Question 6 - A-Level Maths

Edexcel C4 2013 June — Question 6 12 marks

Exam Board	Edexcel
Module	C4 (Core Mathematics 4)
Year	2013
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors 3D & Lines
Type	Reflection and symmetry
Difficulty	Standard +0.3 This is a standard C4 vectors question involving reflection of a point in a line. While it requires multiple steps (finding perpendicular from point to line, calculating foot of perpendicular, then using midpoint property for reflection), these are well-practiced techniques that follow a predictable algorithm. The question appears incomplete but would typically involve finding the reflection of point A or B in line l, which is a routine application of vector methods taught at this level.
Spec	1.10c Magnitude and direction: of vectors 1.10d Vector operations: addition and scalar multiplication 4.04a Line equations: 2D and 3D, cartesian and vector forms

6. Relative to a fixed origin $O$, the point $A$ has position vector $21 \mathbf { i } - 17 \mathbf { j } + 6 \mathbf { k }$ and the point $B$ has position vector $25 \mathbf { i } - 14 \mathbf { j } + 18 \mathbf { k }$. The line $l$ has vector equation $$\mathbf { r } = \left( \begin{array} { r } a \\ b \\ 10 \end{array} \right) + \lambda \left( \begin{array} { r } 6 \\ c \\ - 1 \end{array} \right)$$ where $a , b$ and $c$ are constants and $\lambda$ is a parameter.
Given that the point $A$ lies on the line $l$,

find the value of $a$. Given also that the vector $\overrightarrow { A B }$ is perpendicular to $l$,
find the values of $b$ and $c$,
find the distance $A B$. The image of the point $B$ after reflection in the line $l$ is the point $B ^ { \prime }$.
Find the position vector of the point $B ^ { \prime }$. $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ \section*{Question 6 continued} $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$

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Question 6:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\lambda = 4$ (from $\mathbf{k}$: $10 - \lambda = 6$)	B1	$\lambda = 4$ seen or implied
Substitutes $\lambda$ into $a + 6\lambda = 21$	M1
$a = -3$	A1 cao	Award B1M1A1 if $a=-3$ stated with no working

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\overrightarrow{AB} = \begin{pmatrix}25\\-14\\18\end{pmatrix} - \begin{pmatrix}21\\-17\\6\end{pmatrix} = \begin{pmatrix}4\\3\\12\end{pmatrix}$	M1	Finds difference between $\overrightarrow{OA}$ and $\overrightarrow{OB}$; ignore labelling
$\overrightarrow{AB} \perp l \Rightarrow \overrightarrow{AB}\cdot\mathbf{d}=0 \Rightarrow \begin{pmatrix}4\\3\\12\end{pmatrix}\cdot\begin{pmatrix}6\\c\\-1\end{pmatrix} = 24+3c-12=0 \Rightarrow c=-4$	M1; A1ft	Applies dot product with direction vector set to zero; $c=-4$ or correct ft
$b + c\lambda = -17 \Rightarrow b + (-4)(4) = -17 \Rightarrow b = -1$	ddM1; A1 cso cao	Dependent on both previous M marks

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(\	AB\	= \sqrt{4^2+3^2+12^2}\)
\(\	AB\	= 13\)

Part (d):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\overrightarrow{OB'} = \overrightarrow{OA} + \overrightarrow{BA} = \begin{pmatrix}21\\-17\\6\end{pmatrix}+\begin{pmatrix}-4\\-3\\-12\end{pmatrix} = \begin{pmatrix}17\\-20\\-6\end{pmatrix}$	M1	Full applied method for coordinates of $B'$; M1 for 2 out of 3 correct components
$17\mathbf{i} - 20\mathbf{j} - 6\mathbf{k}$ or $(17,-20,-6)$	A1 cao

Question 6 (Part d - Acceptable Methods):

Answer	Marks	Guidance
Way 1: $\overrightarrow{OB'} = \overrightarrow{OA} + \overrightarrow{BA} = \begin{pmatrix}21\\-17\\6\end{pmatrix} + \begin{pmatrix}-4\\-3\\-12\end{pmatrix}$		using their $\overrightarrow{BA}$
Way 2: $\overrightarrow{OB'} = \overrightarrow{OA} - \overrightarrow{AB} = \begin{pmatrix}21\\-17\\6\end{pmatrix} - \begin{pmatrix}4\\3\\12\end{pmatrix}$		using their $\overrightarrow{AB}$
Way 3: $\overrightarrow{OB'} = \overrightarrow{OB} + 2\overrightarrow{BA} = \begin{pmatrix}25\\-14\\18\end{pmatrix} + 2\begin{pmatrix}-4\\-3\\-12\end{pmatrix}$		using their $\overrightarrow{BA}$
Way 4: $\overrightarrow{OB'} = \overrightarrow{OB} - 2\overrightarrow{AB} = \begin{pmatrix}25\\-14\\18\end{pmatrix} - 2\begin{pmatrix}4\\3\\12\end{pmatrix}$		using their $\overrightarrow{AB}$
Way 5: $\begin{pmatrix}25\\-14\\18\end{pmatrix} \to \begin{pmatrix}\text{Minus }4\\\text{Minus }3\\\text{Minus }12\end{pmatrix} \to \begin{pmatrix}21\\-17\\6\end{pmatrix} \to \begin{pmatrix}\text{Minus }4\\\text{Minus }3\\\text{Minus }12\end{pmatrix} \to \begin{pmatrix}17\\-20\\-6\end{pmatrix}$		so $\overrightarrow{OA}$ + their $\overrightarrow{BA}$

Way 6: $\overrightarrow{OB'} = 2\overrightarrow{OA} - \overrightarrow{OB} = 2\begin{pmatrix}21\\-17\\6\end{pmatrix} - \begin{pmatrix}25\\-14\\18\end{pmatrix}$

Way 7: $\overrightarrow{OB} = 25\mathbf{i} - 14\mathbf{j} + 18\mathbf{k}$, $\overrightarrow{OA} = 21\mathbf{i} - 17\mathbf{j} + 6\mathbf{k}$, $\overrightarrow{OB'} = p\mathbf{i} + q\mathbf{j} + r\mathbf{k}$

$(21,-17,6) = \left(\frac{25+p}{2}, \frac{-14+q}{2}, \frac{18+r}{2}\right)$

Answer	Marks	Guidance
$p = 21(2)-25 = 17$, $q = -17(2)+14 = -20$, $r = 6(2)-18 = -6$	M1	Writing down any two equations correctly and an attempt to find at least two of $p$, $q$ or $r$

# Question 6:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda = 4$ (from $\mathbf{k}$: $10 - \lambda = 6$) | B1 | $\lambda = 4$ seen or implied |
| Substitutes $\lambda$ into $a + 6\lambda = 21$ | M1 | |
| $a = -3$ | A1 cao | Award B1M1A1 if $a=-3$ stated with no working |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix}25\\-14\\18\end{pmatrix} - \begin{pmatrix}21\\-17\\6\end{pmatrix} = \begin{pmatrix}4\\3\\12\end{pmatrix}$ | M1 | Finds difference between $\overrightarrow{OA}$ and $\overrightarrow{OB}$; ignore labelling |
| $\overrightarrow{AB} \perp l \Rightarrow \overrightarrow{AB}\cdot\mathbf{d}=0 \Rightarrow \begin{pmatrix}4\\3\\12\end{pmatrix}\cdot\begin{pmatrix}6\\c\\-1\end{pmatrix} = 24+3c-12=0 \Rightarrow c=-4$ | M1; A1ft | Applies dot product with direction vector set to zero; $c=-4$ or correct ft |
| $b + c\lambda = -17 \Rightarrow b + (-4)(4) = -17 \Rightarrow b = -1$ | ddM1; A1 cso cao | Dependent on both previous M marks |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\|AB\| = \sqrt{4^2+3^2+12^2}$ | M1 | Three-term Pythagoras; square root required |
| $\|AB\| = 13$ | A1 cao | |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OB'} = \overrightarrow{OA} + \overrightarrow{BA} = \begin{pmatrix}21\\-17\\6\end{pmatrix}+\begin{pmatrix}-4\\-3\\-12\end{pmatrix} = \begin{pmatrix}17\\-20\\-6\end{pmatrix}$ | M1 | Full applied method for coordinates of $B'$; M1 for 2 out of 3 correct components |
| $17\mathbf{i} - 20\mathbf{j} - 6\mathbf{k}$ or $(17,-20,-6)$ | A1 cao | |

# Question 6 (Part d - Acceptable Methods):

**Way 1:** $\overrightarrow{OB'} = \overrightarrow{OA} + \overrightarrow{BA} = \begin{pmatrix}21\\-17\\6\end{pmatrix} + \begin{pmatrix}-4\\-3\\-12\end{pmatrix}$ | | using their $\overrightarrow{BA}$

**Way 2:** $\overrightarrow{OB'} = \overrightarrow{OA} - \overrightarrow{AB} = \begin{pmatrix}21\\-17\\6\end{pmatrix} - \begin{pmatrix}4\\3\\12\end{pmatrix}$ | | using their $\overrightarrow{AB}$

**Way 3:** $\overrightarrow{OB'} = \overrightarrow{OB} + 2\overrightarrow{BA} = \begin{pmatrix}25\\-14\\18\end{pmatrix} + 2\begin{pmatrix}-4\\-3\\-12\end{pmatrix}$ | | using their $\overrightarrow{BA}$

**Way 4:** $\overrightarrow{OB'} = \overrightarrow{OB} - 2\overrightarrow{AB} = \begin{pmatrix}25\\-14\\18\end{pmatrix} - 2\begin{pmatrix}4\\3\\12\end{pmatrix}$ | | using their $\overrightarrow{AB}$

**Way 5:** $\begin{pmatrix}25\\-14\\18\end{pmatrix} \to \begin{pmatrix}\text{Minus }4\\\text{Minus }3\\\text{Minus }12\end{pmatrix} \to \begin{pmatrix}21\\-17\\6\end{pmatrix} \to \begin{pmatrix}\text{Minus }4\\\text{Minus }3\\\text{Minus }12\end{pmatrix} \to \begin{pmatrix}17\\-20\\-6\end{pmatrix}$ | | so $\overrightarrow{OA}$ + their $\overrightarrow{BA}$

**Way 6:** $\overrightarrow{OB'} = 2\overrightarrow{OA} - \overrightarrow{OB} = 2\begin{pmatrix}21\\-17\\6\end{pmatrix} - \begin{pmatrix}25\\-14\\18\end{pmatrix}$

**Way 7:** $\overrightarrow{OB} = 25\mathbf{i} - 14\mathbf{j} + 18\mathbf{k}$, $\overrightarrow{OA} = 21\mathbf{i} - 17\mathbf{j} + 6\mathbf{k}$, $\overrightarrow{OB'} = p\mathbf{i} + q\mathbf{j} + r\mathbf{k}$

$(21,-17,6) = \left(\frac{25+p}{2}, \frac{-14+q}{2}, \frac{18+r}{2}\right)$

$p = 21(2)-25 = 17$, $q = -17(2)+14 = -20$, $r = 6(2)-18 = -6$ | M1 | Writing down any two equations correctly and an attempt to find at least two of $p$, $q$ or $r$

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6. Relative to a fixed origin $O$, the point $A$ has position vector $21 \mathbf { i } - 17 \mathbf { j } + 6 \mathbf { k }$ and the point $B$ has position vector $25 \mathbf { i } - 14 \mathbf { j } + 18 \mathbf { k }$.

The line $l$ has vector equation

$$\mathbf { r } = \left( \begin{array} { r } 
a \\
b \\
10
\end{array} \right) + \lambda \left( \begin{array} { r } 
6 \\
c \\
- 1
\end{array} \right)$$

where $a , b$ and $c$ are constants and $\lambda$ is a parameter.\\
Given that the point $A$ lies on the line $l$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $a$.

Given also that the vector $\overrightarrow { A B }$ is perpendicular to $l$,
\item find the values of $b$ and $c$,
\item find the distance $A B$.

The image of the point $B$ after reflection in the line $l$ is the point $B ^ { \prime }$.
\item Find the position vector of the point $B ^ { \prime }$.\\
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\section*{Question 6 continued}
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\hfill \mbox{\textit{Edexcel C4 2013 Q6 [12]}}

This paper (7 questions)

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Q1 4 Q2 7 Q3 8 Q4 9 Q5 11 Q6 12 Q7 12

Answer	Marks	Guidance
Way 1: \(\overrightarrow{OB'} = \overrightarrow{OA} + \overrightarrow{BA} = \begin{pmatrix}21\\-17\\6\end{pmatrix} + \begin{pmatrix}-4\\-3\\-12\end{pmatrix}\)		using their \(\overrightarrow{BA}\)
Way 2: \(\overrightarrow{OB'} = \overrightarrow{OA} - \overrightarrow{AB} = \begin{pmatrix}21\\-17\\6\end{pmatrix} - \begin{pmatrix}4\\3\\12\end{pmatrix}\)		using their \(\overrightarrow{AB}\)
Way 3: \(\overrightarrow{OB'} = \overrightarrow{OB} + 2\overrightarrow{BA} = \begin{pmatrix}25\\-14\\18\end{pmatrix} + 2\begin{pmatrix}-4\\-3\\-12\end{pmatrix}\)		using their \(\overrightarrow{BA}\)
Way 4: \(\overrightarrow{OB'} = \overrightarrow{OB} - 2\overrightarrow{AB} = \begin{pmatrix}25\\-14\\18\end{pmatrix} - 2\begin{pmatrix}4\\3\\12\end{pmatrix}\)		using their \(\overrightarrow{AB}\)
Way 5: \(\begin{pmatrix}25\\-14\\18\end{pmatrix} \to \begin{pmatrix}\text{Minus }4\\\text{Minus }3\\\text{Minus }12\end{pmatrix} \to \begin{pmatrix}21\\-17\\6\end{pmatrix} \to \begin{pmatrix}\text{Minus }4\\\text{Minus }3\\\text{Minus }12\end{pmatrix} \to \begin{pmatrix}17\\-20\\-6\end{pmatrix}\)		so \(\overrightarrow{OA}\) + their \(\overrightarrow{BA}\)