| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 10 |
| 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 C4 vectors question testing routine techniques: equating line equations to find intersection (systematic algebra), then using the scalar product formula for angle between lines. All methods are textbook procedures with no novel insight required, though the arithmetic is slightly more involved than minimal examples. Slightly easier than average due to its predictable structure. |
| Spec | 1.10b Vectors in 3D: i,j,k notation4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Give 2 of these equations and eliminate variable to find \(\lambda\) or \(\mu\), find other | M1 | |
| \(\therefore \mu = -3; \lambda = -2\) | A1 A1 | |
| Check in 3rd equation | B1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Use \(\mu = -3\) or \(\lambda = -2\) to obtain \((3, 1, -2)\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos\theta = \frac{4 \times 7 + 2 \times 1 + 4 \times 5}{\sqrt{4^2 + 2^2 + 4^2}\sqrt{7^2 + 1^2 + 5^2}} = \frac{50}{\sqrt{36}\sqrt{75}}\) | M1 A1 | |
| \(\therefore \cos\theta = \frac{50}{6\sqrt{75}} = \frac{50\sqrt{3}}{90} = \frac{5\sqrt{3}}{9}\) | M1 A1 | (4) |
## (a)
$11 + 4\lambda = 24 + 7\mu$
$5 + 2\lambda = 4 + \mu$
$6 + 4\lambda = 13 + 5\mu$
$5 = 11 + 2\mu$
Give 2 of these equations and eliminate variable to find $\lambda$ or $\mu$, find other | M1 |
$\therefore \mu = -3; \lambda = -2$ | A1 A1 |
Check in 3rd equation | B1 | (4)
## (b)
Use $\mu = -3$ or $\lambda = -2$ to obtain $(3, 1, -2)$ | M1 A1 | (2)
## (c)
$\cos\theta = \frac{4 \times 7 + 2 \times 1 + 4 \times 5}{\sqrt{4^2 + 2^2 + 4^2}\sqrt{7^2 + 1^2 + 5^2}} = \frac{50}{\sqrt{36}\sqrt{75}}$ | M1 A1 |
$\therefore \cos\theta = \frac{50}{6\sqrt{75}} = \frac{50\sqrt{3}}{90} = \frac{5\sqrt{3}}{9}$ | M1 A1 | (4)
**Total: (10 marks)**
---The line $l_1$ has vector equation $\mathbf{r} = \begin{pmatrix} 11 \\ 5 \\ 6 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix}$, where $\lambda$ is a parameter.
The line $l_2$ has vector equation $\mathbf{r} = \begin{pmatrix} 24 \\ 4 \\ 13 \end{pmatrix} + \mu \begin{pmatrix} 7 \\ 1 \\ 5 \end{pmatrix}$, where $\mu$ is a parameter.
\begin{enumerate}[label=(\alph*)]
\item Show that the lines $l_1$ and $l_2$ intersect.
[4]
\item Find the coordinates of their point of intersection.
[2]
\end{enumerate}
Given that $\theta$ is the acute angle between $l_1$ and $l_2$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the value of $\cos \theta$. Give your answer in the form $k\sqrt{3}$, where $k$ is a simplified fraction.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q4 [10]}}