| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Shortest distance from point to line |
| Difficulty | Standard +0.3 This is a standard C4 vectors question involving finding the perpendicular from a point to a line or working with two lines in 3D. It requires routine application of dot product for perpendicularity and vector equation manipulation, which are core A-level techniques. The question appears incomplete but the setup is typical textbook material, slightly easier than average due to being a straightforward application of standard methods. |
| Spec | 1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| (a) i: \(6 - \lambda = -5 + 2\mu\) | M1 | |
| j: \(-3 + 2\lambda = 15 - 3\mu\) | M1 | |
| leading to \(\lambda = 3, \mu = 4\) | M1 A1 | |
| \(\mathbf{r} = \begin{pmatrix}6\\-3\\-2\end{pmatrix} + 3\begin{pmatrix}-1\\2\\3\end{pmatrix} = \begin{pmatrix}3\\3\\7\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}-5\\15\\3\end{pmatrix} + 4\begin{pmatrix}2\\-3\\1\end{pmatrix} = \begin{pmatrix}3\\3\\7\end{pmatrix}\) | M1 A1 | |
| k: LHS \(= -2 + 3(3) = 7\), RHS \(= 3 + 4(1) = 7\) (As LHS = RHS, lines intersect) | B1 | (6) |
| Alternatively for B1, showing that \(\lambda = 3\) and \(\mu = 4\) both give \(\begin{pmatrix}3\\3\\7\end{pmatrix}\) | ||
| (b) \(\begin{pmatrix}-1\\2\\3\end{pmatrix} \cdot \begin{pmatrix}2\\-3\\1\end{pmatrix} = -2 - 6 + 3 = \sqrt{14}\sqrt{14}\cos\theta\) (\(\theta \approx 110.92°\)) | M1 A1 | |
| Acute angle is 69.1° | A1 | awrt 69.1 |
| (c) \(\mathbf{r} = \begin{pmatrix}6\\-3\\-2\end{pmatrix} + t\begin{pmatrix}-1\\2\\3\end{pmatrix} = \begin{pmatrix}5\\-1\\1\end{pmatrix}\) (\(\Rightarrow B\) lies on \(l_1\)) | B1 | (1) |
| (d) Let \(d\) be shortest distance from \(B\) to \(l_2\) | M1 | |
| \(\overrightarrow{AB} = \begin{pmatrix}5\\-1\\1\end{pmatrix} - \begin{pmatrix}3\\-3\\7\end{pmatrix} = \begin{pmatrix}2\\-4\\-6\end{pmatrix}\) | M1 | |
| \( | \overrightarrow{AB} | = \sqrt{2^2 + (-4)^2 + (-6)^2} = \sqrt{56}\) |
| \(\frac{d}{\sqrt{56}} = \sin\theta\) | M1 | |
| \(d = \sqrt{56}\sin 69.1° \approx 6.99\) | A1 | awrt 6.99 |
**(a)** i: $6 - \lambda = -5 + 2\mu$ | M1 |
j: $-3 + 2\lambda = 15 - 3\mu$ | M1 |
leading to $\lambda = 3, \mu = 4$ | M1 A1 |
$\mathbf{r} = \begin{pmatrix}6\\-3\\-2\end{pmatrix} + 3\begin{pmatrix}-1\\2\\3\end{pmatrix} = \begin{pmatrix}3\\3\\7\end{pmatrix}$ or $\mathbf{r} = \begin{pmatrix}-5\\15\\3\end{pmatrix} + 4\begin{pmatrix}2\\-3\\1\end{pmatrix} = \begin{pmatrix}3\\3\\7\end{pmatrix}$ | M1 A1 |
k: LHS $= -2 + 3(3) = 7$, RHS $= 3 + 4(1) = 7$ (As LHS = RHS, lines intersect) | B1 | (6)
Alternatively for B1, showing that $\lambda = 3$ and $\mu = 4$ both give $\begin{pmatrix}3\\3\\7\end{pmatrix}$ |
**(b)** $\begin{pmatrix}-1\\2\\3\end{pmatrix} \cdot \begin{pmatrix}2\\-3\\1\end{pmatrix} = -2 - 6 + 3 = \sqrt{14}\sqrt{14}\cos\theta$ ($\theta \approx 110.92°$) | M1 A1 |
Acute angle is 69.1° | A1 | awrt 69.1 | (3)
**(c)** $\mathbf{r} = \begin{pmatrix}6\\-3\\-2\end{pmatrix} + t\begin{pmatrix}-1\\2\\3\end{pmatrix} = \begin{pmatrix}5\\-1\\1\end{pmatrix}$ ($\Rightarrow B$ lies on $l_1$) | B1 | (1)
**(d)** Let $d$ be shortest distance from $B$ to $l_2$ | M1 |
$\overrightarrow{AB} = \begin{pmatrix}5\\-1\\1\end{pmatrix} - \begin{pmatrix}3\\-3\\7\end{pmatrix} = \begin{pmatrix}2\\-4\\-6\end{pmatrix}$ | M1 |
$|\overrightarrow{AB}| = \sqrt{2^2 + (-4)^2 + (-6)^2} = \sqrt{56}$ | A1 | awrt 7.5 |
$\frac{d}{\sqrt{56}} = \sin\theta$ | M1 |
$d = \sqrt{56}\sin 69.1° \approx 6.99$ | A1 | awrt 6.99 | (4) [14] |
---6. With respect to a fixed origin $O$, the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations
$$l _ { 1 } : \quad \mathbf { r } = \left( \begin{array} { r }
6 \\
- 3 \\
- 2
\end{array} \right) + \lambda \left( \begin{array} { r }
- 1 \\
2 \\
3
\end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r }
- 5 \\
15 \\
3
\end{array} \right) + \mu \left( \begin{array} { r }
2 \\
- 3 \\
1
\end{array} \right)$$
where $\lambda$ and $\mu$ are scalar parameters.
\begin{enumerate}[label=(\alph*)]
\item Show that $l _ { 1 }$ and $l _ { 2 }$ meet and find the position vector of their point of intersection $A$.
\item Find, to the nearest $0.1 ^ { \circ }$, the acute angle between $l _ { 1 }$ and $l _ { 2 }$.
The point $B$ has position vector $\left( \begin{array} { r } 5 \\ - 1 \\ 1 \end{array} \right)$.
\item Show that $B$ lies on $l _ { 1 }$.
\item Find the shortest distance from $B$ to the line $l _ { 2 }$, giving your answer to 3 significant figures.\\
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\section*{Question 6 continued}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2011 Q6 [14]}}