| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Angle between two vectors/lines (direct) |
| Difficulty | Moderate -0.8 This is a straightforward vectors question requiring standard techniques: (i) applying the scalar product formula to find an angle between two position vectors, and (ii) using basic vector arithmetic to find a displacement vector. Both parts are routine calculations with no problem-solving insight required, making this easier than average for A-level. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Scalar product \(= 15-8+3 = 10 = | \mathbf{OA} | |
| \( | \mathbf{OA} | = \sqrt{26}, |
| Angle \(BOA = 71.4\) or \(71.5\) or \(1.25\) radians | M1, M1, M1, A1 [4] | Use of \(x_1x_2 + y_1y_2 + z_1z_2\); Correct magnitude for either; Linking everything correctly; cao |
| Answer | Marks |
|---|---|
| \(-6\mathbf{i} + 5\mathbf{j} + 4\mathbf{k}\) | M1, M1, A2,1,0 [4] |
**(i)** Scalar product $= 15-8+3 = 10 = |\mathbf{OA}||\mathbf{OB}|\cos\theta$
$|\mathbf{OA}| = \sqrt{26}, |\mathbf{OB}| = \sqrt{38}$
Angle $BOA = 71.4$ or $71.5$ or $1.25$ radians | M1, M1, M1, A1 [4] | Use of $x_1x_2 + y_1y_2 + z_1z_2$; Correct magnitude for either; Linking everything correctly; cao
**(ii)** $\mathbf{a}+\frac{1}{2}(\mathbf{b}-\mathbf{a})$ or $\mathbf{b}+\frac{1}{2}(\mathbf{a}-\mathbf{b})$ or $\frac{1}{2}(\mathbf{a}+\mathbf{b})$
$-2\mathbf{b}+$ their $c$ or $oe$
$-6\mathbf{i} + 5\mathbf{j} + 4\mathbf{k}$ | M1, M1, A2,1,0 [4]
---Relative to an origin $O$, the position vectors of points $A$ and $B$ are $\mathbf{3i} + 4\mathbf{j} - \mathbf{k}$ and $5\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}$ respectively.
\begin{enumerate}[label=(\roman*)]
\item Use a scalar product to find angle $BOA$. [4]
\end{enumerate}
The point $C$ is the mid-point of $AB$. The point $D$ is such that $\overrightarrow{OD} = 2\overrightarrow{OB}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find $\overrightarrow{DC}$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2011 Q6 [8]}}