| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line intersection: unknown constant then intersect |
| Difficulty | Standard +0.3 This is a standard C4 vectors question requiring routine application of the scalar product formula for angles and solving simultaneous equations for intersection. The angle calculation is straightforward, and finding the intersection point involves equating components and solving for parameters—a typical textbook exercise with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using \(\begin{pmatrix}-8\\1\\-2\end{pmatrix}\) and \(\begin{pmatrix}-9\\2\\-5\end{pmatrix}\) as the relevant vectors | M1 | i.e. correct direction vectors |
| Using \(\cos\theta = \frac{a \cdot b}{ | a | |
| Method for scalar product of any 2 vectors | M1 | |
| Method for finding magnitude of any vector | M1 | |
| \(15°\) \((15.38\ldots)\), \(0.268\) rad | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Produce (at least) 2 of the 3 eqns in \(t\) and \(s\) | M1 | e.g. \(4-8t=-2-9s\), \(-6-2t=-2-5s\) |
| Solve the \((x)\) and \((z)\) equations | M1 | |
| \(t=3\) or \(s=2\) | A1 | for first value found |
| \(s=2\) or \(t=3\) f.t. | A1\(\sqrt{}\) | for second value found |
| Substituting their \((t,s)\) into \((y)\) equation | M1 | |
| \(a=1\) | A1 | |
| Substituting their \(t\) into \(l_1\) or their \((s,a)\) into \(l_2\) | M1 | |
| \(\begin{pmatrix}-20\\5\\-12\end{pmatrix}\) | A1 | 8 Any format but not \(\begin{pmatrix}\ \end{pmatrix}+\begin{pmatrix}\ \end{pmatrix}\) |
# Question 9:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using $\begin{pmatrix}-8\\1\\-2\end{pmatrix}$ and $\begin{pmatrix}-9\\2\\-5\end{pmatrix}$ as the relevant vectors | M1 | i.e. correct direction vectors |
| Using $\cos\theta = \frac{a \cdot b}{|a||b|}$ AEF for any 2 vectors | M1 | Accept $\cos\theta = \left|\frac{a\cdot b}{|a||b|}\right|$ |
| Method for scalar product of any 2 vectors | M1 | |
| Method for finding magnitude of any vector | M1 | |
| $15°$ $(15.38\ldots)$, $0.268$ rad | A1 | **5** |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Produce (at least) 2 of the 3 eqns in $t$ and $s$ | M1 | e.g. $4-8t=-2-9s$, $-6-2t=-2-5s$ |
| Solve the $(x)$ and $(z)$ equations | M1 | |
| $t=3$ or $s=2$ | A1 | for first value found |
| $s=2$ or $t=3$ f.t. | A1$\sqrt{}$ | for second value found |
| Substituting their $(t,s)$ into $(y)$ equation | M1 | |
| $a=1$ | A1 | |
| Substituting their $t$ into $l_1$ or their $(s,a)$ into $l_2$ | M1 | |
| $\begin{pmatrix}-20\\5\\-12\end{pmatrix}$ | A1 | **8** Any format but not $\begin{pmatrix}\ \end{pmatrix}+\begin{pmatrix}\ \end{pmatrix}$ |9 Two lines have vector equations
$$\mathbf { r } = \left( \begin{array} { r }
4 \\
2 \\
- 6
\end{array} \right) + t \left( \begin{array} { r }
- 8 \\
1 \\
- 2
\end{array} \right) \quad \text { and } \quad \mathbf { r } = \left( \begin{array} { r }
- 2 \\
a \\
- 2
\end{array} \right) + s \left( \begin{array} { r }
- 9 \\
2 \\
- 5
\end{array} \right) ,$$
where $a$ is a constant.\\
(i) Calculate the acute angle between the lines.\\
(ii) Given that these two lines intersect, find $a$ and the point of intersection.
\hfill \mbox{\textit{OCR C4 2006 Q9 [13]}}