| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring dot product calculation, perpendicularity condition (dot product = 0), and angle formula application. Part (i) is routine computation, part (ii) involves showing a quadratic has no real solutions (discriminant check), and part (iii) uses cos(60°) = 1/2 with the dot product formula. All techniques are standard with no novel insight 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 multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((4i + 7j - pk)\cdot(8i - j - pk) = 25 + p^2\) | M1A1 | [2] \(x_1x_2 + y_1y_2 + z_1z_2\) (Not \(25 + (-p)^2\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(25 + p^2 = 0 \Rightarrow\) no real solutions | B1\(\sqrt{}\) | [1] Ft provided equation has no real solutions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos 60 = \frac{OA \cdot OB}{ | OA | |
| \( | OA | = \sqrt{65 + p^2}\) or \( |
| \(\frac{25 + p^2}{65 + p^2} = \frac{1}{2}\) or \(\frac{\text{his scalar (i)}}{65 + p^2} = \frac{1}{2}\) | A1\(\sqrt{}\) | Scalar product \(= 25 + p^2\) can score here if not scored in part (i) |
| \(p = \pm 3.87\) or \(\pm\sqrt{15}\) | A1 | [4] |
## Question 8:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(4i + 7j - pk)\cdot(8i - j - pk) = 25 + p^2$ | M1A1 | [2] $x_1x_2 + y_1y_2 + z_1z_2$ (Not $25 + (-p)^2$) |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $25 + p^2 = 0 \Rightarrow$ no real solutions | B1$\sqrt{}$ | [1] Ft provided equation has no real solutions |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos 60 = \frac{OA \cdot OB}{|OA||OB|}$ used | M1 | $OA \cdot OB$ must be scalar |
| $|OA| = \sqrt{65 + p^2}$ or $|OB| = \sqrt{65 + p^2}$ | M1 | Not $\sqrt{65 - p^2}$ unless follows $\sqrt{65 + (-p)^2}$ |
| $\frac{25 + p^2}{65 + p^2} = \frac{1}{2}$ or $\frac{\text{his scalar (i)}}{65 + p^2} = \frac{1}{2}$ | A1$\sqrt{}$ | Scalar product $= 25 + p^2$ can score here if not scored in part **(i)** |
| $p = \pm 3.87$ or $\pm\sqrt{15}$ | A1 | [4] |
---8 Relative to an origin $O$, the point $A$ has position vector $4 \mathbf { i } + 7 \mathbf { j } - p \mathbf { k }$ and the point $B$ has position vector $8 \mathbf { i } - \mathbf { j } - p \mathbf { k }$, where $p$ is a constant.\\
(i) Find $\overrightarrow { O A } \cdot \overrightarrow { O B }$.\\
(ii) Hence show that there are no real values of $p$ for which $O A$ and $O B$ are perpendicular to each other.\\
(iii) Find the values of $p$ for which angle $A O B = 60 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2011 Q8 [7]}}