| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2007 |
| Session | November |
| 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 testing routine techniques: substituting a line equation into a plane equation to find intersection, using the formula for angle between line and plane (involving direction vector and normal), and finding a line in a plane perpendicular to a given line. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 4.04a 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 plane |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute for \(r\) and expand the given scalar product, or correct equivalent, to obtain an equation in \(s\) | M1 | |
| Solve a linear equation formed by a scalar product for \(s\) | M1 | |
| Obtain \(s = 2\) and position vector \(3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) for \(A\) | A1 | [3] |
| (ii) State or imply a normal vector of \(p\) is \(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\), or equivalent | B1 | |
| Use the correct process for evaluating a relevant scalar product, e.g. \((i - 2j + 2k).(2i - 3j + 6k)\) | M1 | |
| Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate cosine of the result | M1 | |
| Obtain final answer \(72.2°\) or \(1.26\) radians | A1 | [4] |
| (iii) EITHER: Taking the direction vector of the line to be \(a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), state equation \(2a - 3b + 6c = 0\) | B1 | |
| State equation \(a - 2b + 2c = 0\) | B1 | |
| Solve to find one ratio, e.g. \(a : b\) | M1 | |
| Obtain ratio \(a : b : c = 6 : 2 : -1\), or equivalent | A1 | |
| State answer \(\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(6\mathbf{i} + 2\mathbf{j} - \mathbf{k})\), or equivalent | A1∇ | |
| OR1: Attempt to calculate the vector product of a direction vector for the line \(l\) and a normal vector of the plane \(p\), e.g. \((i - 2j + 2k) \times (2i - 3j + 6k)\) | M2 | |
| Obtain two correct components of the product | A1 | |
| Obtain answer \(-6i - 2j + k\), or equivalent | A1 | |
| State answer \(\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(-6\mathbf{i} - 2\mathbf{j} + \mathbf{k})\), or equivalent | A1∇ | |
| OR2: Obtain the equation of the plane containing \(A\) and perpendicular to the line \(l\) | M1 | |
| State answer \(-2y + 2z = 1\), or equivalent | A1∇ | |
| Find position vector of a second point \(B\) on the line of intersection of this plane with the plane \(p\), e.g. \(9\mathbf{i} + 4\mathbf{j}\) | M1 | |
| Obtain a direction vector for this line of intersection, e.g. \(6\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) | A1 | |
| State answer \(\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(6\mathbf{i} + 2\mathbf{j} - \mathbf{k})\), or equivalent | A1 | [5] |
**(i)** Substitute for $r$ and expand the given scalar product, or correct equivalent, to obtain an equation in $s$ | M1 |
Solve a linear equation formed by a scalar product for $s$ | M1 |
Obtain $s = 2$ and position vector $3\mathbf{i} + 2\mathbf{j} + \mathbf{k}$ for $A$ | A1 | [3]
**(ii)** State or imply a normal vector of $p$ is $2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$, or equivalent | B1 |
Use the correct process for evaluating a relevant scalar product, e.g. $(i - 2j + 2k).(2i - 3j + 6k)$ | M1 |
Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate cosine of the result | M1 |
Obtain final answer $72.2°$ or $1.26$ radians | A1 | [4]
**(iii)** **EITHER:** Taking the direction vector of the line to be $a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$, state equation $2a - 3b + 6c = 0$ | B1 |
State equation $a - 2b + 2c = 0$ | B1 |
Solve to find one ratio, e.g. $a : b$ | M1 |
Obtain ratio $a : b : c = 6 : 2 : -1$, or equivalent | A1 |
State answer $\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(6\mathbf{i} + 2\mathbf{j} - \mathbf{k})$, or equivalent | A1∇ |
**OR1:** Attempt to calculate the vector product of a direction vector for the line $l$ and a normal vector of the plane $p$, e.g. $(i - 2j + 2k) \times (2i - 3j + 6k)$ | M2 |
Obtain two correct components of the product | A1 |
Obtain answer $-6i - 2j + k$, or equivalent | A1 |
State answer $\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(-6\mathbf{i} - 2\mathbf{j} + \mathbf{k})$, or equivalent | A1∇ |
**OR2:** Obtain the equation of the plane containing $A$ and perpendicular to the line $l$ | M1 |
State answer $-2y + 2z = 1$, or equivalent | A1∇ |
Find position vector of a second point $B$ on the line of intersection of this plane with the plane $p$, e.g. $9\mathbf{i} + 4\mathbf{j}$ | M1 |
Obtain a direction vector for this line of intersection, e.g. $6\mathbf{i} + 2\mathbf{j} - \mathbf{k}$ | A1 |
State answer $\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(6\mathbf{i} + 2\mathbf{j} - \mathbf{k})$, or equivalent | A1 | [5]
[The f.t. is on $A$.]10 The straight line $l$ has equation $\mathbf { r } = \mathbf { i } + 6 \mathbf { j } - 3 \mathbf { k } + s ( \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } )$. The plane $p$ has equation $( \mathbf { r } - 3 \mathbf { i } ) \cdot ( 2 \mathbf { i } - 3 \mathbf { j } + 6 \mathbf { k } ) = 0$. The line $l$ intersects the plane $p$ at the point $A$.\\
(i) Find the position vector of $A$.\\
(ii) Find the acute angle between $l$ and $p$.\\
(iii) Find a vector equation for the line which lies in $p$, passes through $A$ and is perpendicular to $l$.
\hfill \mbox{\textit{CAIE P3 2007 Q10 [12]}}