| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Moderate -0.3 This is a straightforward multi-part vectors question testing standard techniques: (i) perpendicularity via dot product = 0, (ii) angle formula using scalar product, (iii) magnitude calculation and solving a quadratic. All parts are routine applications of basic vector formulas with no novel problem-solving required, making it slightly easier than average. |
| 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) \((2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \cdot (3\mathbf{i} - 2\mathbf{j} + p\mathbf{k}) = 0\) | M1 | For \(x_1x_2+y_1y_2+z_1z_2\) (in (i) or (ii)) |
| \(\rightarrow 6 - 2 + 2p = 0\) | A1 | co |
| \(\rightarrow p = -2\) | [2] | |
| (ii) \((2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \cdot (3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k})\) | A1 | nb Part (ii) gains 4 marks if (i) missing. co (M1 here if (i) not done) |
| \(\rightarrow 6 - 2 + 12\) allow for \(\pm\) this | M1 | All connected correctly |
| \(= \sqrt{9} \times \sqrt{49}\cos\theta\) | A1 | co [3] |
| \(\rightarrow \theta = 40°\) | ||
| (iii) \(\mathbf{AB} = \mathbf{i} - 3\mathbf{j} + (p - 2)\mathbf{k}\) | B1 | Must be for AB, not BA. |
| \(1^2 + 3^2 + (p - 2)^2 = 3.5^2\) | M1 | Pythagoras (allow if √ wrong once) |
| DM1 | Method of solution. | |
| \(\rightarrow p = 0.5\) or \(3.5\) | A1 | co [4] (use of BA can score the last 3 marks) |
$\mathbf{OA} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}, \quad \mathbf{OB} = 3\mathbf{i} - 2\mathbf{j} + p\mathbf{k}$
**(i)** $(2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \cdot (3\mathbf{i} - 2\mathbf{j} + p\mathbf{k}) = 0$ | M1 | For $x_1x_2+y_1y_2+z_1z_2$ (in (i) or (ii))
$\rightarrow 6 - 2 + 2p = 0$ | A1 | co
$\rightarrow p = -2$ | [2]
**(ii)** $(2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \cdot (3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k})$ | A1 | nb Part (ii) gains 4 marks if (i) missing. co (M1 here if (i) not done)
$\rightarrow 6 - 2 + 12$ allow for $\pm$ this | M1 | All connected correctly
$= \sqrt{9} \times \sqrt{49}\cos\theta$ | A1 | co [3]
$\rightarrow \theta = 40°$ |
**(iii)** $\mathbf{AB} = \mathbf{i} - 3\mathbf{j} + (p - 2)\mathbf{k}$ | B1 | Must be for AB, not BA.
$1^2 + 3^2 + (p - 2)^2 = 3.5^2$ | M1 | Pythagoras (allow if √ wrong once)
| DM1 | Method of solution.
$\rightarrow p = 0.5$ or $3.5$ | A1 | co [4] (use of BA can score the last 3 marks)10 Relative to an origin $O$, the position vectors of points $A$ and $B$ are $2 \mathbf { i } + \mathbf { j } + 2 \mathbf { k }$ and $3 \mathbf { i } - 2 \mathbf { j } + p \mathbf { k }$ respectively.\\
(i) Find the value of $p$ for which $O A$ and $O B$ are perpendicular.\\
(ii) In the case where $p = 6$, use a scalar product to find angle $A O B$, correct to the nearest degree.\\
(iii) Express the vector $\overrightarrow { A B }$ is terms of $p$ and hence find the values of $p$ for which the length of $A B$ is 3.5 units.
\hfill \mbox{\textit{CAIE P1 2008 Q10 [9]}}