| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Point on line satisfying distance or other condition |
| Difficulty | Standard +0.3 This is a standard C4 vectors question requiring dot product for perpendicularity (part i) and solving a quadratic from |OQ|=3 (part ii). Both parts use routine techniques with straightforward algebra, making it slightly easier than average but still requiring proper method application. |
| Spec | 1.10f Distance between points: using position vectors4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1+t\\-t\\2\end{pmatrix}\) | B1 | |
| \(\begin{pmatrix}x\\y\\z\end{pmatrix} \cdot \begin{pmatrix}1\\-1\\0\end{pmatrix} = 0\) | M1 | |
| \(\begin{pmatrix}\frac{1}{2}\\\frac{1}{2}\\2\end{pmatrix}\) or \(\frac{1}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + 2\mathbf{k}\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1+t)^2 + t^2 + 4 = 3^2\) or \(\sqrt{(1+t)^2 + t^2 + 4} = 3\) | M1 | FT from their (i) \(P\). SR: If A0A0 award A1A0 for either value of \(t\) leading to its correct answer |
| \(t = 1\) or \(-2\) | A1 | |
| \(\begin{pmatrix}2\\-1\\2\end{pmatrix}\) and \(\begin{pmatrix}-1\\2\\2\end{pmatrix}\) | A1 | |
| [3] |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1+t\\-t\\2\end{pmatrix}$ | B1 | |
| $\begin{pmatrix}x\\y\\z\end{pmatrix} \cdot \begin{pmatrix}1\\-1\\0\end{pmatrix} = 0$ | M1 | |
| $\begin{pmatrix}\frac{1}{2}\\\frac{1}{2}\\2\end{pmatrix}$ or $\frac{1}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + 2\mathbf{k}$ | A1 | |
| **[3]** | | |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1+t)^2 + t^2 + 4 = 3^2$ or $\sqrt{(1+t)^2 + t^2 + 4} = 3$ | M1 | FT from their (i) $P$. SR: If A0A0 award A1A0 for either value of $t$ leading to its correct answer |
| $t = 1$ or $-2$ | A1 | |
| $\begin{pmatrix}2\\-1\\2\end{pmatrix}$ and $\begin{pmatrix}-1\\2\\2\end{pmatrix}$ | A1 | |
| **[3]** | | |
---7 The equation of a straight line $l$ is
$$\mathbf { r } = \left( \begin{array} { l }
1 \\
0 \\
2
\end{array} \right) + t \left( \begin{array} { r }
1 \\
- 1 \\
0
\end{array} \right) .$$
$O$ is the origin.\\
(i) Find the position vector of the point $P$ on $l$ such that $O P$ is perpendicular to $l$.\\
(ii) A point $Q$ on $l$ is such that the length of $O Q$ is 3 units. Find the two possible position vectors of $Q$. [3]
\hfill \mbox{\textit{OCR C4 2012 Q7 [6]}}