| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Geometric configuration of planes |
| Difficulty | Challenging +1.2 This is a structured FP3 vector geometry question requiring finding direction vectors of lines of intersection using cross products, recognizing parallel lines, and checking consistency of plane equations. While it involves multiple steps and Further Maths content (3D vector geometry), the techniques are standard and the question guides students through the logic systematically. The cross product calculations are routine, and part (iii) reduces to checking if a point satisfies an equation. More challenging than average A-level but typical for FP3. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| \([1, 1, -2] \times [1, -1, 3] = (\pm)[1, -5, -2]\) | M1 | For using \(\times\) of direction vectors |
| A1 | For correct direction | |
| \([1, -1, 3] \times [1, 5, -12] = (\pm)[-3, 15, 6]\) | M1 | For using \(\times\) of direction vectors |
| A1 | For correct direction | |
| \([-3, 15, 6] \propto k[1, -5, -2] \Rightarrow\) parallel | A1 5 | For argument completed AG (\(k = -3\) not essential) |
| Answer | Marks |
|---|---|
| B1 1 | For correct statement |
| Answer | Marks | Guidance |
|---|---|---|
| \(x + y - 2z = 5\) and \(x - y + 3z = 6\) e.g. \(z = 0 \Rightarrow (\frac{11}{2}, -\frac{1}{2}, 0)\) on \(l\) | M1 | For attempt to find points on 2 lines |
| A1 | For a correct point on one line | |
| \(x - y + 3z = 6\) and \(x + 5y - 12z = 12\) e.g. \(z = 0 \Rightarrow (7, 1, 0)\) on \(m\) | A1 | For a correct point on another line |
| \(x + y - 2z = 5\) and \(x + 5y - 12z = 12\) e.g. \(z = 0 \Rightarrow (\frac{13}{4}, \frac{7}{4}, 0)\) on \(l_3\) | A1 | For correct answer |
| Different points \(\Rightarrow\) no common line of intersection | A1 4 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(x + y - 2z = 5\) and \(x - y + 3z = 6\) e.g. \(z = 11 - 2x, \quad y = 27 - 5x\) | M1 | For finding (e.g.) \(y\) and \(z\) in terms of \(x\) OR eliminating one variable |
| A1 | For correct expressions OR equations | |
| LHS of eqn 3 \(=\) \(x + (135 - 25x) - (132 - 24x) = 3 \neq 12\) | A1 | For obtaining a contradiction from 3rd equation |
| \(\Rightarrow\) no common line of intersection | A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| LHS \(H_1 + 3l_1 - 2H_3\) | M2 | For attempt to link 3 equations |
| RHS \(3 \times 5 - 2 \times 6 = 3 \neq 12\) | A1 | For obtaining a contradiction |
| \(\Rightarrow\) no common line of intersection | A1 | For correct answer |
| SR Variations on all methods may gain full credit | SR t.t. may be allowed from relevant working |
## (i)
$[1, 1, -2] \times [1, -1, 3] = (\pm)[1, -5, -2]$ | M1 | For using $\times$ of direction vectors
| A1 | For correct direction
$[1, -1, 3] \times [1, 5, -12] = (\pm)[-3, 15, 6]$ | M1 | For using $\times$ of direction vectors
| A1 | For correct direction
$[-3, 15, 6] \propto k[1, -5, -2] \Rightarrow$ parallel | A1 5 | For argument completed **AG** ($k = -3$ not essential)
## (ii)
| B1 1 | For correct statement
## (iii)
**METHOD 1**
$x + y - 2z = 5$ and $x - y + 3z = 6$ e.g. $z = 0 \Rightarrow (\frac{11}{2}, -\frac{1}{2}, 0)$ on $l$ | M1 | For attempt to find points on 2 lines
| A1 | For a correct point on one line
$x - y + 3z = 6$ and $x + 5y - 12z = 12$ e.g. $z = 0 \Rightarrow (7, 1, 0)$ on $m$ | A1 | For a correct point on another line
$x + y - 2z = 5$ and $x + 5y - 12z = 12$ e.g. $z = 0 \Rightarrow (\frac{13}{4}, \frac{7}{4}, 0)$ on $l_3$ | A1 | For correct answer
Different points $\Rightarrow$ no common line of intersection | A1 4 | For correct answer
**METHOD 2**
$x + y - 2z = 5$ and $x - y + 3z = 6$ e.g. $z = 11 - 2x, \quad y = 27 - 5x$ | M1 | For finding (e.g.) $y$ and $z$ in terms of $x$ OR eliminating one variable
| A1 | For correct expressions OR equations
LHS of eqn 3 $=$ $x + (135 - 25x) - (132 - 24x) = 3 \neq 12$ | A1 | For obtaining a contradiction from 3rd equation
$\Rightarrow$ no common line of intersection | A1 | For correct answer
**METHOD 3**
LHS $H_1 + 3l_1 - 2H_3$ | M2 | For attempt to link 3 equations
RHS $3 \times 5 - 2 \times 6 = 3 \neq 12$ | A1 | For obtaining a contradiction
$\Rightarrow$ no common line of intersection | A1 | For correct answer
SR Variations on all methods may gain full credit | SR t.t. may be allowed from relevant working
---Three planes $\Pi_1$, $\Pi_2$ and $\Pi_3$ have equations
$$\mathbf{r} \cdot (\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = 5, \quad \mathbf{r} \cdot (\mathbf{i} - \mathbf{j} + 3\mathbf{k}) = 6, \quad \mathbf{r} \cdot (\mathbf{i} + 5\mathbf{j} - 12\mathbf{k}) = 12,$$
respectively. Planes $\Pi_1$ and $\Pi_2$ intersect in a line $l$; planes $\Pi_2$ and $\Pi_3$ intersect in a line $m$.
\begin{enumerate}[label=(\roman*)]
\item Show that $l$ and $m$ are in the same direction. [5]
\item Write down what you can deduce about the line of intersection of planes $\Pi_1$ and $\Pi_3$. [1]
\item By considering the cartesian equations of $\Pi_1$, $\Pi_2$ and $\Pi_3$, or otherwise, determine whether or not the three planes have a common line of intersection. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q7 [10]}}