| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Angle with unknown parameter |
| Difficulty | Standard +0.3 This is a straightforward two-part vectors question requiring standard techniques: (i) finding a unit vector by calculating AB, then dividing by its magnitude, and (ii) using the scalar product formula cos θ = (a·b)/(|a||b|) to set up and solve a quadratic equation for p. Both parts are routine applications of A-level vector methods with no novel problem-solving required, 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 multiplication1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}\). Unit vector = \((3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}) \div 7\) | B1, M1 A1✓ [3] | Must be \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\). Divides by modulus. √ on vector AB. |
| (ii) Scalar product = 4 \(= \sqrt{5} \times \sqrt{16 + p^2} \times \cos \theta\) \(\to p = \pm 8\) | M1, M1 M1, A1 [4] | Use of \(v_1v_2 + v_1v_2 + v_1v_2\). For modulus. All three. For modulus, including correct use of \(\cos\theta=1/5\). |
$\overrightarrow{OA} = \mathbf{i} + 2\mathbf{j}$ and $\overrightarrow{OB} = 4\mathbf{i} + p\mathbf{k}$
(i) $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}$. Unit vector = $(3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}) \div 7$ | B1, M1 A1✓ [3] | Must be $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$. Divides by modulus. √ on vector AB.
(ii) Scalar product = 4 $= \sqrt{5} \times \sqrt{16 + p^2} \times \cos \theta$ $\to p = \pm 8$ | M1, M1 M1, A1 [4] | Use of $v_1v_2 + v_1v_2 + v_1v_2$. For modulus. All three. For modulus, including correct use of $\cos\theta=1/5$.4 Relative to an origin $O$, the position vectors of points $A$ and $B$ are given by
$$\overrightarrow { O A } = \mathbf { i } + 2 \mathbf { j } \quad \text { and } \quad \overrightarrow { O B } = 4 \mathbf { i } + p \mathbf { k } .$$
(i) In the case where $p = 6$, find the unit vector in the direction of $\overrightarrow { A B }$.\\
(ii) Find the values of $p$ for which angle $A O B = \cos ^ { - 1 } \left( \frac { 1 } { 5 } \right)$.
\hfill \mbox{\textit{CAIE P1 2013 Q4 [7]}}