| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line of intersection of planes |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard techniques: finding line of intersection requires solving simultaneous equations and expressing parametrically (routine but multi-step), while point-to-plane distance uses a direct formula. Both are textbook exercises with no novel insight required, though the Further Maths context and multi-step nature place it slightly above average difficulty. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}2\\1\\-1\end{pmatrix}\times\begin{pmatrix}3\\5\\2\end{pmatrix}=\begin{pmatrix}7\\-7\\7\end{pmatrix}=7\begin{pmatrix}1\\-1\\1\end{pmatrix}\) | M1 | Requires evidence of method for cross product or at least 2 correct values calculated |
| A1 | ||
| (eg) \(z=0\Rightarrow 2x+y=4, 3x+5y=13\Rightarrow x=1, y=2\) | M1 | Or any valid point e.g. \((0,3,-1)\), \((3,0,2)\) |
| \(\mathbf{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\1\end{pmatrix}\) | A1 | oe vector form. Must have full equation including '\(\mathbf{r}=\)' |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Find one point | M1 | |
| Find a second point and vector between points | M1 | |
| Multiple of \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\) | A1 | |
| \(\mathbf{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\1\end{pmatrix}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Solve simultaneously | M1 | To at least expressions for \(x,y,z\) parametrically, or two relationships between 2 variables |
| M1 | ||
| Point and direction found | A1 | |
| \(\mathbf{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\1\end{pmatrix}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{\lvert 2\times2+5-{-2}-4\rvert}{\sqrt{2^2+1^2+1^2}}=\dfrac{7}{\sqrt{6}}\) | M1 | Condone lack of absolute signs for M1. 2.86 with no workings scores M1 |
| A1 | oe surd form. isw | |
| Alternative: find parameter for perpendicular meets plane and use to find distance | M1 | For complete method with calculation errors. Look for \(\lambda=-7/6\) |
| [2] |
## Question 1:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}2\\1\\-1\end{pmatrix}\times\begin{pmatrix}3\\5\\2\end{pmatrix}=\begin{pmatrix}7\\-7\\7\end{pmatrix}=7\begin{pmatrix}1\\-1\\1\end{pmatrix}$ | M1 | Requires evidence of method for cross product or at least 2 correct values calculated |
| | A1 | |
| (eg) $z=0\Rightarrow 2x+y=4, 3x+5y=13\Rightarrow x=1, y=2$ | M1 | Or any valid point e.g. $(0,3,-1)$, $(3,0,2)$ |
| $\mathbf{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\1\end{pmatrix}$ | A1 | oe vector form. Must have full equation including '$\mathbf{r}=$' |
| **[4]** | | |
**Alternative: Find one point**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Find one point | M1 | |
| Find a second point and vector between points | M1 | |
| Multiple of $\begin{pmatrix}1\\-1\\1\end{pmatrix}$ | A1 | |
| $\mathbf{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\1\end{pmatrix}$ | A1 | |
**Alternative: Solve simultaneously**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solve simultaneously | M1 | To at least expressions for $x,y,z$ parametrically, or two relationships between 2 variables |
| | M1 | |
| Point **and** direction found | A1 | |
| $\mathbf{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\1\end{pmatrix}$ | A1 | |
---
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{\lvert 2\times2+5-{-2}-4\rvert}{\sqrt{2^2+1^2+1^2}}=\dfrac{7}{\sqrt{6}}$ | M1 | Condone lack of absolute signs for M1. 2.86 with no workings scores M1 |
| | A1 | oe surd form. isw |
| **Alternative:** find parameter for perpendicular meets plane and use to find distance | M1 | For complete method with calculation errors. Look for $\lambda=-7/6$ |
| **[2]** | | |
---1 (i) Find a vector equation of the line of intersection of the planes $2 x + y - z = 4$ and $3 x + 5 y + 2 z = 13$.\\
(ii) Find the exact distance of the point $( 2,5 , - 2 )$ from the plane $2 x + y - z = 4$.
\hfill \mbox{\textit{OCR FP3 2014 Q1 [6]}}