Question 8 - A-Level Maths

Edexcel FP3 2016 June — Question 8 12 marks

Exam Board	Edexcel
Module	FP3 (Further Pure Mathematics 3)
Year	2016
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Lines & Planes
Type	Line of intersection of planes
Difficulty	Standard +0.8 This is a multi-part Further Maths question requiring: (a) finding normal vectors and checking perpendicularity via dot product, (b) converting parametric to Cartesian form using cross product, and (c) finding line of intersection in vector product form. While each technique is standard for FP3, the combination of three distinct procedures and the less common vector product form for the line makes this moderately challenging but still within typical Further Maths scope.
Spec	4.04a Line equations: 2D and 3D, cartesian and vector forms 4.04b Plane equations: cartesian and vector forms

The plane $\Pi _ { 1 }$ has equation

$$x - 5 y - 2 z = 3$$ The plane $\Pi _ { 2 }$ has equation $$\mathbf { r } = \mathbf { i } + 2 \mathbf { j } + \mathbf { k } + \lambda ( \mathbf { i } + 4 \mathbf { j } + 3 \mathbf { k } ) + \mu ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } )$$ where $\lambda$ and $\mu$ are scalar parameters.

Show that $\Pi _ { 1 }$ is perpendicular to $\Pi _ { 2 }$
Find a cartesian equation for $\Pi _ { 2 }$
Find an equation for the line of intersection of $\Pi _ { 1 }$ and $\Pi _ { 2 }$ giving your answer in the form $( \mathbf { r } - \mathbf { a } ) \times \mathbf { b } = \mathbf { 0 }$, where $\mathbf { a }$ and $\mathbf { b }$ are constant vectors to be found.
(6)

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Question 8:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&4&3\\2&-1&1\end{vmatrix} = \begin{pmatrix}7\\5\\-9\end{pmatrix}$	M1A1	M1: Attempt cross product between direction vectors or any 2 vectors in the plane. If working not shown or unclear, 2 elements should be correct. A1: Correct vector
$\begin{pmatrix}1\\-5\\-2\end{pmatrix}\cdot\begin{pmatrix}7\\5\\-9\end{pmatrix} = 7-25+18$	M1	Attempts $\begin{pmatrix}1\\-5\\-2\end{pmatrix}\cdot$ their vector product
$\begin{pmatrix}1\\-5\\-2\end{pmatrix}\cdot\begin{pmatrix}7\\5\\-9\end{pmatrix} = 7-25+18 = 0\ \therefore$ perpendicular	A1	Correctly obtains $= 0$ and gives a conclusion. Note: just writing $= 0\ \therefore$ perpendicular scores M1A0; showing $7-25+18=0\ \therefore$ perpendicular scores M1A1. If $\begin{pmatrix}7\\5\\-9\end{pmatrix}$ is incorrect then $\begin{pmatrix}1\\-5\\-2\end{pmatrix}\cdot\begin{pmatrix}a\\b\\c\end{pmatrix} = a-5b-2c$ needs to be seen to score the M mark

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\begin{pmatrix}7\\5\\-9\end{pmatrix}\cdot\begin{pmatrix}1\\2\\1\end{pmatrix} = 8 \Rightarrow 7x+5y-9z=8$	M1A1	M1: Uses $\mathbf{i}+2\mathbf{j}+\mathbf{k}$ and their vector product to find the Cartesian equation of $\Pi_2$. Must be clear that $\mathbf{i}+2\mathbf{j}+\mathbf{k}$ (or a point on the plane) is being used. A1: Correct equation (any multiple or equivalent equation)

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\7&5&-9\\1&-5&-2\end{vmatrix} = \begin{pmatrix}-55\\5\\-40\end{pmatrix}$	M1A1	M1: Attempt cross product of normal vectors. A1: $k(11\mathbf{i}-\mathbf{j}+8\mathbf{k})$
$x=0: \left(0,-\frac{1}{5},-1\right),\quad y=0: \left(-\frac{11}{5},0,-\frac{13}{5}\right),\quad z=0: \left(\frac{11}{8},-\frac{13}{8},0\right)$ Note points on the line satisfy $\left(11t,-\frac{1}{5}-t,-1+8t\right)$	M1A1	M1: Attempt point on the line ($x$, $y$ and $z$). A1: Correct coordinates
$\left(\mathbf{r}-\left(-\frac{1}{5}\mathbf{j}-\mathbf{k}\right)\right)\times(11\mathbf{i}-\mathbf{j}+8\mathbf{k}) = \mathbf{0}$	ddM1A1	ddM1: $(\mathbf{r} - \text{their point})\times\text{their direction} = \mathbf{0}$; not required for this mark. Dependent on both previous method marks. A1: Correct equation (oe)

Question (c): Line of Intersection of Two Planes

Alternatives by Simultaneous Equations

Case 1: Eliminates $y$ then obtains $f(x) = g(y) = z$

Answer	Marks	Guidance
Working	Marks	Guidance
$x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 8x - 11z = 11$
$z = \frac{8x-11}{11},\ x = \frac{11+11z}{8} \Rightarrow \frac{11+11z}{8} - 5y - 2z = 3 \Rightarrow z = \frac{-40y-13}{5}$
Obtains $\frac{8x-11}{11} = \frac{-40y-13}{5} = z$	M1A1	M1: Obtains $f(x) = g(y) = z$; A1: Correct expressions
$\frac{x - \frac{11}{8}}{\frac{11}{8}} = \frac{y + \frac{13}{40}}{-\frac{1}{8}} = \frac{z(−0)}{(1)}$	M1A1	M1: Correct processing on at least one expression (not $z$) to enable identification of position and direction; A1: Correct equations
$\left(\mathbf{r} - \left(\frac{11}{8}\mathbf{i} - \frac{13}{40}\mathbf{j}\right)\right) \times \left(\frac{11}{8}\mathbf{i} - \frac{1}{8}\mathbf{j} + \mathbf{k}\right) = \mathbf{0}$	ddM1A1	ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= 0" not required; Dependent on both previous M marks; A1: Correct equation (oe)

Case 2: Eliminates $x$ then obtains $f(x) = y = g(z)$

Answer	Marks	Guidance
Working	Marks	Guidance
$x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 40y + 5z = -13$
$y = \frac{-13-5z}{40},\ z = \frac{-13-40y}{5} \Rightarrow x - 5y + 2\!\left(\frac{13+40y}{5}\right) = 3 \Rightarrow y = \frac{-5x-11}{55}$
$\frac{-5x-11}{55} = y = \frac{-13-5z}{40}$	M1A1	M1: Obtains $f(x) = y = g(z)$; A1: Correct expressions
$\frac{x + \frac{11}{5}}{-11} = \frac{y(−0)}{(1)} = \frac{z + \frac{13}{5}}{-8}$	M1A1	M1: Correct processing on at least one expression (not $y$) to enable identification of position and direction; A1: Correct equations
$\left(\mathbf{r} - \left(-\frac{11}{5}\mathbf{i} - \frac{13}{5}\mathbf{k}\right)\right) \times \left(-1\mathbf{i} + \mathbf{j} - 8\mathbf{k}\right) = \mathbf{0}$	ddM1A1	ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= 0" not required; Dependent on both previous M marks; A1: Correct equation (oe)

Case 3: Eliminates $z$ then obtains $x = f(y) = g(z)$

Answer	Marks	Guidance
Working	Marks	Guidance
$x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 5x + 55y = -11$
$x = \frac{-55y-11}{5},\ y = \frac{-11-5x}{55} \Rightarrow x + 5\!\left(\frac{11+5x}{55}\right) - 2x = 3 \Rightarrow x = \frac{11z+11}{8}$
$x = \frac{-55y-11}{5} = \frac{11z+11}{8}$	M1A1	M1: Obtains $x = f(y) = g(z)$; A1: Correct expressions
$\frac{x(−0)}{(1)} = \frac{y + \frac{1}{5}}{-\frac{1}{11}} = \frac{z+1}{\frac{8}{11}}$	M1A1	M1: Correct processing on at least one expression (not $z$) to enable identification of position and direction; A1: Correct equations
$\left(\mathbf{r} - \left(-\frac{1}{5}\mathbf{j} - \mathbf{k}\right)\right) \times \left(\mathbf{i} - \frac{1}{11}\mathbf{j} + \frac{8}{11}\mathbf{k}\right) = \mathbf{0}$	ddM1A1	ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= 0" not required; Dependent on both previous M marks; A1: Correct equation (oe)

Alternatives by Parameters

Case 1: Eliminates $x$

Answer	Marks	Guidance
Working	Marks	Guidance
$x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 8x - 11z = 11$
$x = t \Rightarrow z = -1 + \frac{8}{11}t,\ y = -\frac{1}{5} - \frac{1}{11}t$	M1A1	M1: Obtains $x$, $y$ and $z$ in terms of $\lambda$; A1: Correct expressions
$\text{Pos}: -\frac{1}{5}\mathbf{j} - \mathbf{k} \quad \text{Dir}: \mathbf{i} - \frac{1}{11}\mathbf{j} + \frac{8}{11}\mathbf{k}$	M1A1	M1: Uses their equations to obtain position and direction; A1: Correct position and direction
$\left(\mathbf{r} - \left(-\frac{1}{5}\mathbf{j} - \mathbf{k}\right)\right) \times \left(\mathbf{i} - \frac{1}{11}\mathbf{j} + \frac{8}{11}\mathbf{k}\right) = \mathbf{0}$	ddM1A1	ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= 0" not required; Dependent on both previous M marks; A1: Correct equation (oe)

Case 2: Eliminates $y$

Answer	Marks	Guidance
Working	Marks	Guidance
$x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 40y + 15z = -13$
$y = t \Rightarrow z = -\frac{13}{5} - 8t,\ y = -\frac{1}{5} - 11t$	M1A1	M1: Obtains $x$, $y$ and $z$ in terms of $\lambda$; A1: Correct expressions
$\text{Pos}: -\frac{11}{5}\mathbf{i} - \frac{13}{5}\mathbf{k} \quad \text{Dir}: -\frac{11}{5}\mathbf{i} + \mathbf{j} - 8\mathbf{k}$	M1A1	M1: Uses their equations to obtain position and direction; A1: Correct position and direction
$\left(\mathbf{r} - \left(-\frac{11}{5}\mathbf{i} - \frac{13}{5}\mathbf{k}\right)\right) \times \left(-11\mathbf{i} + \mathbf{j} - 8\mathbf{k}\right) = \mathbf{0}$	ddM1A1	ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= 0" not required; Dependent on both previous M marks; A1: Correct equation (oe)

Case 3: Eliminates $z$

Answer	Marks	Guidance
Working	Marks	Guidance
$x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 8x - 11z = 11$
$z = t \Rightarrow x = \frac{11}{8} + \frac{11}{8}t,\ y = -\frac{13}{40} - \frac{1}{8}t$	M1A1	M1: Obtains $x$, $y$ and $z$ in terms of $\lambda$; A1: Correct expressions
$\text{Pos}: \frac{11}{8}\mathbf{i} - \frac{13}{40}\mathbf{j} \quad \text{Dir}: \frac{11}{8}\mathbf{i} - \frac{1}{8}\mathbf{j} + \mathbf{k}$	M1A1	M1: Uses their equations to obtain position and direction; A1: Correct position and direction
$\left(\mathbf{r} - \left(\frac{11}{8}\mathbf{i} - \frac{13}{40}\mathbf{j}\right)\right) \times \left(\frac{11}{8}\mathbf{i} - \frac{1}{8}\mathbf{j} + \mathbf{k}\right) = \mathbf{0}$	ddM1A1	ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= 0" not required; Dependent on both previous M marks; A1: Correct equation (oe)

Alternative by Finding 2 Points on the Line

Answer	Marks	Guidance
Working	Marks	Guidance
$x=0: \left(0, -\frac{1}{5}, -1\right),\quad y=0: \left(-\frac{11}{5}, 0, -\frac{13}{5}\right),\quad z=0: \left(\frac{11}{8}, -\frac{13}{40}, 0\right)$	M1A1	M1: Attempts two points on the line; A1: Two correct coordinates
$\text{Dir}: -\frac{1}{5}\mathbf{j} - \mathbf{k} - \left(-\frac{11}{5}\mathbf{i} - \frac{13}{5}\mathbf{k}\right) = \frac{11}{5}\mathbf{i} - \frac{1}{5}\mathbf{j} + \frac{8}{5}\mathbf{k}$	M1A1	M1: Subtracts to obtain direction; A1: Correct direction
$\left(\mathbf{r} - \left(-\frac{1}{5}\mathbf{j} - \mathbf{k}\right)\right) \times \left(\frac{11}{5}\mathbf{i} - \frac{1}{5}\mathbf{j} + \frac{8}{5}\mathbf{k}\right) = \mathbf{0}$	ddM1A1	ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= 0" not required; Dependent on both previous M marks; A1: Correct equation (oe)

# Question 8:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&4&3\\2&-1&1\end{vmatrix} = \begin{pmatrix}7\\5\\-9\end{pmatrix}$ | M1A1 | M1: Attempt cross product between direction vectors or any 2 vectors in the plane. If working not shown or unclear, 2 elements should be correct. A1: Correct vector |
| $\begin{pmatrix}1\\-5\\-2\end{pmatrix}\cdot\begin{pmatrix}7\\5\\-9\end{pmatrix} = 7-25+18$ | M1 | Attempts $\begin{pmatrix}1\\-5\\-2\end{pmatrix}\cdot$ their vector product |
| $\begin{pmatrix}1\\-5\\-2\end{pmatrix}\cdot\begin{pmatrix}7\\5\\-9\end{pmatrix} = 7-25+18 = 0\ \therefore$ perpendicular | A1 | Correctly obtains $= 0$ and gives a conclusion. Note: just writing $= 0\ \therefore$ perpendicular scores M1A0; showing $7-25+18=0\ \therefore$ perpendicular scores M1A1. If $\begin{pmatrix}7\\5\\-9\end{pmatrix}$ is incorrect then $\begin{pmatrix}1\\-5\\-2\end{pmatrix}\cdot\begin{pmatrix}a\\b\\c\end{pmatrix} = a-5b-2c$ needs to be seen to score the M mark |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}7\\5\\-9\end{pmatrix}\cdot\begin{pmatrix}1\\2\\1\end{pmatrix} = 8 \Rightarrow 7x+5y-9z=8$ | M1A1 | M1: Uses $\mathbf{i}+2\mathbf{j}+\mathbf{k}$ and their vector product to find the Cartesian equation of $\Pi_2$. Must be clear that $\mathbf{i}+2\mathbf{j}+\mathbf{k}$ (or a point on the plane) is being used. A1: Correct equation (any multiple or equivalent equation) |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\7&5&-9\\1&-5&-2\end{vmatrix} = \begin{pmatrix}-55\\5\\-40\end{pmatrix}$ | M1A1 | M1: Attempt cross product of normal vectors. A1: $k(11\mathbf{i}-\mathbf{j}+8\mathbf{k})$ |
| $x=0: \left(0,-\frac{1}{5},-1\right),\quad y=0: \left(-\frac{11}{5},0,-\frac{13}{5}\right),\quad z=0: \left(\frac{11}{8},-\frac{13}{8},0\right)$ Note points on the line satisfy $\left(11t,-\frac{1}{5}-t,-1+8t\right)$ | M1A1 | M1: Attempt point on the line ($x$, $y$ and $z$). A1: Correct coordinates |
| $\left(\mathbf{r}-\left(-\frac{1}{5}\mathbf{j}-\mathbf{k}\right)\right)\times(11\mathbf{i}-\mathbf{j}+8\mathbf{k}) = \mathbf{0}$ | ddM1A1 | ddM1: $(\mathbf{r} - \text{their point})\times\text{their direction} = \mathbf{0}$; not required for this mark. Dependent on both previous method marks. A1: Correct equation (oe) |

# Question (c): Line of Intersection of Two Planes

## Alternatives by Simultaneous Equations

---

### Case 1: Eliminates $y$ then obtains $f(x) = g(y) = z$

| Working | Marks | Guidance |
|---------|-------|----------|
| $x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 8x - 11z = 11$ | | |
| $z = \frac{8x-11}{11},\ x = \frac{11+11z}{8} \Rightarrow \frac{11+11z}{8} - 5y - 2z = 3 \Rightarrow z = \frac{-40y-13}{5}$ | | |
| Obtains $\frac{8x-11}{11} = \frac{-40y-13}{5} = z$ | M1A1 | M1: Obtains $f(x) = g(y) = z$; A1: Correct expressions |
| $\frac{x - \frac{11}{8}}{\frac{11}{8}} = \frac{y + \frac{13}{40}}{-\frac{1}{8}} = \frac{z(−0)}{(1)}$ | M1A1 | M1: Correct processing on at least one expression (not $z$) to enable identification of position and direction; A1: Correct equations |
| $\left(\mathbf{r} - \left(\frac{11}{8}\mathbf{i} - \frac{13}{40}\mathbf{j}\right)\right) \times \left(\frac{11}{8}\mathbf{i} - \frac{1}{8}\mathbf{j} + \mathbf{k}\right) = \mathbf{0}$ | ddM1A1 | ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= **0**" not required; Dependent on both previous M marks; A1: Correct equation (oe) |

---

### Case 2: Eliminates $x$ then obtains $f(x) = y = g(z)$

| Working | Marks | Guidance |
|---------|-------|----------|
| $x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 40y + 5z = -13$ | | |
| $y = \frac{-13-5z}{40},\ z = \frac{-13-40y}{5} \Rightarrow x - 5y + 2\!\left(\frac{13+40y}{5}\right) = 3 \Rightarrow y = \frac{-5x-11}{55}$ | | |
| $\frac{-5x-11}{55} = y = \frac{-13-5z}{40}$ | M1A1 | M1: Obtains $f(x) = y = g(z)$; A1: Correct expressions |
| $\frac{x + \frac{11}{5}}{-11} = \frac{y(−0)}{(1)} = \frac{z + \frac{13}{5}}{-8}$ | M1A1 | M1: Correct processing on at least one expression (not $y$) to enable identification of position and direction; A1: Correct equations |
| $\left(\mathbf{r} - \left(-\frac{11}{5}\mathbf{i} - \frac{13}{5}\mathbf{k}\right)\right) \times \left(-1\mathbf{i} + \mathbf{j} - 8\mathbf{k}\right) = \mathbf{0}$ | ddM1A1 | ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= **0**" not required; Dependent on both previous M marks; A1: Correct equation (oe) |

---

### Case 3: Eliminates $z$ then obtains $x = f(y) = g(z)$

| Working | Marks | Guidance |
|---------|-------|----------|
| $x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 5x + 55y = -11$ | | |
| $x = \frac{-55y-11}{5},\ y = \frac{-11-5x}{55} \Rightarrow x + 5\!\left(\frac{11+5x}{55}\right) - 2x = 3 \Rightarrow x = \frac{11z+11}{8}$ | | |
| $x = \frac{-55y-11}{5} = \frac{11z+11}{8}$ | M1A1 | M1: Obtains $x = f(y) = g(z)$; A1: Correct expressions |
| $\frac{x(−0)}{(1)} = \frac{y + \frac{1}{5}}{-\frac{1}{11}} = \frac{z+1}{\frac{8}{11}}$ | M1A1 | M1: Correct processing on at least one expression (not $z$) to enable identification of position and direction; A1: Correct equations |
| $\left(\mathbf{r} - \left(-\frac{1}{5}\mathbf{j} - \mathbf{k}\right)\right) \times \left(\mathbf{i} - \frac{1}{11}\mathbf{j} + \frac{8}{11}\mathbf{k}\right) = \mathbf{0}$ | ddM1A1 | ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= **0**" not required; Dependent on both previous M marks; A1: Correct equation (oe) |

---

## Alternatives by Parameters

### Case 1: Eliminates $x$

| Working | Marks | Guidance |
|---------|-------|----------|
| $x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 8x - 11z = 11$ | | |
| $x = t \Rightarrow z = -1 + \frac{8}{11}t,\ y = -\frac{1}{5} - \frac{1}{11}t$ | M1A1 | M1: Obtains $x$, $y$ and $z$ in terms of $\lambda$; A1: Correct expressions |
| $\text{Pos}: -\frac{1}{5}\mathbf{j} - \mathbf{k} \quad \text{Dir}: \mathbf{i} - \frac{1}{11}\mathbf{j} + \frac{8}{11}\mathbf{k}$ | M1A1 | M1: Uses their equations to obtain position and direction; A1: Correct position and direction |
| $\left(\mathbf{r} - \left(-\frac{1}{5}\mathbf{j} - \mathbf{k}\right)\right) \times \left(\mathbf{i} - \frac{1}{11}\mathbf{j} + \frac{8}{11}\mathbf{k}\right) = \mathbf{0}$ | ddM1A1 | ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= **0**" not required; Dependent on both previous M marks; A1: Correct equation (oe) |

---

### Case 2: Eliminates $y$

| Working | Marks | Guidance |
|---------|-------|----------|
| $x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 40y + 15z = -13$ | | |
| $y = t \Rightarrow z = -\frac{13}{5} - 8t,\ y = -\frac{1}{5} - 11t$ | M1A1 | M1: Obtains $x$, $y$ and $z$ in terms of $\lambda$; A1: Correct expressions |
| $\text{Pos}: -\frac{11}{5}\mathbf{i} - \frac{13}{5}\mathbf{k} \quad \text{Dir}: -\frac{11}{5}\mathbf{i} + \mathbf{j} - 8\mathbf{k}$ | M1A1 | M1: Uses their equations to obtain position and direction; A1: Correct position and direction |
| $\left(\mathbf{r} - \left(-\frac{11}{5}\mathbf{i} - \frac{13}{5}\mathbf{k}\right)\right) \times \left(-11\mathbf{i} + \mathbf{j} - 8\mathbf{k}\right) = \mathbf{0}$ | ddM1A1 | ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= **0**" not required; Dependent on both previous M marks; A1: Correct equation (oe) |

---

### Case 3: Eliminates $z$

| Working | Marks | Guidance |
|---------|-------|----------|
| $x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 8x - 11z = 11$ | | |
| $z = t \Rightarrow x = \frac{11}{8} + \frac{11}{8}t,\ y = -\frac{13}{40} - \frac{1}{8}t$ | M1A1 | M1: Obtains $x$, $y$ and $z$ in terms of $\lambda$; A1: Correct expressions |
| $\text{Pos}: \frac{11}{8}\mathbf{i} - \frac{13}{40}\mathbf{j} \quad \text{Dir}: \frac{11}{8}\mathbf{i} - \frac{1}{8}\mathbf{j} + \mathbf{k}$ | M1A1 | M1: Uses their equations to obtain position and direction; A1: Correct position and direction |
| $\left(\mathbf{r} - \left(\frac{11}{8}\mathbf{i} - \frac{13}{40}\mathbf{j}\right)\right) \times \left(\frac{11}{8}\mathbf{i} - \frac{1}{8}\mathbf{j} + \mathbf{k}\right) = \mathbf{0}$ | ddM1A1 | ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= **0**" not required; Dependent on both previous M marks; A1: Correct equation (oe) |

---

## Alternative by Finding 2 Points on the Line

| Working | Marks | Guidance |
|---------|-------|----------|
| $x=0: \left(0, -\frac{1}{5}, -1\right),\quad y=0: \left(-\frac{11}{5}, 0, -\frac{13}{5}\right),\quad z=0: \left(\frac{11}{8}, -\frac{13}{40}, 0\right)$ | M1A1 | M1: Attempts two points on the line; A1: Two correct coordinates |
| $\text{Dir}: -\frac{1}{5}\mathbf{j} - \mathbf{k} - \left(-\frac{11}{5}\mathbf{i} - \frac{13}{5}\mathbf{k}\right) = \frac{11}{5}\mathbf{i} - \frac{1}{5}\mathbf{j} + \frac{8}{5}\mathbf{k}$ | M1A1 | M1: Subtracts to obtain direction; A1: Correct direction |
| $\left(\mathbf{r} - \left(-\frac{1}{5}\mathbf{j} - \mathbf{k}\right)\right) \times \left(\frac{11}{5}\mathbf{i} - \frac{1}{5}\mathbf{j} + \frac{8}{5}\mathbf{k}\right) = \mathbf{0}$ | ddM1A1 | ddM1: $(\mathbf{r} - \text{their point}) \times \text{their direction}$; "= **0**" not required; Dependent on both previous M marks; A1: Correct equation (oe) |

Show LaTeX source

\begin{enumerate}
  \item The plane $\Pi _ { 1 }$ has equation
\end{enumerate}

$$x - 5 y - 2 z = 3$$

The plane $\Pi _ { 2 }$ has equation

$$\mathbf { r } = \mathbf { i } + 2 \mathbf { j } + \mathbf { k } + \lambda ( \mathbf { i } + 4 \mathbf { j } + 3 \mathbf { k } ) + \mu ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } )$$

where $\lambda$ and $\mu$ are scalar parameters.\\
(a) Show that $\Pi _ { 1 }$ is perpendicular to $\Pi _ { 2 }$\\
(b) Find a cartesian equation for $\Pi _ { 2 }$\\
(c) Find an equation for the line of intersection of $\Pi _ { 1 }$ and $\Pi _ { 2 }$ giving your answer in the form $( \mathbf { r } - \mathbf { a } ) \times \mathbf { b } = \mathbf { 0 }$, where $\mathbf { a }$ and $\mathbf { b }$ are constant vectors to be found.\\
(6)\\

\begin{center}

\end{center}

\hfill \mbox{\textit{Edexcel FP3 2016 Q8 [12]}}

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Q1 4 Q2 7 Q3 8 Q4 12 Q5 11 Q6 11 Q7 10 Q8 12

Edexcel FP3 2016 June — Question 8 12 marks

Question 8:

Part (a):

Part (b):

Part (c):

Question (c): Line of Intersection of Two Planes

Alternatives by Simultaneous Equations

Case 1: Eliminates \(y\) then obtains \(f(x) = g(y) = z\)

Case 2: Eliminates \(x\) then obtains \(f(x) = y = g(z)\)

Case 3: Eliminates \(z\) then obtains \(x = f(y) = g(z)\)

Alternatives by Parameters

Case 1: Eliminates \(x\)

Case 2: Eliminates \(y\)

Case 3: Eliminates \(z\)

Alternative by Finding 2 Points on the Line

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