| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Find unknown constant then intersection |
| Difficulty | Standard +0.3 This is a standard C4 vectors question testing routine techniques: finding intersection of lines, using dot product for angles, and perpendicular distance from point to line. All parts follow textbook methods with straightforward algebra, though part (d) requires careful application of the distance formula. Slightly easier than average due to the structured, methodical nature with no conceptual surprises. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04c Scalar product: calculate and use for angles4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks |
|---|---|
| \(l_1: \mathbf{r} = \begin{pmatrix} -5 \\ -3 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}\), \(l_2: \mathbf{r} = \begin{pmatrix} 8 \\ 5 \\ -2 \end{pmatrix} + \mu\begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix}\). Let \(\theta =\) acute angle between \(l_1\) and \(l_2\). | |
| Note: You can mark parts (a) and (b) together. |
| Answer | Marks |
|---|---|
| \(l_1 \cap l_2: \{\) \(i\): \(5 = 3 + 3\lambda\) \(j\): \(-3 + \lambda = 5 + 4\mu\) \(\}\) | |
| Finds \(\mu\) and substitutes their \(\mu\) into \(l_2\) | M1 |
| So, \(\{\mathbf{O1}\} = \begin{pmatrix} 8 \\ 5 \\ -2 \end{pmatrix} - 1\begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \\ 3 \end{pmatrix}\) | A1 |
| \(5\mathbf{i} + \mathbf{j} + 3\mathbf{k}\) or \(\begin{pmatrix} 5 \\ 1 \\ 3 \end{pmatrix}\) or \((5, 1, 3)\) | |
| [2] |
| Answer | Marks |
|---|---|
| \(\{j\): \(-3 + \lambda = 5 + 4\mu \Rightarrow -3 + \lambda = 5 + 4(-1) \Rightarrow \lambda = 4\) | M1 |
| \(k\): \(p - 3\lambda = -2 - 5\mu \Rightarrow p - 3(4) = -2 - 5(-1) \Rightarrow p = 15\) | M1 |
| or \(k\): \(p - 3\lambda = 3 \Rightarrow p - 3(4) = 3 \Rightarrow p = 15\) | |
| Equates \(k\) components, substitutes their \(\lambda\) and their \(\mu\) and solves to give \(p = \ldots\) | M1 |
| \(p = 15\) | A1 |
| [3] |
| Answer | Marks |
|---|---|
| \(\mathbf{d}_1 = \begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix}\), \(\mathbf{d}_2 = \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix} \Rightarrow \begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix}\) | M1 |
| Realisation that the dot product is required between \(\pm \mathbf{Ad}_1\) and \(\pm \mathbf{Bd}_2\) | M1 |
| \(\cos\theta = \pm K\frac{0(3)+(1)(4)+(-3)(-5)}{\sqrt{0^2+(1)^2+(-3)^2}\sqrt{(3)^2+(4)^2+(-5)^2}}\) | dM1 (At on ePEN) |
| An attempt to apply the dot product formula between \(\pm \mathbf{Ad}_1\) and \(\pm \mathbf{Bd}_2\) | |
| \(\cos\theta = \frac{19}{\sqrt{10}\sqrt{50}} \Rightarrow \theta = 31.8203116\ldots = 31.82\) (2 dp) | A1 |
| anything that rounds to 31.82 | |
| [3] |
| Answer | Marks |
|---|---|
| \(\overline{OB} = \begin{pmatrix} 11 \\ 9 \\ -7 \end{pmatrix}\); \(\overline{AB} = \begin{pmatrix} 11 \\ 9 \\ -7 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 8 \\ -10 \end{pmatrix}\) or \(\overline{AB} = 2\begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix} = \begin{pmatrix} 8 \\ -10 \end{pmatrix}\) | M1 |
| \(\overline{AB} = \sqrt{6^2+8^2+(-10)^2} = \{10\sqrt{2}\}\) | |
| Writes down a correct trigonometric equation involving the shortest distance, \(d\). Eg: \(\frac{d}{10\sqrt{2}} = \sin\theta\) or \(\frac{d}{\text{their } AB} = \sin\theta\), o.e. | dM1 |
| \(d = 10\sqrt{2}\sin 31.82\ldots \Rightarrow d = 7.456540753\ldots = 7.46\) (3sf) | A1 |
| anything that rounds to 7.46 | |
| [3] |
$l_1: \mathbf{r} = \begin{pmatrix} -5 \\ -3 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}$, $l_2: \mathbf{r} = \begin{pmatrix} 8 \\ 5 \\ -2 \end{pmatrix} + \mu\begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix}$. Let $\theta =$ acute angle between $l_1$ and $l_2$. | |
Note: You can mark parts (a) and (b) together. | |
## Part (a)
$l_1 \cap l_2: \{$ $i$: $5 = 3 + 3\lambda$ $j$: $-3 + \lambda = 5 + 4\mu$ $\}$ | |
Finds $\mu$ and substitutes their $\mu$ into $l_2$ | M1 |
So, $\{\mathbf{O1}\} = \begin{pmatrix} 8 \\ 5 \\ -2 \end{pmatrix} - 1\begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \\ 3 \end{pmatrix}$ | A1 |
$5\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ or $\begin{pmatrix} 5 \\ 1 \\ 3 \end{pmatrix}$ or $(5, 1, 3)$ | |
| [2] |
## Part (b)
$\{j$: $-3 + \lambda = 5 + 4\mu \Rightarrow -3 + \lambda = 5 + 4(-1) \Rightarrow \lambda = 4$ | M1 |
$k$: $p - 3\lambda = -2 - 5\mu \Rightarrow p - 3(4) = -2 - 5(-1) \Rightarrow p = 15$ | M1 |
or $k$: $p - 3\lambda = 3 \Rightarrow p - 3(4) = 3 \Rightarrow p = 15$ | |
Equates $k$ components, substitutes their $\lambda$ and their $\mu$ and solves to give $p = \ldots$ | M1 |
$p = 15$ | A1 |
| [3] |
## Part (c)
$\mathbf{d}_1 = \begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix}$, $\mathbf{d}_2 = \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix} \Rightarrow \begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix}$ | M1 |
Realisation that the dot product is required between $\pm \mathbf{Ad}_1$ and $\pm \mathbf{Bd}_2$ | M1 |
$\cos\theta = \pm K\frac{0(3)+(1)(4)+(-3)(-5)}{\sqrt{0^2+(1)^2+(-3)^2}\sqrt{(3)^2+(4)^2+(-5)^2}}$ | dM1 (At on ePEN) |
An attempt to apply the dot product formula between $\pm \mathbf{Ad}_1$ and $\pm \mathbf{Bd}_2$ | |
$\cos\theta = \frac{19}{\sqrt{10}\sqrt{50}} \Rightarrow \theta = 31.8203116\ldots = 31.82$ (2 dp) | A1 |
anything that rounds to 31.82 | |
| [3] |
## Part (d)
$\overline{OB} = \begin{pmatrix} 11 \\ 9 \\ -7 \end{pmatrix}$; $\overline{AB} = \begin{pmatrix} 11 \\ 9 \\ -7 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 8 \\ -10 \end{pmatrix}$ or $\overline{AB} = 2\begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix} = \begin{pmatrix} 8 \\ -10 \end{pmatrix}$ | M1 |
$\overline{AB} = \sqrt{6^2+8^2+(-10)^2} = \{10\sqrt{2}\}$ | |
Writes down a correct trigonometric equation involving the shortest distance, $d$. Eg: $\frac{d}{10\sqrt{2}} = \sin\theta$ or $\frac{d}{\text{their } AB} = \sin\theta$, o.e. | dM1 |
$d = 10\sqrt{2}\sin 31.82\ldots \Rightarrow d = 7.456540753\ldots = 7.46$ (3sf) | A1 |
anything that rounds to 7.46 | |
| [3] |
---With respect to a fixed origin $O$, the lines $l_1$ and $l_2$ are given by the equations
$$l_1: \mathbf{r} = \begin{pmatrix} 5 \\ -3 \\ p \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix}, \quad l_2: \mathbf{r} = \begin{pmatrix} 8 \\ 5 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix}$$
where $\lambda$ and $\mu$ are scalar parameters and $p$ is a constant.
The lines $l_1$ and $l_2$ intersect at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $A$.
[2]
\item Find the value of the constant $p$.
[3]
\item Find the acute angle between $l_1$ and $l_2$, giving your answer in degrees to 2 decimal places.
[3]
\end{enumerate}
The point $B$ lies on $l_2$ where $\mu = 1$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the shortest distance from the point $B$ to the line $l_1$, giving your answer to 3 significant figures.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2015 Q4 [11]}}