| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Show lines intersect and find intersection point |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question requiring finding intersection of lines (solving simultaneous equations), calculating an angle using dot product, and finding triangle area. All techniques are routine C4 material with straightforward application of formulas, making it slightly easier than average. |
| Spec | 4.03c Matrix multiplication: properties (associative, not commutative)4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | j components: \(3 + 2\lambda = 9 \Rightarrow \lambda = 3\) | M1 A1 |
| Leading to \(C:(5, 9, -1)\) | A1 | accept vector forms |
| (b) | Choosing correct directions or finding \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\) | M1 |
| \(\begin{pmatrix}1\\2\\1\end{pmatrix} \cdot \begin{pmatrix}5\\0\\2\end{pmatrix} = 5 + 2 = \sqrt{6} \times \sqrt{29}\cos \angle ACB\) | M1 A1 | use of scalar product |
| \(\angle ACB = 57.95°\) | A1 | awrt \(57.95°\) |
| (c) | \(A:(2, 3, -4)\), \(B:(-5, 9, -5)\) | |
| \(\overrightarrow{AC} = \begin{pmatrix}3\\6\\3\end{pmatrix}\), \(\overrightarrow{BC} = \begin{pmatrix}10\\0\\4\end{pmatrix}\) | ||
| \(AC^2 = 3^2 + 6^2 + 3^2 \Rightarrow AC = 3\sqrt{6}\) | M1 A1 | |
| \(BC^2 = 10^2 + 4^2 \Rightarrow BC = 2\sqrt{29}\) | A1 | |
| \(\triangle ABC = \frac{1}{2}AC \times BC \sin \angle ACB = \frac{1}{2} \times 3\sqrt{6} \times 2\sqrt{29} \sin \angle ACB \approx 33.5\) | M1 A1 | \(15\sqrt{5}\), awrt 34 |
| Answer | Marks | Guidance |
|---|---|---|
| \(A:(2, 3, -4)\), \(B:(-5, 9, -5)\), \(C:(5, 9, -1)\) | ||
| \(AB^2 = 7^2 + 6^2 + 1^2 = 86\) | ||
| \(AC^2 = 3^2 + 6^2 + 3^2 = 54\) | ||
| \(BC^2 = 10^2 + 0^2 + 4^2 = 116\) | M1 | Finding all three sides |
| \(\cos \angle ACB = \frac{116 + 54 - 86}{2\sqrt{116 \times 54}} = (0.53066\ldots)\) | M1 A1 | |
| \(\angle ACB = 57.95°\) | A1 | awrt \(57.95°\) |
(a) | j components: $3 + 2\lambda = 9 \Rightarrow \lambda = 3$ | M1 A1 | $(\mu = 1)$ |
| Leading to $C:(5, 9, -1)$ | A1 | accept vector forms |
(b) | Choosing correct directions or finding $\overrightarrow{AC}$ and $\overrightarrow{BC}$ | M1 | |
| $\begin{pmatrix}1\\2\\1\end{pmatrix} \cdot \begin{pmatrix}5\\0\\2\end{pmatrix} = 5 + 2 = \sqrt{6} \times \sqrt{29}\cos \angle ACB$ | M1 A1 | use of scalar product |
| $\angle ACB = 57.95°$ | A1 | awrt $57.95°$ |
(c) | $A:(2, 3, -4)$, $B:(-5, 9, -5)$ | | |
| $\overrightarrow{AC} = \begin{pmatrix}3\\6\\3\end{pmatrix}$, $\overrightarrow{BC} = \begin{pmatrix}10\\0\\4\end{pmatrix}$ | | |
| $AC^2 = 3^2 + 6^2 + 3^2 \Rightarrow AC = 3\sqrt{6}$ | M1 A1 | |
| $BC^2 = 10^2 + 4^2 \Rightarrow BC = 2\sqrt{29}$ | A1 | |
| $\triangle ABC = \frac{1}{2}AC \times BC \sin \angle ACB = \frac{1}{2} \times 3\sqrt{6} \times 2\sqrt{29} \sin \angle ACB \approx 33.5$ | M1 A1 | $15\sqrt{5}$, awrt 34 |
**Alternative method for (b) and (c):**
| $A:(2, 3, -4)$, $B:(-5, 9, -5)$, $C:(5, 9, -1)$ | | |
| $AB^2 = 7^2 + 6^2 + 1^2 = 86$ | | |
| $AC^2 = 3^2 + 6^2 + 3^2 = 54$ | | |
| $BC^2 = 10^2 + 0^2 + 4^2 = 116$ | M1 | Finding all three sides |
| $\cos \angle ACB = \frac{116 + 54 - 86}{2\sqrt{116 \times 54}} = (0.53066\ldots)$ | M1 A1 | |
| $\angle ACB = 57.95°$ | A1 | awrt $57.95°$ |
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).
---7. The line $l _ { 1 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 2 \\ 3 \\ - 4 \end{array} \right) + \lambda \left( \begin{array} { l } 1 \\ 2 \\ 1 \end{array} \right)$, where $\lambda$ is a scalar parameter.
The line $l _ { 2 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 0 \\ 9 \\ - 3 \end{array} \right) + \mu \left( \begin{array} { l } 5 \\ 0 \\ 2 \end{array} \right)$, where $\mu$ is a scalar parameter.\\
Given that $l _ { 1 }$ and $l _ { 2 }$ meet at the point $C$, find
\begin{enumerate}[label=(\alph*)]
\item the coordinates of $C$.
The point $A$ is the point on $l _ { 1 }$ where $\lambda = 0$ and the point $B$ is the point on $l _ { 2 }$ where $\mu = - 1$.
\item Find the size of the angle $A C B$. Give your answer in degrees to 2 decimal places.
\item Hence, or otherwise, find the area of the triangle $A B C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2010 Q7 [12]}}