| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Cartesian equation of a plane |
| Difficulty | Standard +0.8 This FP3 question requires finding a plane equation from two intersecting lines (needing cross product of direction vectors and verification of intersection), then calculating distance between parallel planes. While methodical, it demands solid 3D vector geometry skills and careful algebraic manipulation across multiple steps, placing it moderately above average difficulty. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{n} = [2.1, -3] \times [-1, 2, 4] = [10, -5, 5] = k[2, -1, 1]\) | M1 | For using \(\times\) of direction vectors |
| A1 | For correct n | |
| \((1, 3, 4) \Rightarrow 2x - y + z = 3\) | A1 3 | For substituting \((1, 3, 4)\) and obtaining AG ... (Verification only M0) |
| Answer | Marks | Guidance |
|---|---|---|
| distance = \(\frac{ | 2I - 3 | }{ |
| OR \(\frac{ | [1, 3, 4] - [a, b, c] | ·[2, -1, 1] |
| \(= \frac{18}{\sqrt{6}} = 3\sqrt{6}\) | A1 3 | For correct distance AEF |
| Answer | Marks | Guidance |
|---|---|---|
| \([1 + 2t, 3 - t, 4 + t]\) on \(q\) | M1 | For forming and solving an equation in \(t\) |
| \(\Rightarrow 2(1 + 2t) - (3 - t) + (4 + t) = 21 \Rightarrow t = 3\) | B1 | For \( |
| \(\Rightarrow\) distance \(= 3 | \mathbf{n} | = 3\sqrt{6}\) |
| Answer | Marks | Guidance |
|---|---|---|
| As Method 2 to \(t = 3 \Rightarrow (7, 0, 7)\) on \(q\) | M1* | For finding point where normal meets \(q\) |
| M1 (*dep) | For finding distance from \((1, 3, 4)\) | |
| \(= \sqrt{(7-1)^2 + (0-3)^2 + (7-4)^2} = \sqrt{54} = 3\sqrt{6}\) | A1 | For correct distance AEF |
## (i)
$\mathbf{n} = [2.1, -3] \times [-1, 2, 4] = [10, -5, 5] = k[2, -1, 1]$ | M1 | For using $\times$ of direction vectors
| A1 | For correct **n**
$(1, 3, 4) \Rightarrow 2x - y + z = 3$ | A1 3 | For substituting $(1, 3, 4)$ and obtaining **AG** ... (Verification only M0)
## (ii)
**METHOD 1**
distance = $\frac{|2I - 3|}{|\mathbf{n}|}$ OR $\frac{|[1, 3, 4]·[2, -1, 1] - 2|}{|\mathbf{n}|}$ | M1 | For using $\times$ of direction vectors OR $|[1, 3, -4] - [a, b, c]| · [2, -1, 1]$ soi
OR $\frac{|[1, 3, 4] - [a, b, c]|·[2, -1, 1]|}{|\mathbf{n}|}$ where $(a, b, c)$ is on $q$ | B1 | For $|\mathbf{n}| = \sqrt{6}$ soi
$= \frac{18}{\sqrt{6}} = 3\sqrt{6}$ | A1 3 | For correct distance **AEF**
**METHOD 2**
$[1 + 2t, 3 - t, 4 + t]$ on $q$ | M1 | For forming and solving an equation in $t$
$\Rightarrow 2(1 + 2t) - (3 - t) + (4 + t) = 21 \Rightarrow t = 3$ | B1 | For $|\mathbf{n}| = \sqrt{6}$ soi
$\Rightarrow$ distance $= 3|\mathbf{n}| = 3\sqrt{6}$ | A1 | For correct distance **AEF**
**METHOD 3**
As Method 2 to $t = 3 \Rightarrow (7, 0, 7)$ on $q$ | M1* | For finding point where normal meets $q$
| M1 (*dep) | For finding distance from $(1, 3, 4)$
$= \sqrt{(7-1)^2 + (0-3)^2 + (7-4)^2} = \sqrt{54} = 3\sqrt{6}$ | A1 | For correct distance **AEF**
---Two intersecting lines, lying in a plane $p$, have equations
$$\frac{x-1}{2} = \frac{y-3}{1} = \frac{z-4}{-3} \quad \text{and} \quad \frac{x-1}{-1} = \frac{y-3}{2} = \frac{z-4}{4}.$$
\begin{enumerate}[label=(\roman*)]
\item Obtain the equation of $p$ in the form $2x - y + z = 3$. [3]
\item Plane $q$ has equation $2x - y + z = 21$. Find the distance between $p$ and $q$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q2 [6]}}