| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2002 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicularity conditions |
| Difficulty | Moderate -0.8 This is a straightforward two-part question testing basic vector operations: (i) finding angle between vectors using dot product formula (standard recall), and (ii) using perpendicularity condition b·c = 0 to solve for p (direct application). Both parts are routine textbook exercises requiring no problem-solving insight, making this easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(a.b = 4 - 12 + 3 = -5\). \(a.b = \sqrt{9}\sqrt{49}\cos\theta\). \(\theta = 103.8°\) or 1.81 radians | M1, M1, A1 | Use of \(a_1b_1 + a_2b_2 + a_3b_3\); Use of \(a.b\cos\theta + \) Use of \(\sqrt{(a_1^2 + a_2^2 + a_3^2)}\); Correct only |
| (ii) Dot product = \(11p + 3\). Dot product = 0. \(P = -3/11\) | M1, DM1, A1 | Use of \(a_1b_1 + a_2b_2 + a_3b_3\); \(= 0\) used; correct only |
**(i)** $a.b = 4 - 12 + 3 = -5$. $a.b = \sqrt{9}\sqrt{49}\cos\theta$. $\theta = 103.8°$ or 1.81 radians | M1, M1, A1 | Use of $a_1b_1 + a_2b_2 + a_3b_3$; Use of $a.b\cos\theta + $ Use of $\sqrt{(a_1^2 + a_2^2 + a_3^2)}$; Correct only
**(ii)** Dot product = $11p + 3$. Dot product = 0. $P = -3/11$ | M1, DM1, A1 | Use of $a_1b_1 + a_2b_2 + a_3b_3$; $= 0$ used; correct only7 Given that $\mathbf { a } = \left( \begin{array} { r } 2 \\ - 2 \\ 1 \end{array} \right) , \mathbf { b } = \left( \begin{array} { l } 2 \\ 6 \\ 3 \end{array} \right)$ and $\mathbf { c } = \left( \begin{array} { c } p \\ p \\ p + 1 \end{array} \right)$, find\\
(i) the angle between the directions of $\mathbf { a }$ and $\mathbf { b }$,\\
(ii) the value of $p$ for which $\mathbf { b }$ and $\mathbf { c }$ are perpendicular.
\hfill \mbox{\textit{CAIE P1 2002 Q7 [7]}}