| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with plane |
| Difficulty | Standard +0.3 This is a standard three-part vectors question requiring routine techniques: substituting the line equation into the plane to find intersection (straightforward algebra), using the angle formula between line and plane (standard formula application), and finding a plane through a line perpendicular to another plane (requires cross product of normals). All techniques are textbook exercises with no novel insight required, though the multi-part nature and algebraic manipulation place it slightly above average difficulty. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| Express general point of \(l\) in component form, e.g. \((1 + 3\lambda, 2 - 2\lambda, -1 + 2\lambda)\) | B1 | |
| Substitute in given equation of \(p\) and solve for \(\lambda\) | M1 | |
| Obtain final answer \(-\frac{1}{4}\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}\), or equivalent, from \(\lambda = -\frac{1}{2}\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| State or imply a vector normal to the plane, e.g. \(2\mathbf{i} + 3\mathbf{j} - 5\mathbf{k}\) | B1 | |
| Using the correct process, evaluate the scalar product of a direction vector for \(l\) and a normal for \(p\) | M1 | |
| Using the correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse sine or cosine of the result | M1 | |
| Obtain answer 23.2° (or 0.404 radians) | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| EITHER: State \(2a + 3b - 5c = 0\) or \(3a - 2b + 2c = 0\) | B1 | |
| Obtain two relevant equations and solve for one ratio, e.g. \(a : b\) | M1 | |
| Obtain \(a : b : c = 4 : 19 : 13\), or equivalent | A1 | |
| Substitute coordinates of a relevant point in \(4x + 19y + 13z = d\), and evaluate \(d\) | M1 | |
| Obtain answer \(4x + 19y + 13z = 29\), or equivalent | A1 | |
| OR1: Attempt to calculate vector product of relevant vectors, e.g. \((2\mathbf{i} + 3\mathbf{j} - 5\mathbf{k}) \times (3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})\) | M1 | |
| Obtain two correct components of the product | A1 | |
| Obtain correct product, e.g. \(-4\mathbf{i} - 19\mathbf{j} - 13\mathbf{k}\) | A1 | |
| Substitute coordinates of a relevant point in \(4x + 19y + 13z = d\) | M1 | |
| Obtain answer \(4x + 19y + 13z = 29\), or equivalent | A1 | |
| OR2: Attempt to form a 2-parameter equation with relevant vectors, e.g. \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} - \mathbf{k} + \lambda(2\mathbf{i} + 3\mathbf{j} - 5\mathbf{k}) + \mu(3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})\) | A1 | |
| State 3 equations in \(x, y, z, \lambda\) and \(\mu\) | A1 | |
| Eliminate \(\lambda\) and \(\mu\) | M1 | |
| Obtain answer \(4x + 19y + 13z = 29\), or equivalent | A1 | |
| OR3: Using a relevant point and relevant direction vectors, form a determinant equation for the plane | M1 | |
| State a correct equation, e.g. \(\begin{vmatrix} x-1 & y-2 & z+1 \\ 2 & 3 & -5 \\ 3 & -2 & 2 \end{vmatrix} = 0\) | A1 | |
| Attempt to expand the determinant | M1 | |
| Obtain correct values of two cofactors | A1 | |
| Obtain answer \(4x + 19y + 13z = 29\), or equivalent | A1 | 5 marks |
**(i)**
Express general point of $l$ in component form, e.g. $(1 + 3\lambda, 2 - 2\lambda, -1 + 2\lambda)$ | B1 |
Substitute in given equation of $p$ and solve for $\lambda$ | M1 |
Obtain final answer $-\frac{1}{4}\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}$, or equivalent, from $\lambda = -\frac{1}{2}$ | A1 | 3 marks
**(ii)**
State or imply a vector normal to the plane, e.g. $2\mathbf{i} + 3\mathbf{j} - 5\mathbf{k}$ | B1 |
Using the correct process, evaluate the scalar product of a direction vector for $l$ and a normal for $p$ | M1 |
Using the correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse sine or cosine of the result | M1 |
Obtain answer 23.2° (or 0.404 radians) | A1 | 4 marks
**(iii)**
**EITHER:** State $2a + 3b - 5c = 0$ or $3a - 2b + 2c = 0$ | B1 |
Obtain two relevant equations and solve for one ratio, e.g. $a : b$ | M1 |
Obtain $a : b : c = 4 : 19 : 13$, or equivalent | A1 |
Substitute coordinates of a relevant point in $4x + 19y + 13z = d$, and evaluate $d$ | M1 |
Obtain answer $4x + 19y + 13z = 29$, or equivalent | A1 |
**OR1:** Attempt to calculate vector product of relevant vectors, e.g. $(2\mathbf{i} + 3\mathbf{j} - 5\mathbf{k}) \times (3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})$ | M1 |
Obtain two correct components of the product | A1 |
Obtain correct product, e.g. $-4\mathbf{i} - 19\mathbf{j} - 13\mathbf{k}$ | A1 |
Substitute coordinates of a relevant point in $4x + 19y + 13z = d$ | M1 |
Obtain answer $4x + 19y + 13z = 29$, or equivalent | A1 |
**OR2:** Attempt to form a 2-parameter equation with relevant vectors, e.g. $\mathbf{r} = \mathbf{i} + 2\mathbf{j} - \mathbf{k} + \lambda(2\mathbf{i} + 3\mathbf{j} - 5\mathbf{k}) + \mu(3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})$ | A1 |
State 3 equations in $x, y, z, \lambda$ and $\mu$ | A1 |
Eliminate $\lambda$ and $\mu$ | M1 |
Obtain answer $4x + 19y + 13z = 29$, or equivalent | A1 |
**OR3:** Using a relevant point and relevant direction vectors, form a determinant equation for the plane | M1 |
State a correct equation, e.g. $\begin{vmatrix} x-1 & y-2 & z+1 \\ 2 & 3 & -5 \\ 3 & -2 & 2 \end{vmatrix} = 0$ | A1 |
Attempt to expand the determinant | M1 |
Obtain correct values of two cofactors | A1 |
Obtain answer $4x + 19y + 13z = 29$, or equivalent | A1 | 5 marks10 The line $l$ has equation $\mathbf { r } = \mathbf { i } + 2 \mathbf { j } - \mathbf { k } + \lambda ( 3 \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } )$ and the plane $p$ has equation $2 x + 3 y - 5 z = 18$.\\
(i) Find the position vector of the point of intersection of $l$ and $p$.\\
(ii) Find the acute angle between $l$ and $p$.\\
(iii) A second plane $q$ is perpendicular to the plane $p$ and contains the line $l$. Find the equation of $q$, giving your answer in the form $a x + b y + c z = d$.
\hfill \mbox{\textit{CAIE P3 2014 Q10 [12]}}