OCR FP3 2009 June — Question 6 9 marks

Exam BoardOCR
ModuleFP3 (Further Pure Mathematics 3)
Year2009
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine of intersection of planes
DifficultyStandard +0.8 This is a Further Maths question requiring conversion between vector forms of planes and finding line of intersection. Part (i) requires computing a normal vector via cross product, which is routine FP3 technique. Part (ii) requires finding the direction vector (cross product of normals) and a point on both planes (solving simultaneous equations), involving multiple coordinated steps but following standard algorithms. Slightly above average difficulty due to being Further Maths content with computational complexity, but remains a textbook-style exercise.
Spec4.04b Plane equations: cartesian and vector forms

6 The plane \(\Pi _ { 1 }\) has equation \(\mathbf { r } = \left( \begin{array} { l } 2 \\ 2 \\ 1 \end{array} \right) + \lambda \left( \begin{array} { l } 1 \\ 1 \\ 0 \end{array} \right) + \mu \left( \begin{array} { r } 1 \\ - 5 \\ - 2 \end{array} \right)\).
  1. Express the equation of \(\Pi _ { 1 }\) in the form r.n \(= p\). The plane \(\Pi _ { 2 }\) has equation \(\mathbf { r } . \left( \begin{array} { r } 7 \\ 17 \\ - 3 \end{array} \right) = 21\).
  2. Find an equation of the line of intersection of \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\), giving your answer in the form \(\mathbf { r } = \mathbf { a } + t \mathbf { b }\).

Part (i)
AnswerMarks Guidance
\(n_1 = [1, 1, 0] \times [1, -5, -2]\)M1 For attempting to find vector product of the pair of direction vectors
\(= [-2, 2, -6] = k[1, -1, 3]\)A1 For correct \(n_1\)
Use \((2, 2, 1)\)M1 For substituting a point into equation
\(\Rightarrow r \cdot [-2, 2, -6] = -6 \Rightarrow r \cdot [1, -1, 3] = 3\)A1, 4 For correct equation, aef in this form
METHOD 2
AnswerMarks Guidance
\(x = 2 + \lambda + \mu\), \(y = 2\lambda - 5\mu\), \(z = 1 - 2\mu\)M1, M1 For writing as 3 linear equations. For attempting to eliminate \(\lambda\) and \(\mu\)
\(\Rightarrow x - y + 3z = 3\)A1 For correct cartesian equation
\(\Rightarrow r \cdot [1, -1, 3] = 3\)A1 For correct equation, aef in this form
Part (ii)
AnswerMarks
For \(r = a + tb\)
METHOD 1
AnswerMarks Guidance
\(b = [1, -1, 3] \times [7, 17, -3]\)M1 For attempting to find \(n_1 \times n_2\)
\(= k[2, -1, -1]\)A1∇ For a correct vector. If from \(n_1\) in (i). SR a correct vector may be stated without working. SR for \(a = [2, 2, 1]\) stated award M0
e.g. \(x, y\) or \(z = 0\) in \(\begin{cases} x - y + 3z = 3 \\ 7x + 17y - 3z = 21 \end{cases}\)M1 For attempting to find a point on the line
\(\Rightarrow a = \left[0, \frac{3}{2}, \frac{3}{2}\right]\) OR \([3, 0, 0]\) OR \([1, 1, 1]\)A1∇ For a correct vector. If from equation in (i). SR a correct vector may be stated without working. SR for \(a = [2, 2, 1]\) stated award M0
Line is (e.g.) \(r = [1, 1, 1] + t[2, -1, -1]\)A1∇, 5 For stating equation of line. If from \(a\) and \(b\). SR \(a = [2, 2, 1]\) stated award M0
METHOD 2
AnswerMarks Guidance
Solve \(\begin{cases} x - y + 3z = 3 \\ 7x + 17y - 3z = 21 \end{cases}\) by eliminating one variable (e.g. \(z\)). Use parameter for another variable (e.g. \(x\)) to find other variables in terms of \(t\)M1, M1 For attempting to solve equations. For attempting to find parametric solution
(eg) \(y = \frac{3}{2} - \frac{1}{2}t\), \(z = \frac{3}{2} - \frac{1}{2}t\)A1∇, A1∇ For correct expression for one variable. For correct expression for the other variable. If from equation in (i) for both
Line is (eg) \(r = \left[0, \frac{3}{2}, \frac{3}{2}\right] + t[2, -1, -1]\)A1∇ For stating equation of line. If from parametric solutions
METHOD 3
AnswerMarks Guidance
eg \(x, y\) or \(z = 0\) in \(\begin{cases} x - y + 3z = 3 \\ 7x + 17y - 3z = 21 \end{cases}\)M1 For attempting to find a point on the line
\(\Rightarrow a = \left[0, \frac{3}{2}, \frac{3}{2}\right]\) OR \([3, 0, 0]\) OR \([1, 1, 1]\)A1∇ For a correct vector. If from equation in (i). SR a correct vector may be stated without working. SR for \(a = [2, 2, 1]\) stated award M0
eg \([3, 0, 0] - [1, 1, 1]\)M1 For finding another point on the line and using it with the one already found to find \(b\)
\(b = k[2, -1, -1]\)A1∇ For a correct vector. If from equation in (i)
Line is (eg) \(r = [1, 1, 1] + t[2, -1, -1]\)A1∇ For stating equation of line. If from \(a\) and \(b\)
METHOD 4
AnswerMarks Guidance
A point on \(\Pi_1\) is \([2 + \lambda + \mu, 2\lambda - 5\mu, 1 - 2\mu]\)M1 For using parametric form for \(\Pi_1\) and substituting into \(\Pi_2\)
On \(\Pi_2 \Rightarrow [2 + \lambda + \mu, 2\lambda - 5\mu, 1 - 2\mu] \cdot [7, 17, -3] = 21\)A1 For correct unsimplified equation
\(\Rightarrow \lambda - 3\mu = -1\)A1 For correct equation
Line is (e.g.) \(r = [2, 2, 1] + (3\mu - 1)[1, 1, 0] + \mu[1, -5, -2]\)M1 For substituting into \(\Pi_1\) for \(\lambda\) or \(\mu\)
\(\Rightarrow r = [1, 1, 1]\) or \(\left[\frac{3}{2}, \frac{3}{2}, \frac{3}{2}\right] + t[2, -1, -1]\)A1 For stating equation of line
**Part (i)**

| $n_1 = [1, 1, 0] \times [1, -5, -2]$ | M1 | For attempting to find vector product of the pair of direction vectors |
|---|---|---|
| $= [-2, 2, -6] = k[1, -1, 3]$ | A1 | For correct $n_1$ |
| Use $(2, 2, 1)$ | M1 | For substituting a point into equation |
| $\Rightarrow r \cdot [-2, 2, -6] = -6 \Rightarrow r \cdot [1, -1, 3] = 3$ | A1, 4 | For correct equation, aef in this form |

**METHOD 2**

| $x = 2 + \lambda + \mu$, $y = 2\lambda - 5\mu$, $z = 1 - 2\mu$ | M1, M1 | For writing as 3 linear equations. For attempting to eliminate $\lambda$ and $\mu$ |
|---|---|---|
| $\Rightarrow x - y + 3z = 3$ | A1 | For correct cartesian equation |
| $\Rightarrow r \cdot [1, -1, 3] = 3$ | A1 | For correct equation, aef in this form |

**Part (ii)**

| For $r = a + tb$ | | |
|---|---|---|

**METHOD 1**

| $b = [1, -1, 3] \times [7, 17, -3]$ | M1 | For attempting to find $n_1 \times n_2$ |
|---|---|---|
| $= k[2, -1, -1]$ | A1∇ | For a correct vector. If from $n_1$ in (i). SR a correct vector may be stated without working. SR for $a = [2, 2, 1]$ stated award M0 |
| e.g. $x, y$ or $z = 0$ in $\begin{cases} x - y + 3z = 3 \\ 7x + 17y - 3z = 21 \end{cases}$ | M1 | For attempting to find a point on the line |
| $\Rightarrow a = \left[0, \frac{3}{2}, \frac{3}{2}\right]$ OR $[3, 0, 0]$ OR $[1, 1, 1]$ | A1∇ | For a correct vector. If from equation in (i). SR a correct vector may be stated without working. SR for $a = [2, 2, 1]$ stated award M0 |
| Line is (e.g.) $r = [1, 1, 1] + t[2, -1, -1]$ | A1∇, 5 | For stating equation of line. If from $a$ and $b$. SR $a = [2, 2, 1]$ stated award M0 |

**METHOD 2**

| Solve $\begin{cases} x - y + 3z = 3 \\ 7x + 17y - 3z = 21 \end{cases}$ by eliminating one variable (e.g. $z$). Use parameter for another variable (e.g. $x$) to find other variables in terms of $t$ | M1, M1 | For attempting to solve equations. For attempting to find parametric solution |
|---|---|---|
| (eg) $y = \frac{3}{2} - \frac{1}{2}t$, $z = \frac{3}{2} - \frac{1}{2}t$ | A1∇, A1∇ | For correct expression for one variable. For correct expression for the other variable. If from equation in (i) for both |
| Line is (eg) $r = \left[0, \frac{3}{2}, \frac{3}{2}\right] + t[2, -1, -1]$ | A1∇ | For stating equation of line. If from parametric solutions |

**METHOD 3**

| eg $x, y$ or $z = 0$ in $\begin{cases} x - y + 3z = 3 \\ 7x + 17y - 3z = 21 \end{cases}$ | M1 | For attempting to find a point on the line |
|---|---|---|
| $\Rightarrow a = \left[0, \frac{3}{2}, \frac{3}{2}\right]$ OR $[3, 0, 0]$ OR $[1, 1, 1]$ | A1∇ | For a correct vector. If from equation in (i). SR a correct vector may be stated without working. SR for $a = [2, 2, 1]$ stated award M0 |
| eg $[3, 0, 0] - [1, 1, 1]$ | M1 | For finding another point on the line and using it with the one already found to find $b$ |
| $b = k[2, -1, -1]$ | A1∇ | For a correct vector. If from equation in (i) |
| Line is (eg) $r = [1, 1, 1] + t[2, -1, -1]$ | A1∇ | For stating equation of line. If from $a$ and $b$ |

**METHOD 4**

| A point on $\Pi_1$ is $[2 + \lambda + \mu, 2\lambda - 5\mu, 1 - 2\mu]$ | M1 | For using parametric form for $\Pi_1$ and substituting into $\Pi_2$ |
|---|---|---|
| On $\Pi_2 \Rightarrow [2 + \lambda + \mu, 2\lambda - 5\mu, 1 - 2\mu] \cdot [7, 17, -3] = 21$ | A1 | For correct unsimplified equation |
| $\Rightarrow \lambda - 3\mu = -1$ | A1 | For correct equation |
| Line is (e.g.) $r = [2, 2, 1] + (3\mu - 1)[1, 1, 0] + \mu[1, -5, -2]$ | M1 | For substituting into $\Pi_1$ for $\lambda$ or $\mu$ |
| $\Rightarrow r = [1, 1, 1]$ or $\left[\frac{3}{2}, \frac{3}{2}, \frac{3}{2}\right] + t[2, -1, -1]$ | A1 | For stating equation of line |

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6 The plane $\Pi _ { 1 }$ has equation $\mathbf { r } = \left( \begin{array} { l } 2 \\ 2 \\ 1 \end{array} \right) + \lambda \left( \begin{array} { l } 1 \\ 1 \\ 0 \end{array} \right) + \mu \left( \begin{array} { r } 1 \\ - 5 \\ - 2 \end{array} \right)$.\\
(i) Express the equation of $\Pi _ { 1 }$ in the form r.n $= p$.

The plane $\Pi _ { 2 }$ has equation $\mathbf { r } . \left( \begin{array} { r } 7 \\ 17 \\ - 3 \end{array} \right) = 21$.\\
(ii) Find an equation of the line of intersection of $\Pi _ { 1 }$ and $\Pi _ { 2 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + t \mathbf { b }$.

\hfill \mbox{\textit{OCR FP3 2009 Q6 [9]}}