| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Foot of perpendicular from general external point to line |
| Difficulty | Standard +0.3 This is a standard 3D vectors question testing routine techniques: writing a line equation (trivial), finding the foot of perpendicular using scalar product (standard method with 4-5 steps), and finding a plane equation from three points using cross product. All methods are textbook exercises requiring no novel insight, though the multi-step nature and calculation care push it slightly above average difficulty. |
| 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 point4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| State \(r = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}\), or equivalent | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Express \(\overrightarrow{BN}\) in terms of \(\lambda\), e.g. \(\begin{pmatrix} -1+3\lambda \\ 3-\lambda \\ 5-4\lambda \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}\), or equivalent | B1 | |
| Equate its scalar product with \(\begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}\) to zero and solve for \(\lambda\) | M1 | |
| Obtain \(\lambda = 2\) | A1 | |
| Obtain \(\overrightarrow{ON} = \begin{pmatrix} 5 \\ 1 \\ -3 \end{pmatrix}\), or equivalent | A1♦ | |
| Carry out method for calculating \(BN\), i.e. \(\ | 2i + 2j + k\ | \) |
| Obtain the given answer \(BN = 3\) correctly | A1 | 6 |
| Answer | Marks |
|---|---|
| Use scalar product to obtain a relevant equation in \(a, b\) and \(c\), e.g. \(3a - b - 4c = 0\) or \(2a + 2b + c = 0\) | M1 |
| State two correct equations in \(a, b, c\) | A1♦ |
| Solve simultaneous equations to obtain one ratio, e.g. \(a : b\) | M1 |
| Obtain \(a : b : c = 7 : -11 : 8\), or equivalent | A1 |
| Obtain equation \(7x - 11y + 8z = 0\), or equivalent | A1 |
| Answer | Marks |
|---|---|
| Substitute for \(A, B\) and \(\lambda\) in equations of plane and state 3 equations in \(a, b, c, d\) | B1 |
| Eliminate one unknown, e.g. \(d\) entirely and obtain 2 equations in 3 unknowns | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Calculate vector product of two vectors parallel to the plane, e.g. \(3i - j - 4k)(2i + 2j + k)\) | M1 | |
| Obtain 2 correct components of the product | A1♦ | |
| Obtain correct product c.g. \(7i - 11j - 8k\), or equivalent | A1 | |
| Substitute \(B, 7i - 11j - 8k\) into form correctly a 2-parameter equation for the plane | M1 | |
| Obtain equation in any correct form e.g. \(r = -i + 3j + 5k + \mu(3i - j - 4k) + \nu(2i + 2j + k)\) | A1♦ | |
| State 3 equations in \(x, y, z, \lambda\) and \(\mu\) | A1 | |
| Eliminate \(\lambda\) and \(\mu\) | M1 | |
| Obtain equations \(7x - 11y + 8z = 0\), or equivalent | A1 | 5 |
**(i)**
| State $r = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}$, or equivalent | B1 | 1 |
**(ii)**
| Express $\overrightarrow{BN}$ in terms of $\lambda$, e.g. $\begin{pmatrix} -1+3\lambda \\ 3-\lambda \\ 5-4\lambda \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}$, or equivalent | B1 |
| Equate its scalar product with $\begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}$ to zero and solve for $\lambda$ | M1 |
| Obtain $\lambda = 2$ | A1 |
| Obtain $\overrightarrow{ON} = \begin{pmatrix} 5 \\ 1 \\ -3 \end{pmatrix}$, or equivalent | A1♦ |
| Carry out method for calculating $BN$, i.e. $\|2i + 2j + k\|$ | M1 |
| Obtain the given answer $BN = 3$ correctly | A1 | 6 |
**(iii) EITHER:**
| Use scalar product to obtain a relevant equation in $a, b$ and $c$, e.g. $3a - b - 4c = 0$ or $2a + 2b + c = 0$ | M1 |
| State two correct equations in $a, b, c$ | A1♦ |
| Solve simultaneous equations to obtain one ratio, e.g. $a : b$ | M1 |
| Obtain $a : b : c = 7 : -11 : 8$, or equivalent | A1 |
| Obtain equation $7x - 11y + 8z = 0$, or equivalent | A1 |
**OR:**
| Substitute for $A, B$ and $\lambda$ in equations of plane and state 3 equations in $a, b, c, d$ | B1 |
| Eliminate one unknown, e.g. $d$ entirely and obtain 2 equations in 3 unknowns | M1 |
**OR:**
| Calculate vector product of two vectors parallel to the plane, e.g. $3i - j - 4k)(2i + 2j + k)$ | M1 |
| Obtain 2 correct components of the product | A1♦ |
| Obtain correct product c.g. $7i - 11j - 8k$, or equivalent | A1 |
| Substitute $B, 7i - 11j - 8k$ into form correctly a 2-parameter equation for the plane | M1 |
| Obtain equation in any correct form e.g. $r = -i + 3j + 5k + \mu(3i - j - 4k) + \nu(2i + 2j + k)$ | A1♦ |
| State 3 equations in $x, y, z, \lambda$ and $\mu$ | A1 |
| Eliminate $\lambda$ and $\mu$ | M1 |
| Obtain equations $7x - 11y + 8z = 0$, or equivalent | A1 | 5 |The points $A$ and $B$ have position vectors, relative to the origin $O$, given by
$$\overrightarrow{OA} = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix} \quad \text{and} \quad \overrightarrow{OB} = \begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}.$$
The line $l$ passes through $A$ and is parallel to $OB$. The point $N$ is the foot of the perpendicular from $B$ to $l$.
\begin{enumerate}[label=(\roman*)]
\item State a vector equation for the line $l$. [1]
\item Find the position vector of $N$ and show that $BN = 3$. [6]
\item Find the equation of the plane containing $A$, $B$ and $N$, giving your answer in the form $ax + by + cz = d$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2006 Q10 [12]}}