| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| 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 two-part question requiring basic vector subtraction to find position vectors, then direct application of the scalar product formula (cos θ = a·b/|a||b|) to find an angle. Both parts are standard textbook exercises with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors |
| Answer | Marks |
|---|---|
| \(\overrightarrow{DA} = 6\mathbf{i} - 4\mathbf{k}\) | B1 |
| \(\overrightarrow{CA} = 6\mathbf{i} - 5\mathbf{j} - 4\mathbf{k}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Method marks awarded only for their vectors \(\pm\overrightarrow{CA}\) & \(\pm\overrightarrow{DA}\) | Full marks can be obtained using \(\overrightarrow{AC}\) & \(\overrightarrow{AD}\) | |
| \(\overrightarrow{CA} \cdot \overrightarrow{DA} = 36 + 16\ (= 52)\) | M1 | Using \(x_1x_2 + y_1y_2 + z_1z_2\) |
| \( | \overrightarrow{DA} | = \sqrt{52}\), \( |
| \(52 = \sqrt{77}\sqrt{52}\cos C\hat{A}D\) | M1 | All linked correctly |
| \(\cos C\hat{A}D = 0.82178... \rightarrow C\hat{A}D = 34.7°\) or \(0.606^c\) awrt | A1 | Answer must come from +ve cosine ratio |
## Question 5(i):
$\overrightarrow{DA} = 6\mathbf{i} - 4\mathbf{k}$ | B1 |
$\overrightarrow{CA} = 6\mathbf{i} - 5\mathbf{j} - 4\mathbf{k}$ | B1 |
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## Question 5(ii):
Method marks awarded only for **their** vectors $\pm\overrightarrow{CA}$ & $\pm\overrightarrow{DA}$ | | Full marks can be obtained using $\overrightarrow{AC}$ & $\overrightarrow{AD}$
$\overrightarrow{CA} \cdot \overrightarrow{DA} = 36 + 16\ (= 52)$ | M1 | Using $x_1x_2 + y_1y_2 + z_1z_2$
$|\overrightarrow{DA}| = \sqrt{52}$, $|\overrightarrow{CA}| = \sqrt{77}$ | M1 | Uses modulus twice
$52 = \sqrt{77}\sqrt{52}\cos C\hat{A}D$ | M1 | All linked correctly
$\cos C\hat{A}D = 0.82178... \rightarrow C\hat{A}D = 34.7°$ or $0.606^c$ awrt | A1 | Answer must come from +ve cosine ratio
---(i) Express each of the vectors $\overrightarrow { D A }$ and $\overrightarrow { C A }$ in terms of $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$.\\
(ii) Use a scalar product to find angle $C A D$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q5 [6]}}