| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Session | Specimen |
| Marks | 9 |
| 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 dividing by magnitude, and (ii) using the scalar product formula cos θ = (a·b)/(|a||b|) to solve for an unknown parameter. 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 vectors1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{PM} = 2\mathbf{i} - 10\mathbf{k} + \frac{1}{2}(6\mathbf{j} + 8\mathbf{k})\) oe | M1 | Any valid method |
| \(\mathbf{PM} = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\) | A1 | |
| \(\div\sqrt{4+9+36}\) | M1 | |
| Unit vector \(= \frac{1}{7}(2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k})\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{AT} = 6\mathbf{j} + 8\mathbf{k}\), \(\mathbf{PT} = a\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}\) soi (or \(\mathbf{TA}\) and \(\mathbf{TP}\)) | B1 | Allow 1 vector reversed at this stage |
| \((\cos ATP) = \dfrac{(6\mathbf{j}+8\mathbf{k}).(a\mathbf{i}+6\mathbf{j}-2\mathbf{k})}{\sqrt{36+64}\sqrt{a^2+36+4}}\) | M1 | (AM or MT could be used for AT) |
| \(= \dfrac{36-16}{\sqrt{36+64}\sqrt{a^2+36+4}} = \dfrac{20}{10\sqrt{a^2+40}}\) | A1\(\checkmark\) | Ft from their AT and PT |
| \(\dfrac{2}{\sqrt{a^2+40}} = \dfrac{2}{7}\) oe and attempt to solve | M1 | |
| \(a = 3\) | A1 | Withheld if only 1 vector reversed |
| Total: 5 | ||
| Alternative Method (Cosine Rule): | ||
| Alt (Cosine Rule) Vectors (AT, PT etc.) | (B1) | |
| \(\cos ATP = \dfrac{a^2+36+4+36+64-(100+a^2)}{2\sqrt{(a^2+40)}\sqrt{100}}\) then as above | (M1A1) |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{PM} = 2\mathbf{i} - 10\mathbf{k} + \frac{1}{2}(6\mathbf{j} + 8\mathbf{k})$ oe | M1 | Any valid method |
| $\mathbf{PM} = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}$ | A1 | |
| $\div\sqrt{4+9+36}$ | M1 | |
| Unit vector $= \frac{1}{7}(2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k})$ | A1 | |
| **Total: 4** | | |
---
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{AT} = 6\mathbf{j} + 8\mathbf{k}$, $\mathbf{PT} = a\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}$ soi (or $\mathbf{TA}$ and $\mathbf{TP}$) | B1 | Allow 1 vector reversed at this stage |
| $(\cos ATP) = \dfrac{(6\mathbf{j}+8\mathbf{k}).(a\mathbf{i}+6\mathbf{j}-2\mathbf{k})}{\sqrt{36+64}\sqrt{a^2+36+4}}$ | M1 | (**AM** or **MT** could be used for **AT**) |
| $= \dfrac{36-16}{\sqrt{36+64}\sqrt{a^2+36+4}} = \dfrac{20}{10\sqrt{a^2+40}}$ | A1$\checkmark$ | Ft from their **AT** and **PT** |
| $\dfrac{2}{\sqrt{a^2+40}} = \dfrac{2}{7}$ oe and attempt to solve | M1 | |
| $a = 3$ | A1 | Withheld if only 1 vector reversed |
| **Total: 5** | | |
| **Alternative Method (Cosine Rule):** | | |
| Alt (Cosine Rule) Vectors (AT, PT etc.) | (B1) | |
| $\cos ATP = \dfrac{a^2+36+4+36+64-(100+a^2)}{2\sqrt{(a^2+40)}\sqrt{100}}$ then as above | (M1A1) | |(i) For the case where $a = 2$, find the unit vector in the direction of $\overrightarrow { P M }$.\\
(ii) For the case where angle $A T P = \cos ^ { - 1 } \left( \frac { 2 } { 7 } \right)$, find the value of $a$.\\
\hfill \mbox{\textit{CAIE P1 Q10 [9]}}