| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2003 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Angles between vectors |
| Difficulty | Standard +0.3 This is a standard 3D vectors question requiring coordinate setup, vector arithmetic, and dot product application to find an angle. While it involves multiple steps and 3D visualization, the techniques are routine for Further Maths students: finding coordinates from geometric constraints, expressing vectors in component form, and using the scalar product formula. The prism setup is clearly described and the calculations are straightforward once the coordinate system is established. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Height \(= 4\) | B1 [1] | Pythagoras or guess – anywhere, 4k ok. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\vec{MN} = 6\mathbf{j} - 4\mathbf{k}\) | B2, 1 ∇; B1, ∇ [3] | \(\surd\) for "4". Special case B1 for "\(3\mathbf{i}+6\mathbf{j}+4\mathbf{k}\)" on "4". Accept column vectors. (nb if (i) incorrect, but answers are correct in (iii) allow feedback). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\rightarrow \theta = 111°\) | M1, A1, ∇, M1, A1 [4] | Use of \(x_1y_1+x_2y_2+x_3y_3\). \(\surd\) on MC and MN. Product of two moduli and cos \(\theta\). Co. Nb If both MC and MN "reversed", allow \(111°\) for full marks. |
**(i)** Height $= 4$ | B1 [1] | Pythagoras or guess – anywhere, 4k ok.
**(ii)** $\vec{MC} = 3\mathbf{i}+6\mathbf{j}-4\mathbf{k}$
$\vec{MN} = 6\mathbf{j} - 4\mathbf{k}$ | B2, 1 ∇; B1, ∇ [3] | $\surd$ for "4". Special case B1 for "$3\mathbf{i}+6\mathbf{j}+4\mathbf{k}$" on "4". Accept column vectors. (nb if (i) incorrect, but answers are correct in (iii) allow feedback).
**(iii)** $\vec{MC}.\vec{MN} = -36+16 = -20$
$\vec{MC}.\vec{MN} = \sqrt{61}\sqrt{52}\cos\theta$
$\rightarrow \theta = 111°$ | M1, A1, ∇, M1, A1 [4] | Use of $x_1y_1+x_2y_2+x_3y_3$. $\surd$ on MC and MN. Product of two moduli and cos $\theta$. Co. Nb If both MC and MN "reversed", allow $111°$ for full marks.7\\
\includegraphics[max width=\textwidth, alt={}, center]{1cf37a58-8a7f-4dc8-9e35-2e8badf3eb83-3_636_1047_1153_550}
The diagram shows a triangular prism with a horizontal rectangular base $A D F C$, where $C F = 12$ units and $D F = 6$ units. The vertical ends $A B C$ and $D E F$ are isosceles triangles with $A B = B C = 5$ units. The mid-points of $B E$ and $D F$ are $M$ and $N$ respectively. The origin $O$ is at the mid-point of $A C$.
Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O C , O N$ and $O B$ respectively.\\
(i) Find the length of $O B$.\\
(ii) Express each of the vectors $\overrightarrow { M C }$ and $\overrightarrow { M N }$ in terms of $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$.\\
(iii) Evaluate $\overrightarrow { M C } \cdot \overrightarrow { M N }$ and hence find angle $C M N$, giving your answer correct to the nearest degree.
\hfill \mbox{\textit{CAIE P1 2003 Q7 [8]}}