| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2004 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Vector operations and magnitudes |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic vector operations: finding a unit vector (routine calculation), using dot product for perpendicularity (standard application), and using magnitude formula to solve for an unknown (algebraic manipulation). All three parts are textbook exercises requiring only direct application of standard formulas with no problem-solving insight needed. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks |
|---|---|
| M1 M1 A1 [3] | Use of \(\mathbf{b}-\mathbf{a}\), rather than \(\mathbf{b}+\mathbf{a}\) or \(\mathbf{a}-\mathbf{b}\); Dividing by the modulus of his \(\overrightarrow{AB}\); Co (allow – for candidates using \(\mathbf{a}-\mathbf{b}\)) |
| Answer | Marks |
|---|---|
| M1 DM1 A1 [3] | Use of \(x_1x_2 + y_1y_2 + z_1z_2\); Setting to 0 + attempt to solve; co |
| Answer | Marks |
|---|---|
| M1 A1 | Correct method for length with \(\pm\mathbf{d}-\mathbf{a}\), \(\mathbf{d}+\mathbf{a}\); Correct quadratic equation |
| Answer | Marks |
|---|---|
| DM1 A1 or B1 B1 [4] | Correct method of solution. Both correct. Or B1 for each if \((q+1)^2=36\), \(q=5\) only. |
$OA = \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix}$, $OB = \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}$, $OC = \begin{pmatrix} 4 \\ 2 \\ p \end{pmatrix}$, $OD = \begin{pmatrix} -1 \\ 0 \\ q \end{pmatrix}$
**Condone notation throughout. Allow column vectors or i,j,k throughout.**
**(i)** $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}$ Unit vector $(2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}) \sqrt(2^2+4^2+4^2) = \pm(2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k})/6$
| M1 M1 A1 [3] | Use of $\mathbf{b}-\mathbf{a}$, rather than $\mathbf{b}+\mathbf{a}$ or $\mathbf{a}-\mathbf{b}$; Dividing by the modulus of his $\overrightarrow{AB}$; Co (allow – for candidates using $\mathbf{a}-\mathbf{b}$) |
**(ii)** $\overrightarrow{OA} \cdot \overrightarrow{OC} = 4 + 6 - p = 0$ for $90°$ → $p = 10$
| M1 DM1 A1 [3] | Use of $x_1x_2 + y_1y_2 + z_1z_2$; Setting to 0 + attempt to solve; co |
**(iii)** $(-2)^2 + 3^2 + (q+1)^2 = 7^2$ → $(q+1)^2 = 36$ or $q^2 + 2q = 35$
| M1 A1 | Correct method for length with $\pm\mathbf{d}-\mathbf{a}$, $\mathbf{d}+\mathbf{a}$; Correct quadratic equation |
$q = 5$ and $q = -7$
| DM1 A1 or B1 B1 [4] | Correct method of solution. Both correct. Or B1 for each if $(q+1)^2=36$, $q=5$ only. |
---9 Relative to an origin $O$, the position vectors of the points $A , B , C$ and $D$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { r }
1 \\
3 \\
- 1
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r }
3 \\
- 1 \\
3
\end{array} \right) , \quad \overrightarrow { O C } = \left( \begin{array} { l }
4 \\
2 \\
p
\end{array} \right) \quad \text { and } \quad \overrightarrow { O D } = \left( \begin{array} { r }
- 1 \\
0 \\
q
\end{array} \right) ,$$
where $p$ and $q$ are constants. Find\\
(i) the unit vector in the direction of $\overrightarrow { A B }$,\\
(ii) the value of $p$ for which angle $A O C = 90 ^ { \circ }$,\\
(iii) the values of $q$ for which the length of $\overrightarrow { A D }$ is 7 units.
\hfill \mbox{\textit{CAIE P1 2004 Q9 [10]}}