| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Angle between two vectors/lines (direct) |
| Difficulty | Moderate -0.5 This is a straightforward vectors question requiring standard techniques: finding a vector between two points, calculating a unit vector (magnitude then division), finding a midpoint, and using the scalar product formula for angles. All steps are routine applications of formulas with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\overrightarrow{CD} = -3\begin{pmatrix}1\\2\end{pmatrix} + 2\begin{pmatrix}4\\6\end{pmatrix} = \begin{pmatrix}-3\\-6\end{pmatrix}\) | B1 | |
| Unit vector \(= \frac{1}{7}\begin{pmatrix}-2\\-3\\6\end{pmatrix}\) | M1A1 | Allow M1A1 for \(\frac{1}{7}\begin{pmatrix}-2\\-3\\6\end{pmatrix}\) or \((\frac{-2}{7}, \frac{-3}{7})\) etc |
| (ii) \(\overrightarrow{OE} = \begin{pmatrix}6\\6\end{pmatrix} + (\begin{pmatrix}1\\-1\frac{1}{2}\end{pmatrix}) = \begin{pmatrix}7\\1\frac{1}{2}\end{pmatrix}\) | M1A1 | or equivalent method |
| \(\overrightarrow{OE} \cdot \overrightarrow{OD} = 56 + 0 + 108 = 164\) | M1 | Use of \(x_1x_2 + y_1y_2 + z_1z_2\) |
| \( | \overrightarrow{OE} | = \sqrt{132.25}(= 11.5)\), \( |
| \(164 = \sqrt{132.25} \times \sqrt{208} \times \cos\theta\) | M1 | All connected correctly. Dependent on \(\overrightarrow{OE}, \overrightarrow{OD}, \overrightarrow{DO}\) used |
| \(\theta = 8.6°\) cao | A1 | [6] |
**(i)** $\overrightarrow{CD} = -3\begin{pmatrix}1\\2\end{pmatrix} + 2\begin{pmatrix}4\\6\end{pmatrix} = \begin{pmatrix}-3\\-6\end{pmatrix}$ | B1 |
Unit vector $= \frac{1}{7}\begin{pmatrix}-2\\-3\\6\end{pmatrix}$ | M1A1 | Allow M1A1 for $\frac{1}{7}\begin{pmatrix}-2\\-3\\6\end{pmatrix}$ or $(\frac{-2}{7}, \frac{-3}{7})$ etc | [3]
**(ii)** $\overrightarrow{OE} = \begin{pmatrix}6\\6\end{pmatrix} + (\begin{pmatrix}1\\-1\frac{1}{2}\end{pmatrix}) = \begin{pmatrix}7\\1\frac{1}{2}\end{pmatrix}$ | M1A1 | or equivalent method
$\overrightarrow{OE} \cdot \overrightarrow{OD} = 56 + 0 + 108 = 164$ | M1 | Use of $x_1x_2 + y_1y_2 + z_1z_2$
$|\overrightarrow{OE}| = \sqrt{132.25}(= 11.5)$, $|\overrightarrow{OD}| = \sqrt{208}$ | M1 | Correct method for moduli
$164 = \sqrt{132.25} \times \sqrt{208} \times \cos\theta$ | M1 | All connected correctly. Dependent on $\overrightarrow{OE}, \overrightarrow{OD}, \overrightarrow{DO}$ used
$\theta = 8.6°$ cao | A1 | [6]9 The position vectors of points $A$ and $B$ relative to an origin $O$ are $\mathbf { a }$ and $\mathbf { b }$ respectively. The position vectors of points $C$ and $D$ relative to $O$ are $3 \mathbf { a }$ and $2 \mathbf { b }$ respectively. It is given that
$$\mathbf { a } = \left( \begin{array} { l }
2 \\
1 \\
2
\end{array} \right) \quad \text { and } \quad \mathbf { b } = \left( \begin{array} { l }
4 \\
0 \\
6
\end{array} \right) .$$
(i) Find the unit vector in the direction of $\overrightarrow { C D }$.\\
(ii) The point $E$ is the mid-point of $C D$. Find angle $E O D$.
\hfill \mbox{\textit{CAIE P1 2012 Q9 [9]}}