| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line-plane intersection and related angle/perpendicularity |
| Difficulty | Standard +0.3 This is a standard three-part vectors question requiring routine techniques: substituting a line equation into a plane to find intersection (part i), using the formula for angle between line and plane involving dot product (part ii), and finding a plane through a line perpendicular to another plane using cross product of direction vectors (part iii). All methods are textbook procedures with no novel insight 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 point |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Express general point of the line in component form, e.g. \((2 + \lambda, -1 + 2\lambda, -4 + 2\lambda)\) | B1 | — |
| Substitute in plane equation and solve for \(\lambda\) | M1 | — |
| Obtain position vector \(4\mathbf{i} + 3\mathbf{j}\), or equivalent | A1 | [3] |
| (ii) State or imply a correct vector normal to the plane, e.g. \(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) | 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 evaluate the inverse sine of the result | M1 | — |
| Obtain answer \(26.5°\) (or 0.462 radians) | A1 | [4] |
| (iii) EITHER: State \(a + 2b + 2c = 0\) or \(3a - b + 2c = 0\) | B1 | — |
| Obtain two relevant equations and solve for one ratio, e.g. \(a : b\) | M1 | — |
| Obtain \(a : b : c = 6 : 4 : -7\), or equivalent | A1 | — |
| Substitute coordinates of a relevant point in \(6x + 4y - 7z = d\) and evaluate \(d\) | M1 | — |
| Obtain answer \(6x + 4y - 7z = 36\), or equivalent | A1 | — |
| OR1: Attempt to calculate vector product of relevant vectors, e.g. \((i + 2\mathbf{j} + 2\mathbf{k}) \times (3\mathbf{i} - \mathbf{j} + 2\mathbf{k})\) | M1 | — |
| Obtain two correct components of the product | A1 | — |
| Obtain correct product, e.g. \(6\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}\) | A1 | — |
| Substitute coordinates of a relevant point in \(6x + 4y - 7z = d\) and evaluate \(d\) | M1 | — |
| Obtain answer \(6x + 4y - 7z = 36\), or equivalent | A1 | — |
| OR2: Attempt to form 2-parameter equation with relevant vectors | M1 | — |
| State a correct equation, e.g. \(\mathbf{r} = 2\mathbf{i} - \mathbf{j} - 4\mathbf{k} + \lambda(i + 2\mathbf{j} + 2\mathbf{k}) + \mu(3\mathbf{i} - \mathbf{j} + 2\mathbf{k})\) | A1 | — |
| State three equations in \(x, y, z, \lambda, \mu\) | A1 | — |
| Eliminate \(\lambda\) and \(\mu\) | M1 | — |
| Obtain answer \(6x + 4y - 7z = 36\), or equivalent | A1 | [5] |
**(i)** Express general point of the line in component form, e.g. $(2 + \lambda, -1 + 2\lambda, -4 + 2\lambda)$ | B1 | —
Substitute in plane equation and solve for $\lambda$ | M1 | —
Obtain position vector $4\mathbf{i} + 3\mathbf{j}$, or equivalent | A1 | [3]
**(ii)** State or imply a correct vector normal to the plane, e.g. $3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ | 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 evaluate the inverse sine of the result | M1 | —
Obtain answer $26.5°$ (or 0.462 radians) | A1 | [4]
**(iii)** **EITHER:** State $a + 2b + 2c = 0$ or $3a - b + 2c = 0$ | B1 | —
Obtain two relevant equations and solve for one ratio, e.g. $a : b$ | M1 | —
Obtain $a : b : c = 6 : 4 : -7$, or equivalent | A1 | —
Substitute coordinates of a relevant point in $6x + 4y - 7z = d$ and evaluate $d$ | M1 | —
Obtain answer $6x + 4y - 7z = 36$, or equivalent | A1 | —
**OR1:** Attempt to calculate vector product of relevant vectors, e.g. $(i + 2\mathbf{j} + 2\mathbf{k}) \times (3\mathbf{i} - \mathbf{j} + 2\mathbf{k})$ | M1 | —
Obtain two correct components of the product | A1 | —
Obtain correct product, e.g. $6\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}$ | A1 | —
Substitute coordinates of a relevant point in $6x + 4y - 7z = d$ and evaluate $d$ | M1 | —
Obtain answer $6x + 4y - 7z = 36$, or equivalent | A1 | —
**OR2:** Attempt to form 2-parameter equation with relevant vectors | M1 | —
State a correct equation, e.g. $\mathbf{r} = 2\mathbf{i} - \mathbf{j} - 4\mathbf{k} + \lambda(i + 2\mathbf{j} + 2\mathbf{k}) + \mu(3\mathbf{i} - \mathbf{j} + 2\mathbf{k})$ | A1 | —
State three equations in $x, y, z, \lambda, \mu$ | A1 | —
Eliminate $\lambda$ and $\mu$ | M1 | —
Obtain answer $6x + 4y - 7z = 36$, or equivalent | A1 | [5]The straight line $l$ has equation $\mathbf{r} = 2\mathbf{i} - \mathbf{j} - 4\mathbf{k} + \lambda(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$. The plane $p$ has equation $3x - y + 2z = 9$. The line $l$ intersects the plane $p$ at the point $A$.
\begin{enumerate}[label=(\roman*)]
\item Find the position vector of $A$. [3]
\item Find the acute angle between $l$ and $p$. [4]
\item Find an equation for the plane which contains $l$ and is perpendicular to $p$, giving your answer in the form $ax + by + cz = d$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2010 Q10 [12]}}