| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| 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 basic 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/Working | Marks | Guidance |
| \(PM = 2\mathbf{i} - 10\mathbf{k} + \frac{1}{2}(6\mathbf{j} + 8\mathbf{k})\) | M1 | Any valid method |
| \(PM = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\) | A1 | |
| \(\div \sqrt{4 + 9 + 36}\) | M1 | |
| Unit vector \(= \frac{1}{7}(2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k})\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(AT = 6\mathbf{j} + 8\mathbf{k}\), \(PT = a\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}\) | B1 | Allow 1 vector reversed at this stage. (\(AM\) or \(MT\) could be used for \(AT\)) |
| \(\cos ATP = \frac{(6\mathbf{j}+8\mathbf{k})\cdot(a\mathbf{i}+6\mathbf{j}-2\mathbf{k})}{\sqrt{36+64}\sqrt{a^2+36+4}}\) | M1 | |
| \(= \frac{36-16}{\sqrt{36+64}\sqrt{a^2+36+4}}\) | ||
| \(= \frac{20}{10\sqrt{a^2+40}}\) | A1↑ | Ft from their \(AT\) and \(PT\) |
| \(\frac{2}{\sqrt{a^2+40}} = \frac{2}{7}\) oe and attempt to solve | M1 | |
| \(a = 3\) | A1 [5] | Withheld if only 1 vector reversed |
| ALT: \(\cos ATP = \frac{a^2+36+4+36+64-(100+a^2)}{2\sqrt{(a^2+40)}\sqrt{100}}\) | B1, M1A1 | then as above |
## Question 10:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $PM = 2\mathbf{i} - 10\mathbf{k} + \frac{1}{2}(6\mathbf{j} + 8\mathbf{k})$ | M1 | Any valid method |
| $PM = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}$ | A1 | |
| $\div \sqrt{4 + 9 + 36}$ | M1 | |
| Unit vector $= \frac{1}{7}(2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k})$ | A1 | [4] |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $AT = 6\mathbf{j} + 8\mathbf{k}$, $PT = a\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}$ | B1 | Allow 1 vector reversed at this stage. ($AM$ or $MT$ could be used for $AT$) |
| $\cos ATP = \frac{(6\mathbf{j}+8\mathbf{k})\cdot(a\mathbf{i}+6\mathbf{j}-2\mathbf{k})}{\sqrt{36+64}\sqrt{a^2+36+4}}$ | M1 | |
| $= \frac{36-16}{\sqrt{36+64}\sqrt{a^2+36+4}}$ | | |
| $= \frac{20}{10\sqrt{a^2+40}}$ | A1↑ | Ft from their $AT$ and $PT$ |
| $\frac{2}{\sqrt{a^2+40}} = \frac{2}{7}$ oe and attempt to solve | M1 | |
| $a = 3$ | A1 [5] | Withheld if only 1 vector reversed |
| **ALT:** $\cos ATP = \frac{a^2+36+4+36+64-(100+a^2)}{2\sqrt{(a^2+40)}\sqrt{100}}$ | B1, M1A1 | then as above |
---(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 2015 Q10 [9]}}