Question 7 - A-Level Maths

AQA C4 2006 January — Question 7 10 marks

Exam Board	AQA
Module	C4 (Core Mathematics 4)
Year	2006
Session	January
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors 3D & Lines
Type	Parallel and perpendicular lines
Difficulty	Moderate -0.3 This is a straightforward multi-part vectors question requiring standard techniques: finding direction vectors, verifying parallelism via scalar multiples, substituting points into line equations, writing vector equations, and calculating angles using the dot product formula. All parts are routine applications with no problem-solving insight required, making it slightly easier than average.
Spec	1.10b Vectors in 3D: i,j,k notation 1.10f Distance between points: using position vectors 4.04a Line equations: 2D and 3D, cartesian and vector forms 4.04c Scalar product: calculate and use for angles

7 The quadrilateral \(A B C D\) has vertices \(A ( 2,1,3 ) , B ( 6,5,3 ) , C ( 6,1 , - 1 )\) and \(D ( 2 , - 3 , - 1 )\).
The line \(l _ { 1 }\) has vector equation \(\mathbf { r } = \left[ \begin{array} { r } 6 \\ 1 \\ - 1 \end{array} \right] + \lambda \left[ \begin{array} { l } 1 \\ 1 \\ 0 \end{array} \right]\).

1. Find the vector \(\overrightarrow { A B }\).
2. Show that the line \(A B\) is parallel to \(l _ { 1 }\).
3. Verify that \(D\) lies on \(l _ { 1 }\).
The line \(l _ { 2 }\) passes through \(D ( 2 , - 3 , - 1 )\) and \(M ( 4,1,1 )\).
1. Find the vector equation of \(l _ { 2 }\).
2. Find the angle between \(l _ { 2 }\) and \(A C\).

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Question 7:

Part (a)(i)

Answer	Marks	Guidance
Answer	Marks	Guidance
\(\overrightarrow{AB} = \begin{pmatrix}6\\5\\3\end{pmatrix} - \begin{pmatrix}2\\1\\3\end{pmatrix} = \begin{pmatrix}4\\4\\0\end{pmatrix}\)	M1, A1	Penalise use of co-ordinates at first occurrence only

Part (a)(ii)

Answer	Marks	Guidance
Answer	Marks	Guidance
\(\begin{pmatrix}4\\4\\0\end{pmatrix} = 4\begin{pmatrix}1\\1\\0\end{pmatrix} \Rightarrow\) parallel	E1	Needs comment "same direction" or "same gradient"; Or by scalar product

Part (a)(iii)

Answer	Marks	Guidance
Answer	Marks	Guidance
\(\begin{pmatrix}2\\-3\\-1\end{pmatrix} = \begin{pmatrix}6\\1\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\0\end{pmatrix}\), satisfied by \(\lambda = -4\)	M1, A1	\(\lambda = -4\) satisfies 2 equations

Part (b)(i)

Answer	Marks	Guidance
Answer	Marks	Guidance
\(l_2\) has equation \(r = \begin{pmatrix}4\\1\\1\end{pmatrix} + \lambda\left(\begin{pmatrix}4\\1\\1\end{pmatrix} - \begin{pmatrix}2\\-3\\-1\end{pmatrix}\right) = \begin{pmatrix}4\\1\\1\end{pmatrix} + \lambda\begin{pmatrix}2\\4\\2\end{pmatrix}\)	M1A1	Or \(r = \begin{pmatrix}2\\-3\\-1\end{pmatrix} + t\begin{pmatrix}2\\4\\2\end{pmatrix}\); M1 calculate and use direction vector; A1 all correct

Part (b)(ii)

Answer	Marks	Guidance
Answer	Marks	Guidance
\(\begin{pmatrix}1\\2\\1\end{pmatrix} \cdot \begin{pmatrix}4\\0\\-4\end{pmatrix} = 4 - 4 = 0\)	M1A1	Clear attempt to use directions of \(AC\) and \(l_2\) in scalar product
\(\Rightarrow 90°\) (or perpendicular)	A1F	Accept a correct ft value of \(\cos\theta\)

## Question 7:

### Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = \begin{pmatrix}6\\5\\3\end{pmatrix} - \begin{pmatrix}2\\1\\3\end{pmatrix} = \begin{pmatrix}4\\4\\0\end{pmatrix}$ | M1, A1 | Penalise use of co-ordinates at first occurrence only |

### Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}4\\4\\0\end{pmatrix} = 4\begin{pmatrix}1\\1\\0\end{pmatrix} \Rightarrow$ parallel | E1 | Needs comment "same direction" or "same gradient"; Or by scalar product |

### Part (a)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}2\\-3\\-1\end{pmatrix} = \begin{pmatrix}6\\1\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\0\end{pmatrix}$, satisfied by $\lambda = -4$ | M1, A1 | $\lambda = -4$ satisfies 2 equations |

### Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $l_2$ has equation $r = \begin{pmatrix}4\\1\\1\end{pmatrix} + \lambda\left(\begin{pmatrix}4\\1\\1\end{pmatrix} - \begin{pmatrix}2\\-3\\-1\end{pmatrix}\right) = \begin{pmatrix}4\\1\\1\end{pmatrix} + \lambda\begin{pmatrix}2\\4\\2\end{pmatrix}$ | M1A1 | Or $r = \begin{pmatrix}2\\-3\\-1\end{pmatrix} + t\begin{pmatrix}2\\4\\2\end{pmatrix}$; M1 calculate and use direction vector; A1 all correct |

### Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}1\\2\\1\end{pmatrix} \cdot \begin{pmatrix}4\\0\\-4\end{pmatrix} = 4 - 4 = 0$ | M1A1 | Clear attempt to use directions of $AC$ and $l_2$ in scalar product |
| $\Rightarrow 90°$ (or perpendicular) | A1F | Accept a correct ft value of $\cos\theta$ |

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Show LaTeX source

7 The quadrilateral $A B C D$ has vertices $A ( 2,1,3 ) , B ( 6,5,3 ) , C ( 6,1 , - 1 )$ and $D ( 2 , - 3 , - 1 )$.\\
The line $l _ { 1 }$ has vector equation $\mathbf { r } = \left[ \begin{array} { r } 6 \\ 1 \\ - 1 \end{array} \right] + \lambda \left[ \begin{array} { l } 1 \\ 1 \\ 0 \end{array} \right]$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the vector $\overrightarrow { A B }$.
\item Show that the line $A B$ is parallel to $l _ { 1 }$.
\item Verify that $D$ lies on $l _ { 1 }$.
\end{enumerate}\item The line $l _ { 2 }$ passes through $D ( 2 , - 3 , - 1 )$ and $M ( 4,1,1 )$.
\begin{enumerate}[label=(\roman*)]
\item Find the vector equation of $l _ { 2 }$.
\item Find the angle between $l _ { 2 }$ and $A C$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C4 2006 Q7 [10]}}

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