Question 6 - A-Level Maths

AQA C4 2007 January — Question 6 13 marks

Exam Board	AQA
Module	C4 (Core Mathematics 4)
Year	2007
Session	January
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors 3D & Lines
Type	Triangle and parallelogram problems
Difficulty	Moderate -0.3 This is a straightforward multi-part vectors question testing standard techniques: finding vectors from coordinates, using the scalar product formula for angles, verifying a point lies on a line, checking parallelism, and using the parallelogram property. All parts are routine applications of Core 4 material with no novel problem-solving required, making it slightly easier than average.
Spec	1.10b Vectors in 3D: i,j,k notation 1.10c Magnitude and direction: of vectors 1.10f Distance between points: using position vectors 4.04c Scalar product: calculate and use for angles

6 The points \(A , B\) and \(C\) have coordinates \(( 3 , - 2,4 ) , ( 5,4,0 )\) and \(( 11,6 , - 4 )\) respectively.

1. Find the vector \(\overrightarrow { B A }\).
2. Show that the size of angle \(A B C\) is \(\cos ^ { - 1 } \left( - \frac { 5 } { 7 } \right)\).
The line \(l\) has equation \(\mathbf { r } = \left[ \begin{array} { r } 8 \\ - 3 \\ 2 \end{array} \right] + \lambda \left[ \begin{array} { r } 1 \\ 3 \\ - 2 \end{array} \right]\).
1. Verify that \(C\) lies on \(l\).
2. Show that \(A B\) is parallel to \(l\).
The quadrilateral \(A B C D\) is a parallelogram. Find the coordinates of \(D\).

Show mark scheme Show mark scheme source

Question 6(a)(i):

Answer	Marks	Guidance
\(\overrightarrow{BA} = \begin{bmatrix} 3 \\ -2 \\ 4 \end{bmatrix} - \begin{bmatrix} 5 \\ 4 \\ 0 \end{bmatrix} = \begin{bmatrix} -2 \\ -6 \\ 4 \end{bmatrix}\)	M1A1 (2 marks)	Attempt \(\pm\overrightarrow{BA}\) (\(OA - OB\) or \(OB - OA\))

Question 6(a)(ii):

Answer	Marks	Guidance
\(\overrightarrow{BC} = \begin{bmatrix} 6 \\ 2 \\ -4 \end{bmatrix}\)	B1	Allow \(\overrightarrow{CB}\); or \(\begin{bmatrix} -6 \\ -2 \\ 4 \end{bmatrix} = \overrightarrow{BC}\) or \(\overrightarrow{CB} = \begin{bmatrix} 6 \\ 2 \\ -4 \end{bmatrix}\). May not see explicitly.
\(	\overrightarrow{BA}	= \sqrt{(-2)^2 + (-6)^2 + (4)^2} = \sqrt{56}\)
\(\overrightarrow{BA} \cdot \overrightarrow{BC} = \begin{bmatrix} -2 \\ -6 \\ 4 \end{bmatrix} \cdot \begin{bmatrix} 6 \\ 2 \\ -4 \end{bmatrix} = -12 - 12 - 16\)	M1	Attempt at \(\overrightarrow{BA} \cdot \overrightarrow{BC}\) with numerical answer; or \(\overrightarrow{AB} \cdot \overrightarrow{CB}\)
A1	for \(-40\), or correct if done with multiples of vectors
\(\cos ABC = \dfrac{-40}{\sqrt{56}\sqrt{56}} = -\dfrac{5}{7}\)	A1 (5 marks)	AG (convincingly obtained). Cosine rule alternative: M1 attempt to find 3 sides, A1 lengths of sides, M1 cosine rule, A1F correct, A1 rearrange to get \(\cos ABC = \frac{-5}{7}\)

Question 6(b)(i):

Answer	Marks	Guidance
\(\begin{bmatrix} 8 \\ -3 \\ 2 \end{bmatrix} + \lambda \begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix} = \begin{bmatrix} 11 \\ 6 \\ -4 \end{bmatrix} \quad (\lambda = 3)\)	M1A1 (2 marks)	\(\lambda = 3\) verified in three equations: \(11 = 8 + \lambda\), \(6 = -3 + 3\lambda\), \(-4 = 2 - 2\lambda\). A1 for \(\lambda = 3\) shown for all three equations. SC: \(\lambda = 3\) written and nothing else: SC1

Question 6(b)(ii):

\(\begin{bmatrix} 2 \\ 6 \\ -4 \end{bmatrix} = 2\begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}\)

Answer	Marks
\(\therefore\) same direction or same gradient or parallel	E1 (1 mark)

Question 6(c):

Answer	Marks	Guidance
\(\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{BA}\)	B1	PI; \(\overrightarrow{OD}\) = correct vector expression which may involve \(\overrightarrow{AD}\)
\(= \begin{bmatrix} 11 \\ 6 \\ -4 \end{bmatrix} + \begin{bmatrix} -2 \\ -6 \\ 4 \end{bmatrix} = \begin{bmatrix} 9 \\ 0 \\ 0 \end{bmatrix}\) \(D\) is \((9,0,0)\)	M1A1 (3 marks)	M1 for substituting into vector expression for \(\overrightarrow{OD}\). NMS 3/3

# Question 6(a)(i):

$\overrightarrow{BA} = \begin{bmatrix} 3 \\ -2 \\ 4 \end{bmatrix} - \begin{bmatrix} 5 \\ 4 \\ 0 \end{bmatrix} = \begin{bmatrix} -2 \\ -6 \\ 4 \end{bmatrix}$ | M1A1 (2 marks) | Attempt $\pm\overrightarrow{BA}$ ($OA - OB$ or $OB - OA$)

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# Question 6(a)(ii):

$\overrightarrow{BC} = \begin{bmatrix} 6 \\ 2 \\ -4 \end{bmatrix}$ | B1 | Allow $\overrightarrow{CB}$; or $\begin{bmatrix} -6 \\ -2 \\ 4 \end{bmatrix} = \overrightarrow{BC}$ or $\overrightarrow{CB} = \begin{bmatrix} 6 \\ 2 \\ -4 \end{bmatrix}$. May not see explicitly.

$|\overrightarrow{BA}| = \sqrt{(-2)^2 + (-6)^2 + (4)^2} = \sqrt{56}$ | B1F | Calculate modulus of $\overrightarrow{BA}$ or $\overrightarrow{BC}$; for finding modulus of one of the vectors they have used

$\overrightarrow{BA} \cdot \overrightarrow{BC} = \begin{bmatrix} -2 \\ -6 \\ 4 \end{bmatrix} \cdot \begin{bmatrix} 6 \\ 2 \\ -4 \end{bmatrix} = -12 - 12 - 16$ | M1 | Attempt at $\overrightarrow{BA} \cdot \overrightarrow{BC}$ with numerical answer; or $\overrightarrow{AB} \cdot \overrightarrow{CB}$

| A1 | for $-40$, or correct if done with multiples of vectors

$\cos ABC = \dfrac{-40}{\sqrt{56}\sqrt{56}} = -\dfrac{5}{7}$ | A1 (5 marks) | AG (convincingly obtained). Cosine rule alternative: M1 attempt to find 3 sides, A1 lengths of sides, M1 cosine rule, A1F correct, A1 rearrange to get $\cos ABC = \frac{-5}{7}$

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# Question 6(b)(i):

$\begin{bmatrix} 8 \\ -3 \\ 2 \end{bmatrix} + \lambda \begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix} = \begin{bmatrix} 11 \\ 6 \\ -4 \end{bmatrix} \quad (\lambda = 3)$ | M1A1 (2 marks) | $\lambda = 3$ verified in three equations: $11 = 8 + \lambda$, $6 = -3 + 3\lambda$, $-4 = 2 - 2\lambda$. A1 for $\lambda = 3$ shown for all three equations. SC: $\lambda = 3$ written and nothing else: SC1

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# Question 6(b)(ii):

$\begin{bmatrix} 2 \\ 6 \\ -4 \end{bmatrix} = 2\begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}$

$\therefore$ same direction or same gradient or parallel | E1 (1 mark) |

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# Question 6(c):

$\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{BA}$ | B1 | PI; $\overrightarrow{OD}$ = correct vector expression which may involve $\overrightarrow{AD}$

$= \begin{bmatrix} 11 \\ 6 \\ -4 \end{bmatrix} + \begin{bmatrix} -2 \\ -6 \\ 4 \end{bmatrix} = \begin{bmatrix} 9 \\ 0 \\ 0 \end{bmatrix}$ $D$ is $(9,0,0)$ | M1A1 (3 marks) | M1 for substituting into vector expression for $\overrightarrow{OD}$. NMS 3/3

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Show LaTeX source

6 The points $A , B$ and $C$ have coordinates $( 3 , - 2,4 ) , ( 5,4,0 )$ and $( 11,6 , - 4 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the vector $\overrightarrow { B A }$.
\item Show that the size of angle $A B C$ is $\cos ^ { - 1 } \left( - \frac { 5 } { 7 } \right)$.
\end{enumerate}\item The line $l$ has equation $\mathbf { r } = \left[ \begin{array} { r } 8 \\ - 3 \\ 2 \end{array} \right] + \lambda \left[ \begin{array} { r } 1 \\ 3 \\ - 2 \end{array} \right]$.
\begin{enumerate}[label=(\roman*)]
\item Verify that $C$ lies on $l$.
\item Show that $A B$ is parallel to $l$.
\end{enumerate}\item The quadrilateral $A B C D$ is a parallelogram. Find the coordinates of $D$.
\end{enumerate}

\hfill \mbox{\textit{AQA C4 2007 Q6 [13]}}

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Q1 11 Q2 6 Q3 9 Q4 10 Q5 7 Q6 13 Q7 6 Q8 13