| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line intersection: unknown constant then intersect |
| Difficulty | Standard +0.3 This is a structured, multi-part question with clear signposting where each part builds on the previous one. Part (a) is routine substitution, (b) is a standard dot product calculation, (c) requires equating components and solving simultaneous equations, and (d) follows directly from (c). While it involves multiple steps, the techniques are all standard Further Maths content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| \(3+3\lambda = -3 \Rightarrow 3\lambda = -6 \Rightarrow \lambda = -2\); \(1-4\lambda = 9 \Rightarrow -4\lambda = 8 \Rightarrow \lambda = -2\); \(-2+\lambda = -4 \Rightarrow \lambda = -2\) | M1 | Calculates a value of \(\lambda\) for point \(P\); or writes correct equation for \(l_1\) in Cartesian form and substitutes at least one of \(x=-3\), \(y=9\), \(z=-4\) |
| All three values of \(\lambda\) are the same so \(P\) lies on \(l_1\) | R1 | Completes argument; accept three correct calculations using \(\lambda=-2\) leading to \((-3, 9, -4)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{bmatrix}3\\-4\\1\end{bmatrix}\cdot\begin{bmatrix}3\\2\\-1\end{bmatrix} = 3\times3 + (-4)\times2 + 1\times(-1) = 9-8-1 = 0\) | M1 | Writes the scalar product of the two direction vectors |
| \(\therefore l_1\) and \(l_2\) are perpendicular | R1 | Completes argument; accept \(3\times3+(-4)\times2+1\times(-1)=0\) so perpendicular; condone \(9-8-1=0\) with reference to scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| \(3+3\lambda = -12+3\mu \Rightarrow \lambda = \mu-5\); \(-2+\lambda = -3-\mu \Rightarrow \lambda = -1-\mu\) | M1 | Selects method; equates \(i\) or \(k\) component to form at least one equation in \(\lambda\) and \(\mu\); 3.1a |
| \(\mu - 5 = -1-\mu \Rightarrow 2\mu = 4 \Rightarrow \mu = 2\) | A1 | Forms two correct equations in \(\lambda\) and \(\mu\); PI by correct value of \(a\) or correct value of \(\lambda\) or \(\mu\) |
| \(\lambda = -1-2 = -3\) | A1 | Calculates correct values of \(\lambda\) and \(\mu\); PI by correct value of \(a\) |
| \(1-4\lambda = a+2\mu \Rightarrow 1-4\times(-3) = a+2\times2 \Rightarrow a=9\) | A1F | Obtains correct value of \(a\); FT their \(\lambda\) and \(\mu\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{bmatrix}3+(-3)\times3\\1+(-3)\times(-4)\\-2+(-3)\times1\end{bmatrix} = \begin{bmatrix}-6\\13\\-5\end{bmatrix}\); Point of intersection \(= (-6, 13, -5)\) | B1F | Obtains correct coordinates of point of intersection; condone answer of \(\begin{bmatrix}-6\\13\\-5\end{bmatrix}\); FT their \(\lambda\) or their \(a\) and \(\mu\) |
## Question 7(a):
$3+3\lambda = -3 \Rightarrow 3\lambda = -6 \Rightarrow \lambda = -2$; $1-4\lambda = 9 \Rightarrow -4\lambda = 8 \Rightarrow \lambda = -2$; $-2+\lambda = -4 \Rightarrow \lambda = -2$ | M1 | Calculates a value of $\lambda$ for point $P$; or writes correct equation for $l_1$ in Cartesian form and substitutes at least one of $x=-3$, $y=9$, $z=-4$
All three values of $\lambda$ are the same so $P$ lies on $l_1$ | R1 | Completes argument; accept three correct calculations using $\lambda=-2$ leading to $(-3, 9, -4)$
---
## Question 7(b):
$\begin{bmatrix}3\\-4\\1\end{bmatrix}\cdot\begin{bmatrix}3\\2\\-1\end{bmatrix} = 3\times3 + (-4)\times2 + 1\times(-1) = 9-8-1 = 0$ | M1 | Writes the scalar product of the two direction vectors
$\therefore l_1$ and $l_2$ are perpendicular | R1 | Completes argument; accept $3\times3+(-4)\times2+1\times(-1)=0$ so perpendicular; condone $9-8-1=0$ with reference to scalar product
---
## Question 7(c):
$3+3\lambda = -12+3\mu \Rightarrow \lambda = \mu-5$; $-2+\lambda = -3-\mu \Rightarrow \lambda = -1-\mu$ | M1 | Selects method; equates $i$ or $k$ component to form at least one equation in $\lambda$ and $\mu$; 3.1a
$\mu - 5 = -1-\mu \Rightarrow 2\mu = 4 \Rightarrow \mu = 2$ | A1 | Forms two correct equations in $\lambda$ and $\mu$; PI by correct value of $a$ or correct value of $\lambda$ or $\mu$
$\lambda = -1-2 = -3$ | A1 | Calculates correct values of $\lambda$ and $\mu$; PI by correct value of $a$
$1-4\lambda = a+2\mu \Rightarrow 1-4\times(-3) = a+2\times2 \Rightarrow a=9$ | A1F | Obtains correct value of $a$; FT their $\lambda$ and $\mu$
---
## Question 7(d):
$\begin{bmatrix}3+(-3)\times3\\1+(-3)\times(-4)\\-2+(-3)\times1\end{bmatrix} = \begin{bmatrix}-6\\13\\-5\end{bmatrix}$; Point of intersection $= (-6, 13, -5)$ | B1F | Obtains correct coordinates of point of intersection; condone answer of $\begin{bmatrix}-6\\13\\-5\end{bmatrix}$; FT their $\lambda$ or their $a$ and $\mu$
---7 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations
$$\begin{aligned}
& l _ { 1 } : \mathbf { r } = \left[ \begin{array} { c }
3 \\
1 \\
- 2
\end{array} \right] + \lambda \left[ \begin{array} { c }
3 \\
- 4 \\
1
\end{array} \right] \\
& l _ { 2 } : \mathbf { r } = \left[ \begin{array} { c }
- 12 \\
a \\
- 3
\end{array} \right] + \mu \left[ \begin{array} { c }
3 \\
2 \\
- 1
\end{array} \right]
\end{aligned}$$
7
\begin{enumerate}[label=(\alph*)]
\item Show that the point $P ( - 3,9 , - 4 )$ lies on $l _ { 1 }$\\
7
\item Show that $l _ { 1 }$ is perpendicular to $l _ { 2 }$\\
7
\item Given that the lines $l _ { 1 }$ and $l _ { 2 }$ intersect, calculate the value of the constant $a$
7
\item Hence, find the coordinates of the point of intersection of $l _ { 1 }$ and $l _ { 2 }$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2022 Q7 [9]}}