| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2005 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with plane |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring standard techniques: finding a line equation, substituting into a plane equation to find intersection, calculating angle between line and plane using dot product, and computing perpendicular distance using the standard formula. All methods are routine for Further Maths students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point4.04j Shortest distance: between a point and a plane |
| Answer | Marks |
|---|---|
| State or imply a direction vector of \(AB\) is \(\mathbf{i} + 2\mathbf{j} + \mathbf{k}\), or equivalent | B1 |
| State equation of \(AB\) is \(\mathbf{r} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(-\mathbf{i} + 2\mathbf{j} + \mathbf{k})\), or equivalent | B1√ |
| Substitute in equation of \(p\) and solve for \(\lambda\) | M1 |
| Obtain \(4\mathbf{i} - 2\mathbf{j} - \mathbf{k}\) as position vector of \(C\) | A1 |
| Answer | Marks |
|---|---|
| State or imply a normal vector of \(p\) is \(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\), or equivalent | B1 |
| Carry out correct process for evaluating the scalar product of two relevant vectors, e.g. \((-\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \cdot (-\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})\) | M1 |
| Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine or inverse sine of the result | M1 |
| Obtain answer \(24.1°\) | A1 |
| Answer | Marks |
|---|---|
| Obtain \(AC (= \sqrt{24})\) in any correct form | B1√ |
| Use trig to obtain length of perpendicular from \(A\) to \(p\) | M1 |
| Obtain given answer correctly | A1 |
| Answer | Marks |
|---|---|
| State or imply \(\overrightarrow{AC}\) is \(2\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}\), or equivalent | B1√ |
| Use scalar product of \(\overrightarrow{AC}\) and a unit normal of \(p\) to calculate the perpendicular | M1 |
| Obtain given answer correctly | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Use plane perpendicular formula to find perpendicular from \(A\) to \(p\) | M1 | |
| Obtain a correct unsimplified numerical expression, e.g. \(\frac{ | 2 - 2(2) + 2(1) - 6 | }{\sqrt{(1^2 + (-2)^2 + 2^2)}}\) |
| Obtain given answer correctly | A1 |
**(i)**
State or imply a direction vector of $AB$ is $\mathbf{i} + 2\mathbf{j} + \mathbf{k}$, or equivalent | B1 |
State equation of $AB$ is $\mathbf{r} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(-\mathbf{i} + 2\mathbf{j} + \mathbf{k})$, or equivalent | B1√ |
Substitute in equation of $p$ and solve for $\lambda$ | M1 |
Obtain $4\mathbf{i} - 2\mathbf{j} - \mathbf{k}$ as position vector of $C$ | A1 |
**Total for (i): [4]**
**(ii)**
State or imply a normal vector of $p$ is $\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$, or equivalent | B1 |
Carry out correct process for evaluating the scalar product of two relevant vectors, e.g. $(-\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \cdot (-\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})$ | M1 |
Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine or inverse sine of the result | M1 |
Obtain answer $24.1°$ | A1 |
**Total for (ii): [4]**
**(iii)**
**EITHER:**
Obtain $AC (= \sqrt{24})$ in any correct form | B1√ |
Use trig to obtain length of perpendicular from $A$ to $p$ | M1 |
Obtain given answer correctly | A1 |
**OR:**
State or imply $\overrightarrow{AC}$ is $2\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}$, or equivalent | B1√ |
Use scalar product of $\overrightarrow{AC}$ and a unit normal of $p$ to calculate the perpendicular | M1 |
Obtain given answer correctly | A1 |
**OR:**
Use plane perpendicular formula to find perpendicular from $A$ to $p$ | M1 |
Obtain a correct unsimplified numerical expression, e.g. $\frac{|2 - 2(2) + 2(1) - 6|}{\sqrt{(1^2 + (-2)^2 + 2^2)}}$ | A1 |
Obtain given answer correctly | A1 |
**Total for (iii): [3]**10 The straight line $l$ passes through the points $A$ and $B$ with position vectors
$$2 \mathbf { i } + 2 \mathbf { j } + \mathbf { k } \quad \text { and } \quad \mathbf { i } + 4 \mathbf { j } + 2 \mathbf { k }$$
respectively. This line intersects the plane $p$ with equation $x - 2 y + 2 z = 6$ at the point $C$.\\
(i) Find the position vector of $C$.\\
(ii) Find the acute angle between $l$ and $p$.\\
(iii) Show that the perpendicular distance from $A$ to $p$ is equal to 2 .
\hfill \mbox{\textit{CAIE P3 2005 Q10 [11]}}