| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2003 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Parallel and perpendicular lines |
| Difficulty | Moderate -0.3 This is a straightforward application of standard vector techniques: computing direction vectors, using scalar product to verify perpendicularity (routine calculation), and checking parallelism by showing vectors are scalar multiples. The question is slightly easier than average because it's highly structured with clear instructions and involves only basic vector operations without requiring problem-solving insight or proof construction. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\overrightarrow{BA} = a - b = i + 2j - 3k\); \(\overrightarrow{BC} = c - b = -2i + 4j + 2k\) | M1, M1A1∨ [4] | Knowing how to use position vector for \(\overrightarrow{BA}\) or \(\overrightarrow{BC}\) – not for \(\overrightarrow{AB}\) or \(\overrightarrow{CB}\). Knowing how to use \(x_1y_1 + x_2y_2 + x_3y_3\). Co. Correct deduction. Beware fortuitous (uses \(\overrightarrow{AB}\) or \(\overrightarrow{CB}\) – can get 3 out of 4) |
| Dot product = -2 + 8 - 6 = 0 \(\rightarrow\) Perpendicular | A1 [4] | |
| (ii) \(\overrightarrow{BC} = c - b = -2i + 4j + 2k\); \(\overrightarrow{AD} = d - a = -5i + 10j + 5k\) | M1 [2] | Knowing how to get one of these |
| These are in the same ratio \ parallel | M1 [2] | Both correct + conclusion. Could be dot product = 60 \(\rightarrow\) angle = 0° |
| Ratio = 2:5 (or \(\sqrt{24}\): \(\sqrt{150}\)) | M1A1 [4] | Knowing what to do. Co. Allow 5:2 |
**(i)** $\overrightarrow{BA} = a - b = i + 2j - 3k$; $\overrightarrow{BC} = c - b = -2i + 4j + 2k$ | M1, M1A1∨ [4] | Knowing how to use position vector for $\overrightarrow{BA}$ or $\overrightarrow{BC}$ – not for $\overrightarrow{AB}$ or $\overrightarrow{CB}$. Knowing how to use $x_1y_1 + x_2y_2 + x_3y_3$. Co. Correct deduction. Beware fortuitous (uses $\overrightarrow{AB}$ or $\overrightarrow{CB}$ – can get 3 out of 4)
Dot product = -2 + 8 - 6 = 0 $\rightarrow$ Perpendicular | A1 [4]
**(ii)** $\overrightarrow{BC} = c - b = -2i + 4j + 2k$; $\overrightarrow{AD} = d - a = -5i + 10j + 5k$ | M1 [2] | Knowing how to get one of these
These are in the same ratio \ parallel | M1 [2] | Both correct + conclusion. Could be dot product = 60 $\rightarrow$ angle = 0°
Ratio = 2:5 (or $\sqrt{24}$: $\sqrt{150}$) | M1A1 [4] | Knowing what to do. Co. Allow 5:28 The points $A , B , C$ and $D$ have position vectors $3 \mathbf { i } + 2 \mathbf { k } , 2 \mathbf { i } - 2 \mathbf { j } + 5 \mathbf { k } , 2 \mathbf { j } + 7 \mathbf { k }$ and $- 2 \mathbf { i } + 10 \mathbf { j } + 7 \mathbf { k }$ respectively.\\
(i) Use a scalar product to show that $B A$ and $B C$ are perpendicular.\\
(ii) Show that $B C$ and $A D$ are parallel and find the ratio of the length of $B C$ to the length of $A D$.
\hfill \mbox{\textit{CAIE P1 2003 Q8 [8]}}