| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Distance between parallel planes or line and parallel plane |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring understanding of line-plane relationships, perpendicular distances, and geometric reasoning. While the individual techniques (checking parallelism, finding perpendicular lines, closest points) are standard FP3 material, the three-part structure with the 'hence' connections and the final part requiring geometric insight about closest parallel lines elevates it above routine exercises. It's moderately challenging for Further Maths but not exceptionally difficult. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(l \parallel \begin{pmatrix}2\\3\\5\end{pmatrix}\), \(\Pi \perp \begin{pmatrix}4\\-1\\-1\end{pmatrix}\), so \(\begin{pmatrix}2\\3\\5\end{pmatrix} \cdot \begin{pmatrix}4\\-1\\-1\end{pmatrix} = 0 \Rightarrow l \parallel \Pi\) | M1 | dot product of correct vectors \(= 0\) |
| \((1, -2, 7)\) on \(l\) but \(4\times1 - -2 - 7 = -1 \neq 8\) so not in \(\Pi\); hence \(l\) not in \(\Pi\) | M1 | Substitute point on line into \(\Pi\) and calculate d |
| A1 | Full argument includes key components; Argument can be about a general point on line | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{r} = \begin{pmatrix}1\\-2\\7\end{pmatrix} + \lambda\begin{pmatrix}4\\-1\\-1\end{pmatrix}\) | B1 | |
| Closest point where meets \(\Pi\): \(4(1+4\lambda)-(-2-\lambda)-(7-\lambda)=8\) | M1 | Parametric form of \((x, y, z)\) substituted into plane |
| \(\Rightarrow \lambda = \frac{1}{2}\) | A1ft | |
| \(\Rightarrow \mathbf{r} = \begin{pmatrix}3\\-\frac{5}{2}\\\frac{13}{2}\end{pmatrix}\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{r} = \begin{pmatrix}3\\-\frac{5}{2}\\\frac{13}{2}\end{pmatrix} + \lambda\begin{pmatrix}2\\3\\5\end{pmatrix}\) | B1ft | oe; must have "\(\mathbf{r} =\)" |
| [1] |
# Question 6:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $l \parallel \begin{pmatrix}2\\3\\5\end{pmatrix}$, $\Pi \perp \begin{pmatrix}4\\-1\\-1\end{pmatrix}$, so $\begin{pmatrix}2\\3\\5\end{pmatrix} \cdot \begin{pmatrix}4\\-1\\-1\end{pmatrix} = 0 \Rightarrow l \parallel \Pi$ | M1 | dot product of correct vectors $= 0$ |
| $(1, -2, 7)$ on $l$ but $4\times1 - -2 - 7 = -1 \neq 8$ so not in $\Pi$; hence $l$ not in $\Pi$ | M1 | Substitute point on line into $\Pi$ and calculate d |
| | A1 | Full argument includes key components; Argument can be about a general point on line |
| **[3]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r} = \begin{pmatrix}1\\-2\\7\end{pmatrix} + \lambda\begin{pmatrix}4\\-1\\-1\end{pmatrix}$ | B1 | |
| Closest point where meets $\Pi$: $4(1+4\lambda)-(-2-\lambda)-(7-\lambda)=8$ | M1 | Parametric form of $(x, y, z)$ substituted into plane |
| $\Rightarrow \lambda = \frac{1}{2}$ | A1ft | |
| $\Rightarrow \mathbf{r} = \begin{pmatrix}3\\-\frac{5}{2}\\\frac{13}{2}\end{pmatrix}$ | A1 | |
| **[4]** | | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r} = \begin{pmatrix}3\\-\frac{5}{2}\\\frac{13}{2}\end{pmatrix} + \lambda\begin{pmatrix}2\\3\\5\end{pmatrix}$ | B1ft | oe; must have "$\mathbf{r} =$" |
| **[1]** | | |
---6 The line $l$ has equations $\frac { x - 1 } { 2 } = \frac { y + 2 } { 3 } = \frac { z - 7 } { 5 }$. The plane $\Pi$ has equation $4 x - y - z = 8$.\\
(i) Show that $l$ is parallel to $\Pi$ but does not lie in $\Pi$.\\
(ii) The point $A ( 1 , - 2,7 )$ is on $l$. Write down a vector equation of the line through $A$ which is perpendicular to $\Pi$. Hence find the position vector of the point on $\Pi$ which is closest to $A$.\\
(iii) Hence write down a vector equation of the line in $\Pi$ which is parallel to $l$ and closest to it.
\hfill \mbox{\textit{OCR FP3 2014 Q6 [8]}}