| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| 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 requiring standard techniques: distance formula, substitution to verify a point on a line, finding intersection of two lines, and comparing distances. All methods are routine C4 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10f Distance between points: using position vectors4.04a Line equations: 2D and 3D, cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| \((AB^2 =)(4-3)^2 + (0--2)^2 + (1-5)^2\) | M1 | Condone one sign error in one bracket |
| \(AB = \sqrt{21}\) | A1 (2 marks) | Accept 4.58 or better |
| Answer | Marks | Guidance |
|---|---|---|
| \(4 = 6 + 2\lambda \Rightarrow \lambda = -1\); \(0 = -1 + (-1)\times(-1)\); \(1 = 5 + (-1)\times 4\) | M1 | \(\lambda = -1\) |
| \(\lambda = -1\) confirmed in other two equations | A1 (2 marks) | Accept for M1A1: \(\begin{bmatrix}6\\-1\\5\end{bmatrix} + \begin{bmatrix}2\\-1\\4\end{bmatrix} = \begin{bmatrix}4\\0\\1\end{bmatrix}\); M1 condone 1 slip |
| Special case: \(\begin{bmatrix}6\\-1\\5\end{bmatrix} + \lambda\begin{bmatrix}2\\-1\\4\end{bmatrix} = \begin{bmatrix}4\\0\\1\end{bmatrix}\), \(\lambda\begin{bmatrix}2\\-1\\4\end{bmatrix} = \begin{bmatrix}-2\\1\\-4\end{bmatrix}\), \(\lambda = -1\) | (B2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{bmatrix}3\\-2\\5\end{bmatrix} + \mu\begin{bmatrix}-1\\3\\8\end{bmatrix} = \begin{bmatrix}6\\-1\\5\end{bmatrix} + \lambda\begin{bmatrix}2\\-1\\4\end{bmatrix}\) | M1 | Equate vector equations; PI by two equations in \(\lambda\) or \(\mu\) |
| \(3 - \mu = 6 + 2\lambda\); \(-2 + 3\mu = -1 - \lambda\); eliminate \(\lambda\) or \(\mu\) | m1 | Form (any) two simultaneous equations and solve for \(\lambda\) or \(\mu\) |
| \(\lambda = -2\) or \(\mu = 1\) | A1 | |
| \(C\) has coordinates \((2, 1, -3)\) | A1 | CAO; condone \(\begin{bmatrix}2\\1\\-3\end{bmatrix}\) |
| \(BC^2 = (2-4)^2 + (0-1)^2 + (1--3)^2\); \(BC = \sqrt{21}\) | M1 | Use \(C\) to find \(BC\) or \(AC\) or to find two angles |
| \(AB = BC\ (= \sqrt{21})\) | A1 (6 marks) | \(AB = BC\) or \(\angle A = \angle C\ (= 20.2°)\) stated |
# Question 7(a):
$(AB^2 =)(4-3)^2 + (0--2)^2 + (1-5)^2$ | M1 | Condone one sign error in one bracket
$AB = \sqrt{21}$ | A1 (2 marks) | Accept 4.58 or better
# Question 7(b):
$4 = 6 + 2\lambda \Rightarrow \lambda = -1$; $0 = -1 + (-1)\times(-1)$; $1 = 5 + (-1)\times 4$ | M1 | $\lambda = -1$
$\lambda = -1$ confirmed in other two equations | A1 (2 marks) | Accept for M1A1: $\begin{bmatrix}6\\-1\\5\end{bmatrix} + \begin{bmatrix}2\\-1\\4\end{bmatrix} = \begin{bmatrix}4\\0\\1\end{bmatrix}$; M1 condone 1 slip
**Special case:** $\begin{bmatrix}6\\-1\\5\end{bmatrix} + \lambda\begin{bmatrix}2\\-1\\4\end{bmatrix} = \begin{bmatrix}4\\0\\1\end{bmatrix}$, $\lambda\begin{bmatrix}2\\-1\\4\end{bmatrix} = \begin{bmatrix}-2\\1\\-4\end{bmatrix}$, $\lambda = -1$ | (B2) |
# Question 7(c):
$\begin{bmatrix}3\\-2\\5\end{bmatrix} + \mu\begin{bmatrix}-1\\3\\8\end{bmatrix} = \begin{bmatrix}6\\-1\\5\end{bmatrix} + \lambda\begin{bmatrix}2\\-1\\4\end{bmatrix}$ | M1 | Equate vector equations; PI by two equations in $\lambda$ or $\mu$
$3 - \mu = 6 + 2\lambda$; $-2 + 3\mu = -1 - \lambda$; eliminate $\lambda$ or $\mu$ | m1 | Form (any) two simultaneous equations and solve for $\lambda$ or $\mu$
$\lambda = -2$ or $\mu = 1$ | A1 |
$C$ has coordinates $(2, 1, -3)$ | A1 | CAO; condone $\begin{bmatrix}2\\1\\-3\end{bmatrix}$
$BC^2 = (2-4)^2 + (0-1)^2 + (1--3)^2$; $BC = \sqrt{21}$ | M1 | Use $C$ to find $BC$ or $AC$ or to find two angles
$AB = BC\ (= \sqrt{21})$ | A1 (6 marks) | $AB = BC$ or $\angle A = \angle C\ (= 20.2°)$ stated
---7 The points $A$ and $B$ have coordinates ( $3 , - 2,5$ ) and ( $4,0,1$ ) respectively.
The line $l _ { 1 }$ has equation $\mathbf { r } = \left[ \begin{array} { r } 6 \\ - 1 \\ 5 \end{array} \right] + \lambda \left[ \begin{array} { r } 2 \\ - 1 \\ 4 \end{array} \right]$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance between the points $A$ and $B$.
\item Verify that $B$ lies on $l _ { 1 }$.\\
(2 marks)
\item The line $l _ { 2 }$ passes through $A$ and has equation $\mathbf { r } = \left[ \begin{array} { r } 3 \\ - 2 \\ 5 \end{array} \right] + \mu \left[ \begin{array} { r } - 1 \\ 3 \\ - 8 \end{array} \right]$.
The lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $C$. Show that the points $A , B$ and $C$ form an isosceles triangle.\\
(6 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q7 [10]}}