| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2015 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with plane |
| Difficulty | Standard +0.3 This is a standard vectors question requiring routine application of line-plane intersection, angle calculation using dot product, and position vector manipulation. All techniques are textbook exercises with no novel insight required, though it involves multiple steps across three parts. Slightly easier than average due to straightforward computational nature. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| Express a general point on the line in single component form, e.g. \((2, 2 - 3\lambda, -8 + 4\lambda)\) | M1 | |
| Substitute in equation of plane and solve for \(\lambda\) | M1 | |
| Obtain \(\lambda = 3\) | A1 | |
| Obtain \((3, -7, 4)\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| State or imply normal vector to plane is \(4\mathbf{i} - \mathbf{j} + 5\mathbf{k}\) | B1 | |
| Carry out process for evaluating scalar product of two relevant vectors | M1 | |
| Using the correct process for finding moduli, divide the scalar product by the product of the moduli and evaluate \(\sin^{-1}\) or \(\cos^{-1}\) of the result | M1 | |
| Obtain \(54.8°\) or \(0.956\) radians | A1 | [4] |
| Answer | Marks |
|---|---|
| Find at least one position of \(C\) by translating by appropriate multiple of direction vector \(\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}\) from \(A\) or \(B\) | M1 |
| Obtain \((-3, 11, -20)\) | A1 |
| Obtain \((9, -25, 28)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Form quadratic equation in \(\lambda\) by considering \(BC^2 = 4AB^2\) | M1 | |
| Obtain \(26\lambda^2 - 1562 - 702 = 0\) or equivalent and hence \(\lambda = -3, \lambda = 9\) | A1 | |
| Obtain \((-3, 11, -20)\) and \((9, -25, 28)\) | A1 | [3] |
**(i)**
Express a general point on the line in single component form, e.g. $(2, 2 - 3\lambda, -8 + 4\lambda)$ | M1 |
Substitute in equation of plane and solve for $\lambda$ | M1 |
Obtain $\lambda = 3$ | A1 |
Obtain $(3, -7, 4)$ | A1 | [3]
**(ii)**
State or imply normal vector to plane is $4\mathbf{i} - \mathbf{j} + 5\mathbf{k}$ | B1 |
Carry out process for evaluating scalar product of two relevant vectors | M1 |
Using the correct process for finding moduli, divide the scalar product by the product of the moduli and evaluate $\sin^{-1}$ or $\cos^{-1}$ of the result | M1 |
Obtain $54.8°$ or $0.956$ radians | A1 | [4]
**(iii)**
**Either**
Find at least one position of $C$ by translating by appropriate multiple of direction vector $\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ from $A$ or $B$ | M1 |
Obtain $(-3, 11, -20)$ | A1 |
Obtain $(9, -25, 28)$ | A1 |
**Or**
Form quadratic equation in $\lambda$ by considering $BC^2 = 4AB^2$ | M1 |
Obtain $26\lambda^2 - 1562 - 702 = 0$ or equivalent and hence $\lambda = -3, \lambda = 9$ | A1 |
Obtain $(-3, 11, -20)$ and $(9, -25, 28)$ | A1 | [3]8 A plane has equation $4 x - y + 5 z = 39$. A straight line is parallel to the vector $\mathbf { i } - 3 \mathbf { j } + 4 \mathbf { k }$ and passes through the point $A ( 0,2 , - 8 )$. The line meets the plane at the point $B$.\\
(i) Find the coordinates of $B$.\\
(ii) Find the acute angle between the line and the plane.\\
(iii) The point $C$ lies on the line and is such that the distance between $C$ and $B$ is twice the distance between $A$ and $B$. Find the coordinates of each of the possible positions of the point $C$.
\hfill \mbox{\textit{CAIE P3 2015 Q8 [10]}}