| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2002 |
| Session | June |
| 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 straightforward application of standard vector methods: finding a line equation from two points, substituting into a plane equation to find intersection, then using cross product of direction vectors to find a perpendicular plane. All steps are routine techniques covered in P3 with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply a simplified direction vector of \(l\) is \(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\), or equivalent | B1 | |
| State equation of \(l\) is \(\mathbf{r} = \mathbf{i} + k(3\mathbf{i} - \mathbf{j} + 2\mathbf{k})\), or \(\frac{x-1}{3} = \frac{y}{-1} = \frac{z-1}{2}\), or equivalent | B1 | |
| Substitute in equation of \(p\) and solve for \(\lambda\), or one of \(x, y, z\) | M1 | |
| Obtain point of intersection \(-2\mathbf{i} + \mathbf{j} - \mathbf{k}\) | A1 | |
| Guidance: Any notation is acceptable. | Total: 4 marks |
| Answer | Marks |
|---|---|
| State or imply a normal vector of \(p\) is \(\mathbf{n} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k}\) | B1 |
| Use scalar product to obtain \(a + 3b - 2c = 0\) | M1 |
| Use points on \(l\) to obtain two equations in \(a, b, c\) e.g. \(a + c = 1\), \(4a - b + 3c = 1\) | B1 |
| Solve simultaneous equations, obtaining one unknown | M1 |
| Obtain one correct unknown e.g. \(a = -\frac{4}{9}\) | A1 |
| Obtain the other unknowns e.g. \(b = \frac{5}{9}, c = \frac{5}{9}\) | A1 |
| Answer | Marks |
|---|---|
| Use scalar product to obtain \(a + 3b - 2c = 0\) | M1 |
| Use scalar product to obtain \(3a - b + 2c = 0\) | B1 |
| Solve simultaneous equations to obtain one ratio e.g. \(a : b\) | M1 |
| Obtain \(a : b : c = 2 : -4 : -5\), or equivalent | A1 |
| Obtain \(a = -\frac{2}{3}, b = \frac{4}{3}, c = \frac{5}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| State or imply a correct equation of the plane e.g. \(\mathbf{r} = \lambda(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + \mu(\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}) + \mathbf{i} + \mathbf{k}\) | M1 | |
| State 3 equations in \(x, y, z, \lambda, \mu\), e.g. \(x = 3\lambda + \mu + 1, y = -\lambda + 3\mu, z = 2\lambda - 2\mu + 1\) | A1 | |
| Eliminate \(\lambda\) and \(\mu\) | M1 | |
| Obtain equation \(-4x + 8y + 10z = 6\), or equivalent | A1 | |
| Obtain \(a = -\frac{2}{3}, b = \frac{4}{3}, c = \frac{5}{3}\) | A1 | Total: 6 marks |
| Guidance: SR: condone the use of \(\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) for \(\mathbf{i} + 0\mathbf{j} + 2\mathbf{k}\) in the EITHER scheme and the first OR scheme. |
**(i)** State or imply a simplified direction vector of $l$ is $3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$, or equivalent | B1 |
State equation of $l$ is $\mathbf{r} = \mathbf{i} + k(3\mathbf{i} - \mathbf{j} + 2\mathbf{k})$, or $\frac{x-1}{3} = \frac{y}{-1} = \frac{z-1}{2}$, or equivalent | B1 |
Substitute in equation of $p$ and solve for $\lambda$, or one of $x, y, z$ | M1 |
Obtain point of intersection $-2\mathbf{i} + \mathbf{j} - \mathbf{k}$ | A1 |
**Guidance:** Any notation is acceptable. | | **Total: 4 marks**
**(ii) EITHER route:**
State or imply a normal vector of $p$ is $\mathbf{n} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k}$ | B1 |
Use scalar product to obtain $a + 3b - 2c = 0$ | M1 |
Use points on $l$ to obtain two equations in $a, b, c$ e.g. $a + c = 1$, $4a - b + 3c = 1$ | B1 |
Solve simultaneous equations, obtaining one unknown | M1 |
Obtain one correct unknown e.g. $a = -\frac{4}{9}$ | A1 |
Obtain the other unknowns e.g. $b = \frac{5}{9}, c = \frac{5}{9}$ | A1 |
**OR route:**
Use scalar product to obtain $a + 3b - 2c = 0$ | M1 |
Use scalar product to obtain $3a - b + 2c = 0$ | B1 |
Solve simultaneous equations to obtain one ratio e.g. $a : b$ | M1 |
Obtain $a : b : c = 2 : -4 : -5$, or equivalent | A1 |
Obtain $a = -\frac{2}{3}, b = \frac{4}{3}, c = \frac{5}{3}$ | A1 |
**OR route:**
State or imply a correct equation of the plane e.g. $\mathbf{r} = \lambda(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + \mu(\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}) + \mathbf{i} + \mathbf{k}$ | M1 |
State 3 equations in $x, y, z, \lambda, \mu$, e.g. $x = 3\lambda + \mu + 1, y = -\lambda + 3\mu, z = 2\lambda - 2\mu + 1$ | A1 |
Eliminate $\lambda$ and $\mu$ | M1 |
Obtain equation $-4x + 8y + 10z = 6$, or equivalent | A1 |
Obtain $a = -\frac{2}{3}, b = \frac{4}{3}, c = \frac{5}{3}$ | A1 | **Total: 6 marks**
**Guidance:** SR: condone the use of $\mathbf{i} + \mathbf{j} + 2\mathbf{k}$ for $\mathbf{i} + 0\mathbf{j} + 2\mathbf{k}$ in the EITHER scheme and the first OR scheme. | |
---8 The straight line $l$ passes through the points $A$ and $B$ whose position vectors are $\mathbf { i } + \mathbf { k }$ and $4 \mathbf { i } - \mathbf { j } + 3 \mathbf { k }$ respectively. The plane $p$ has equation $x + 3 y - 2 z = 3$.\\
(i) Given that $l$ intersects $p$, find the position vector of the point of intersection.\\
(ii) Find the equation of the plane which contains $l$ and is perpendicular to $p$, giving your answer in the form $a x + b y + c z = 1$.
\hfill \mbox{\textit{CAIE P3 2002 Q8 [10]}}