + 2 = \sqrt{6}\sqrt{29}\cos\angle ACB\) use of scalar product | M1 A1 |
| \(\angle ACB = 57.95^\circ\) awrt \(57.95^\circ\) | A1 |
| (4) |
Alternative method for (b) and (c)
| \(A{:}\ (2, 3, -4)\quad B{:}\ (-5, 9, -5)\quad C{:}\ (5, 9, -1)\) | |
| \(AB^2 = 7^2 + 6^2 + 1^2 = 86\) \(AC^2 = 3^2 + 6^2 + 3^2 = 54\) \(BC^2 = 10^2 + 0^2 + 4^2 = 116\) Finding all three sides | M1 |
| \(\cos\angle ACB = \dfrac{116 + 54 - 86}{2\sqrt{116}\sqrt{54}}\quad (= 0.530\,66\ \ldots)\) | M1 A1 |
| \(\angle ACB = 57.95^\circ\) awrt \(57.95^\circ\) | A1 |
| (4) |
If this method is used some of the working may gain credit in part (c) and appropriate marks may be awarded if there is an attempt at part (c).
| Scheme | Marks |
|---|---|
| \(A{:}\ (2, 3, -4)\quad B{:}\ (-5, 9, -5)\) | |
| \(\overrightarrow{AC} = \begin{pmatrix}3\\6\\3\end{pmatrix},\quad \overrightarrow{BC} = \begin{pmatrix}10\\0\\4\end{pmatrix}\) | |
| \(AC^2 = 3^2 + 6^2 + 3^2 \Rightarrow AC = 3\sqrt{6}\) | M1 A1 |
| \(BC^2 = 10^2 + 4^2 \Rightarrow BC = 2\sqrt{29}\) | A1 |
| \(\triangle ABC = \dfrac{1}{2}AC \times BC\sin\angle ACB\) \(= \dfrac{1}{2}3\sqrt{6} \times 2\sqrt{29}\sin\angle ACB \approx 33.5\) \(15\sqrt{5}\), awrt 34 | M1 A1 |
| (5) | |
| (12 marks) |