| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Parallel and perpendicular lines |
| Difficulty | Standard +0.3 This is a straightforward 3D vectors question requiring basic operations: finding direction vectors, computing a scalar product to verify perpendicularity, and using the parallel condition with a given magnitude to find an unknown position vector. All techniques are standard and the multi-step nature is typical for this topic, 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 multiplication1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((\mathbf{b}-\mathbf{a}).(\mathbf{b}-\mathbf{c}) = \begin{pmatrix}-2\\-1\\2\end{pmatrix} \begin{pmatrix}3\\2\\4\end{pmatrix}\) → \(-6-2+8=0\) → \(90°\) | M1 M1 A1 [3] | AB = b − a once (a − b is ok) Use of x₁y₁z₁ with AB and CB All correct |
| (ii) Unit vector = \(\frac{1}{3}\begin{pmatrix}2\\1\\-2\end{pmatrix}\) | M1 | Method for unit vector. |
| CD = 12 × unit vector = \(\pm\begin{pmatrix}8\\4\\-8\end{pmatrix}\) | M1 | Knows to multiply by 12 or ±4BA |
| OD = OC + CD = \(\begin{pmatrix}12\\9\\-2\end{pmatrix}\) | M1 A1 [4] | Correct method. co |
**(i)** $(\mathbf{b}-\mathbf{a}).(\mathbf{b}-\mathbf{c}) = \begin{pmatrix}-2\\-1\\2\end{pmatrix} \begin{pmatrix}3\\2\\4\end{pmatrix}$ → $-6-2+8=0$ → $90°$ | M1 M1 A1 [3] | **AB = b − a** once (**a − b** is ok) Use of x₁y₁z₁ with AB and CB All correct
**(ii)** Unit vector = $\frac{1}{3}\begin{pmatrix}2\\1\\-2\end{pmatrix}$ | M1 | Method for unit vector.
**CD = 12 × unit vector** = $\pm\begin{pmatrix}8\\4\\-8\end{pmatrix}$ | M1 | Knows to multiply by 12 or ±4BA
**OD = OC + CD** = $\begin{pmatrix}12\\9\\-2\end{pmatrix}$ | M1 A1 [4] | Correct method. co7\\
\includegraphics[max width=\textwidth, alt={}, center]{1a4ddaa9-1ec2-4138-bfcb-a482fe6c942f-3_394_750_260_699}
The diagram shows a trapezium $A B C D$ in which $B A$ is parallel to $C D$. The position vectors of $A , B$ and $C$ relative to an origin $O$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { l }
3 \\
4 \\
0
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { l }
1 \\
3 \\
2
\end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { l }
4 \\
5 \\
6
\end{array} \right)$$
(i) Use a scalar product to show that $A B$ is perpendicular to $B C$.\\
(ii) Given that the length of $C D$ is 12 units, find the position vector of $D$.
\hfill \mbox{\textit{CAIE P1 2014 Q7 [7]}}