| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Show lines intersect and find intersection point |
| Difficulty | Standard +0.3 This is a standard Further Maths line intersection question requiring conversion between Cartesian and vector forms, solving simultaneous equations to find intersection, and calculating angle between direction vectors using dot product. While it has multiple parts and requires careful algebraic manipulation, all techniques are routine for Further Maths students with no novel problem-solving required. Slightly easier than average due to straightforward methodology. |
| 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 |
|---|---|---|
| Forms one equation by substituting two of \(x = 2+5\lambda\), \(y = -1+3\lambda\), \(z = 4-2\lambda\) into second line equation | M1 | Or introduces second parameter \(\mu\) to form at least two equations in \(\lambda\) and \(\mu\) |
| \(\frac{2+5\lambda-5}{4} = \frac{-1+3\lambda}{2}\), giving \(2(5\lambda-3) = 4(3\lambda-1)\), \(-2\lambda = 2\), \(\lambda = -1\) | A1 | Correct value of \(\lambda\) (or \(\mu\)) |
| Same value of \(\lambda\), \(\therefore\) lines intersect; or shows \(\lambda=-1\) satisfies another equation; or shows \(\lambda=-1\) and \(\mu=-2\) satisfy a third equation; or shows \((-3,-4,6)\) lies on both lines | R1 | Completes correct argument to conclude lines intersect |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \begin{bmatrix}2\\-1\\4\end{bmatrix} - 1\begin{bmatrix}5\\3\\-2\end{bmatrix} = \begin{bmatrix}-3\\-4\\6\end{bmatrix}\) | B1 | Accept \(-3\mathbf{i} - 4\mathbf{j} + 6\mathbf{k}\); condone \((-3,-4,6)\) |
| Answer | Marks | Guidance |
|---|---|---|
| The submarines might not be at the intersection point at the same time. | B1 | Correct explanation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{bmatrix}5\\3\\-2\end{bmatrix} \cdot \begin{bmatrix}4\\2\\-1\end{bmatrix} = 5\times4 + 3\times2 + (-2)\times(-1) = 28\) | M1 | Calculates scalar product; allow one error in each direction vector |
| \(\sqrt{5^2+3^2+(-2)^2} \times \sqrt{4^2+2^2+(-1)^2} = \sqrt{38}\times\sqrt{21} = \sqrt{798}\) | M1 | Calculates product of direction vector magnitudes |
| \(\cos\theta = \frac{28}{\sqrt{798}}\), \(\theta = 7.6°\) (1dp) | A1 | Condone 0.13 radians; accept 7.61° or better |
## Question 15(a)(i):
Forms one equation by substituting two of $x = 2+5\lambda$, $y = -1+3\lambda$, $z = 4-2\lambda$ into second line equation | M1 | Or introduces second parameter $\mu$ to form at least two equations in $\lambda$ and $\mu$
$\frac{2+5\lambda-5}{4} = \frac{-1+3\lambda}{2}$, giving $2(5\lambda-3) = 4(3\lambda-1)$, $-2\lambda = 2$, $\lambda = -1$ | A1 | Correct value of $\lambda$ (or $\mu$)
Same value of $\lambda$, $\therefore$ lines intersect; or shows $\lambda=-1$ satisfies another equation; or shows $\lambda=-1$ and $\mu=-2$ satisfy a third equation; or shows $(-3,-4,6)$ lies on both lines | R1 | Completes correct argument to conclude lines intersect
---
## Question 15(a)(ii):
$r = \begin{bmatrix}2\\-1\\4\end{bmatrix} - 1\begin{bmatrix}5\\3\\-2\end{bmatrix} = \begin{bmatrix}-3\\-4\\6\end{bmatrix}$ | B1 | Accept $-3\mathbf{i} - 4\mathbf{j} + 6\mathbf{k}$; condone $(-3,-4,6)$
---
## Question 15(b):
The submarines might not be at the intersection point at the same time. | B1 | Correct explanation
---
## Question 15(c):
$\begin{bmatrix}5\\3\\-2\end{bmatrix} \cdot \begin{bmatrix}4\\2\\-1\end{bmatrix} = 5\times4 + 3\times2 + (-2)\times(-1) = 28$ | M1 | Calculates scalar product; allow one error in each direction vector
$\sqrt{5^2+3^2+(-2)^2} \times \sqrt{4^2+2^2+(-1)^2} = \sqrt{38}\times\sqrt{21} = \sqrt{798}$ | M1 | Calculates product of direction vector magnitudes
$\cos\theta = \frac{28}{\sqrt{798}}$, $\theta = 7.6°$ (1dp) | A1 | Condone 0.13 radians; accept 7.61° or better
---15 Two submarines are travelling on different straight lines.
The two lines are described by the equations
$$\mathbf { r } = \left[ \begin{array} { c }
2 \\
- 1 \\
4
\end{array} \right] + \lambda \left[ \begin{array} { c }
5 \\
3 \\
- 2
\end{array} \right] \quad \text { and } \quad \frac { x - 5 } { 4 } = \frac { y } { 2 } = 4 - z$$
15
\begin{enumerate}[label=(\alph*)]
\item (i) Show that the two lines intersect.\\[0pt]
[3 marks]\\
15 (a) (ii) Find the position vector of the point of intersection.\\
15
\item Tracey says that the submarines will collide because there is a common point on the two lines.
Explain why Tracey is not necessarily correct.
15
\item Calculate the acute angle between the lines
$$\mathbf { r } = \left[ \begin{array} { c }
2 \\
- 1 \\
4
\end{array} \right] + \lambda \left[ \begin{array} { c }
5 \\
3 \\
- 2
\end{array} \right] \quad \text { and } \quad \frac { x - 5 } { 4 } = \frac { y } { 2 } = 4 - z$$
Give your angle to the nearest $0.1 ^ { \circ }$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2021 Q15 [8]}}