| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Moderate -0.3 This is a straightforward multi-part vectors question testing standard techniques: dot product for perpendicularity, magnitude comparison, and section formula. All parts require routine application of formulas with no novel insight, making it slightly easier than average for A-level. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(6 \times 3 + -2 \times k + -6 \times -3 = 0\) giving \((18 - 2k + 18 = 0)\) | M1 | Use of scalar product = 0. Could be \(\overrightarrow{AO} \cdot \overrightarrow{OB}\), \(\overrightarrow{AO} \cdot \overrightarrow{BO}\) or \(\overrightarrow{OA} \cdot \overrightarrow{BO}\) |
| \(k = 18\) | A1 | |
| Alternative: \(76 + 18 + k^2 = 18 + (k+2)^2\) | M1 | Use of Pythagoras with appropriate lengths |
| \(k = 18\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(36 + 4 + 36 = 9 + k^2 + 9\) | M1 | Use of modulus leading to an equation and solve to \(k=\) or \(k^2=\) |
| \(k = \pm\sqrt{58}\) or \(\pm 7.62\) | A1 | Accept exact or decimal answers. Allow decimals to greater accuracy. |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}-3\\6\\3\end{pmatrix} \rightarrow \overrightarrow{AC} = \begin{pmatrix}-2\\4\\2\end{pmatrix}\) then \(\overrightarrow{OA} + \overrightarrow{AC}\) | M1 | Complete method using \(\overrightarrow{AC} = \pm\frac{2}{3}\overrightarrow{AB}\) and then \(\overrightarrow{OA} + their\ \overrightarrow{AC}\) |
| \(\overrightarrow{OC} = \begin{pmatrix}4\\2\\-4\end{pmatrix}\) | A1 | |
| \(\div \sqrt{(their\ 4)^2 + (their\ 2)^2 + (their\ {-4})^2}\) | M1 | Divides by modulus of their \(\overrightarrow{OC}\) |
| \(= \frac{1}{6}\begin{pmatrix}4\\2\\-4\end{pmatrix}\) or \(\frac{1}{6}(4i + 2j - 4k)\) | A1 | |
| Alternative: Let \(\overrightarrow{OC} = \begin{pmatrix}p\\q\\r\end{pmatrix} \rightarrow \overrightarrow{AC} = \begin{pmatrix}p-6\\q+2\\r+6\end{pmatrix}\) & \(\overrightarrow{CB} = \begin{pmatrix}3-p\\4-q\\-3-r\end{pmatrix}\) | M1 | Correct method. Equates coefficients leading to values for \(p, q, r\) |
| \(p-6 = 2(3-p);\ q+2 = 2(4-q);\ r+6 = 2(-3-r)\) giving \(p=4,\ q=2,\ r=-4\) | A1 | |
| \(\div \sqrt{(their\ 4)^2 + (their\ 2)^2 + (their\ {-4})^2}\) | M1 | Divides by modulus of their \(\overrightarrow{OC}\) |
| \(= \frac{1}{6}\begin{pmatrix}4\\2\\-4\end{pmatrix}\) or \(\frac{1}{6}(4i+2j-4k)\) | A1 | |
| Alternative: \(\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC}\ \therefore 2(\overrightarrow{OB}-\overrightarrow{OC}) = \overrightarrow{OC} - \overrightarrow{OA}\) giving \(2\overrightarrow{OB} + \overrightarrow{OA} = 3\overrightarrow{OC}\ \therefore 3\overrightarrow{OC} = \begin{pmatrix}12\\6\\-12\end{pmatrix}\) | M1 | Correct method. Gets to a numerical expression for \(k\overrightarrow{OC}\) from \(\overrightarrow{OA}\) & \(\overrightarrow{OB}\) |
| \(\overrightarrow{OC} = \begin{pmatrix}4\\2\\-4\end{pmatrix}\) | A1 | |
| \(\div \sqrt{(their\ 4)^2 + (their\ 2)^2 + (their\ {-4})^2}\) | M1 | Divides by modulus of their \(\overrightarrow{OC}\) |
| \(= \frac{1}{6}\begin{pmatrix}4\\2\\-4\end{pmatrix}\) or \(\frac{1}{6}(4i+2j-4k)\) | A1 | |
| 4 |
## Question 8(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $6 \times 3 + -2 \times k + -6 \times -3 = 0$ giving $(18 - 2k + 18 = 0)$ | M1 | Use of scalar product = 0. Could be $\overrightarrow{AO} \cdot \overrightarrow{OB}$, $\overrightarrow{AO} \cdot \overrightarrow{BO}$ or $\overrightarrow{OA} \cdot \overrightarrow{BO}$ |
| $k = 18$ | A1 | |
| **Alternative:** $76 + 18 + k^2 = 18 + (k+2)^2$ | M1 | Use of Pythagoras with appropriate lengths |
| $k = 18$ | A1 | |
| | **2** | |
## Question 8(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $36 + 4 + 36 = 9 + k^2 + 9$ | M1 | Use of modulus leading to an equation and solve to $k=$ or $k^2=$ |
| $k = \pm\sqrt{58}$ or $\pm 7.62$ | A1 | Accept exact or decimal answers. Allow decimals to greater accuracy. |
| | **2** | |
## Question 8(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\overrightarrow{AB} = \begin{pmatrix}-3\\6\\3\end{pmatrix} \rightarrow \overrightarrow{AC} = \begin{pmatrix}-2\\4\\2\end{pmatrix}$ then $\overrightarrow{OA} + \overrightarrow{AC}$ | M1 | Complete method using $\overrightarrow{AC} = \pm\frac{2}{3}\overrightarrow{AB}$ and then $\overrightarrow{OA} + their\ \overrightarrow{AC}$ |
| $\overrightarrow{OC} = \begin{pmatrix}4\\2\\-4\end{pmatrix}$ | A1 | |
| $\div \sqrt{(their\ 4)^2 + (their\ 2)^2 + (their\ {-4})^2}$ | M1 | Divides by modulus of their $\overrightarrow{OC}$ |
| $= \frac{1}{6}\begin{pmatrix}4\\2\\-4\end{pmatrix}$ or $\frac{1}{6}(4i + 2j - 4k)$ | A1 | |
| **Alternative:** Let $\overrightarrow{OC} = \begin{pmatrix}p\\q\\r\end{pmatrix} \rightarrow \overrightarrow{AC} = \begin{pmatrix}p-6\\q+2\\r+6\end{pmatrix}$ & $\overrightarrow{CB} = \begin{pmatrix}3-p\\4-q\\-3-r\end{pmatrix}$ | M1 | Correct method. Equates coefficients leading to values for $p, q, r$ |
| $p-6 = 2(3-p);\ q+2 = 2(4-q);\ r+6 = 2(-3-r)$ giving $p=4,\ q=2,\ r=-4$ | A1 | |
| $\div \sqrt{(their\ 4)^2 + (their\ 2)^2 + (their\ {-4})^2}$ | M1 | Divides by modulus of their $\overrightarrow{OC}$ |
| $= \frac{1}{6}\begin{pmatrix}4\\2\\-4\end{pmatrix}$ or $\frac{1}{6}(4i+2j-4k)$ | A1 | |
| **Alternative:** $\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC}\ \therefore 2(\overrightarrow{OB}-\overrightarrow{OC}) = \overrightarrow{OC} - \overrightarrow{OA}$ giving $2\overrightarrow{OB} + \overrightarrow{OA} = 3\overrightarrow{OC}\ \therefore 3\overrightarrow{OC} = \begin{pmatrix}12\\6\\-12\end{pmatrix}$ | M1 | Correct method. Gets to a numerical expression for $k\overrightarrow{OC}$ from $\overrightarrow{OA}$ & $\overrightarrow{OB}$ |
| $\overrightarrow{OC} = \begin{pmatrix}4\\2\\-4\end{pmatrix}$ | A1 | |
| $\div \sqrt{(their\ 4)^2 + (their\ 2)^2 + (their\ {-4})^2}$ | M1 | Divides by modulus of their $\overrightarrow{OC}$ |
| $= \frac{1}{6}\begin{pmatrix}4\\2\\-4\end{pmatrix}$ or $\frac{1}{6}(4i+2j-4k)$ | A1 | |
| | **4** | |8 The position vectors of points $A$ and $B$, relative to an origin $O$, are given by
$$\overrightarrow { O A } = \left( \begin{array} { r }
6 \\
- 2 \\
- 6
\end{array} \right) \quad \text { and } \quad \overrightarrow { O B } = \left( \begin{array} { r }
3 \\
k \\
- 3
\end{array} \right)$$
where $k$ is a constant.\\
(i) Find the value of $k$ for which angle $A O B$ is $90 ^ { \circ }$.\\
(ii) Find the values of $k$ for which the lengths of $O A$ and $O B$ are equal.\\
The point $C$ is such that $\overrightarrow { A C } = 2 \overrightarrow { C B }$.\\
(iii) In the case where $k = 4$, find the unit vector in the direction of $\overrightarrow { O C }$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q8 [8]}}