| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to line |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question covering routine techniques: showing lines intersect by solving simultaneous equations, finding angles using dot product, verifying a point lies on a line, and finding perpendicular distance. All methods are textbook exercises requiring careful calculation but no novel insight. Slightly easier than average due to straightforward setup and clear signposting of methods. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Write any two equations: \(12+5\lambda=2\), \(-4-4\lambda=2+6\mu\), \(5+2\lambda=3\mu\) | M1 | Any two correct equations, condoning slips |
| Full method to find \(\lambda\) or \(\mu\) | M1 | |
| \(\lambda = -2\), \(\mu = \frac{1}{3}\) (both required) | A1 | |
| Check in 3rd equation: \(5+2(-2) = 3\left(\frac{1}{3}\right)\) ✓ | B1 | Must substitute into both sides; minimal statement required |
| Position vector of intersection: \(\begin{pmatrix}2\\4\\1\end{pmatrix}\) | M1, A1 | Substitute \(\lambda\) into \(l_1\) or \(\mu\) into \(l_2\); accept \((2,4,1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos\theta = \dfrac{\begin{pmatrix}5\\-4\\2\end{pmatrix}\cdot\begin{pmatrix}0\\6\\3\end{pmatrix}}{\sqrt{5^2+(-4)^2+2^2}\,\sqrt{6^2+3^2}} = \dfrac{-18}{45} = -0.4\) | M1 A1 | Use \(\mathbf{a}\cdot\mathbf{b} = |
| Acute angle is \(66.4°\) | A1 | cao, awrt \(66.4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(\lambda = -1\): point is \(\begin{pmatrix}7\\0\\3\end{pmatrix}\), so \(B\) lies on \(l_1\) | B1 | State or use \(\lambda = -1\) and check all 3 coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Way 1: \(AB = \sqrt{45}\) | M1 A1 | Correct method for distance; \((AB) = 3\sqrt{5}\), awrt \(6.71\) |
| \(h = \sqrt{45} \times \sin 66.4°\) | M1 | Use their \(AB\) and \(\sin\theta\) |
| \(h = 6.15\) | A1 | awrt \(6.15\); exact answer \(\frac{3}{5}\sqrt{105}\) acceptable |
| Way 2: \(\overrightarrow{XB} = \pm\begin{pmatrix}5\\-2-6\mu\\3-3\mu\end{pmatrix}\) | M1 A1 | \(X=(2,2+6\mu,3\mu)\), \(B=(7,0,3)\) |
| \(\overrightarrow{XB}\cdot\begin{pmatrix}0\\6\\3\end{pmatrix} = 0 \Rightarrow \mu = -\frac{1}{15}\), then find \( | \overrightarrow{XB} | \) by Pythagoras |
| \(h = 6.15\) | A1 |
# Question 12:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Write any two equations: $12+5\lambda=2$, $-4-4\lambda=2+6\mu$, $5+2\lambda=3\mu$ | M1 | Any two correct equations, condoning slips |
| Full method to find $\lambda$ or $\mu$ | M1 | |
| $\lambda = -2$, $\mu = \frac{1}{3}$ (both required) | A1 | |
| Check in 3rd equation: $5+2(-2) = 3\left(\frac{1}{3}\right)$ ✓ | B1 | Must substitute into both sides; minimal statement required |
| Position vector of intersection: $\begin{pmatrix}2\\4\\1\end{pmatrix}$ | M1, A1 | Substitute $\lambda$ into $l_1$ or $\mu$ into $l_2$; accept $(2,4,1)$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\theta = \dfrac{\begin{pmatrix}5\\-4\\2\end{pmatrix}\cdot\begin{pmatrix}0\\6\\3\end{pmatrix}}{\sqrt{5^2+(-4)^2+2^2}\,\sqrt{6^2+3^2}} = \dfrac{-18}{45} = -0.4$ | M1 A1 | Use $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ with direction vectors; allow $\cos\theta = \frac{0-24+6}{\sqrt{45}\sqrt{45}}$ |
| Acute angle is $66.4°$ | A1 | cao, awrt $66.4$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $\lambda = -1$: point is $\begin{pmatrix}7\\0\\3\end{pmatrix}$, so $B$ lies on $l_1$ | B1 | State or use $\lambda = -1$ and check all 3 coordinates |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Way 1:** $AB = \sqrt{45}$ | M1 A1 | Correct method for distance; $(AB) = 3\sqrt{5}$, awrt $6.71$ |
| $h = \sqrt{45} \times \sin 66.4°$ | M1 | Use their $AB$ and $\sin\theta$ |
| $h = 6.15$ | A1 | awrt $6.15$; exact answer $\frac{3}{5}\sqrt{105}$ acceptable |
| **Way 2:** $\overrightarrow{XB} = \pm\begin{pmatrix}5\\-2-6\mu\\3-3\mu\end{pmatrix}$ | M1 A1 | $X=(2,2+6\mu,3\mu)$, $B=(7,0,3)$ |
| $\overrightarrow{XB}\cdot\begin{pmatrix}0\\6\\3\end{pmatrix} = 0 \Rightarrow \mu = -\frac{1}{15}$, then find $|\overrightarrow{XB}|$ by Pythagoras | M1 | |
| $h = 6.15$ | A1 | |
---\begin{enumerate}
\item With respect to a fixed origin $O$, the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations
\end{enumerate}
$$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r }
12 \\
- 4 \\
5
\end{array} \right) + \lambda \left( \begin{array} { r }
5 \\
- 4 \\
2
\end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { l }
2 \\
2 \\
0
\end{array} \right) + \mu \left( \begin{array} { l }
0 \\
6 \\
3
\end{array} \right)$$
where $\lambda$ and $\mu$ are scalar parameters.\\
(a) Show that $l _ { 1 }$ and $l _ { 2 }$ meet, and find the position vector of their point of intersection $A$.\\
(b) Find, to the nearest $0.1 ^ { \circ }$, the acute angle between $l _ { 1 }$ and $l _ { 2 }$
The point $B$ has position vector $\left( \begin{array} { l } 7 \\ 0 \\ 3 \end{array} \right)$.\\
(c) Show that $B$ lies on $l _ { 1 }$\\
(d) Find the shortest distance from $B$ to the line $l _ { 2 }$, giving your answer to 3 significant figures.
\hfill \mbox{\textit{Edexcel C34 2016 Q12 [14]}}