| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Angle between two lines |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question covering routine techniques: showing lines intersect by solving simultaneous equations, finding angle between lines using dot product formula, verifying a point lies on a line, and using vector arithmetic to find a point. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sets up any two of: \(13+3\lambda=7+2\mu\), \(15+3\lambda=-6-3\mu\), \(-8-4\lambda=14+2\mu\) | M1 | Attempt to set coordinates equal; condone one slip across the two equations |
| Full method to find either \(\lambda\) or \(\mu\) | M1 | — |
| \(\mu = -3\) and \(\lambda = -4\) (both required) | A1 | — |
| Checks values in 3rd equation: \(-8-4(-4)=14+6=8\) ✓ | B1 | — |
| Substitutes to find position vector: \(\begin{pmatrix}1\\3\\8\end{pmatrix}\) | dM1 A1 | dM1 for substituting their \(\lambda\) or \(\mu\); A1 for correct position vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos\theta = \dfrac{\begin{pmatrix}3\\3\\-4\end{pmatrix}\cdot\begin{pmatrix}2\\-3\\2\end{pmatrix}}{\sqrt{3^2+3^2+(-4)^2}\,\sqrt{2^2+(-3)^2+2^2}} = \dfrac{-11}{17\sqrt{2}}\) | M1 A1 | — |
| Acute angle \(\approx 62.8°\) or \(\approx 1.10\) radians | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(\lambda = -6\): gives \(\begin{pmatrix}-5\\-3\\16\end{pmatrix}\), so \(A\) lies on \(l_1\) | B1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AB} = 6\mathbf{i}+6\mathbf{j}-8\mathbf{k}\) (vector approach) OR at \(C\): \(\lambda=-2\) (bus-stop approach) | M1 | — |
| \(\overrightarrow{BC} = 6\mathbf{i}+6\mathbf{j}-8\mathbf{k}\) and \(\mathbf{c} = \mathbf{b} + \overrightarrow{BC}\) | M1 | — |
| \(\overrightarrow{OC} = 7\mathbf{i}+9\mathbf{j}\) | A1 | — |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sets up any two of: $13+3\lambda=7+2\mu$, $15+3\lambda=-6-3\mu$, $-8-4\lambda=14+2\mu$ | M1 | Attempt to set coordinates equal; condone one slip across the two equations |
| Full method to find either $\lambda$ or $\mu$ | M1 | — |
| $\mu = -3$ and $\lambda = -4$ (both required) | A1 | — |
| Checks values in 3rd equation: $-8-4(-4)=14+6=8$ ✓ | B1 | — |
| Substitutes to find position vector: $\begin{pmatrix}1\\3\\8\end{pmatrix}$ | dM1 A1 | dM1 for substituting their $\lambda$ or $\mu$; A1 for correct position vector |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos\theta = \dfrac{\begin{pmatrix}3\\3\\-4\end{pmatrix}\cdot\begin{pmatrix}2\\-3\\2\end{pmatrix}}{\sqrt{3^2+3^2+(-4)^2}\,\sqrt{2^2+(-3)^2+2^2}} = \dfrac{-11}{17\sqrt{2}}$ | M1 A1 | — |
| Acute angle $\approx 62.8°$ or $\approx 1.10$ radians | A1 | — |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $\lambda = -6$: gives $\begin{pmatrix}-5\\-3\\16\end{pmatrix}$, so $A$ lies on $l_1$ | B1 | — |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = 6\mathbf{i}+6\mathbf{j}-8\mathbf{k}$ (vector approach) OR at $C$: $\lambda=-2$ (bus-stop approach) | M1 | — |
| $\overrightarrow{BC} = 6\mathbf{i}+6\mathbf{j}-8\mathbf{k}$ and $\mathbf{c} = \mathbf{b} + \overrightarrow{BC}$ | M1 | — |
| $\overrightarrow{OC} = 7\mathbf{i}+9\mathbf{j}$ | A1 | — |7. With respect to a fixed origin $O$, the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations
$$\begin{aligned}
& l _ { 1 } : \mathbf { r } = ( 13 \mathbf { i } + 15 \mathbf { j } - 8 \mathbf { k } ) + \lambda ( 3 \mathbf { i } + 3 \mathbf { j } - 4 \mathbf { k } ) \\
& l _ { 2 } : \mathbf { r } = ( 7 \mathbf { i } - 6 \mathbf { j } + 14 \mathbf { k } ) + \mu ( 2 \mathbf { i } - 3 \mathbf { j } + 2 \mathbf { k } )
\end{aligned}$$
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, $B$.
\item Find the acute angle between the lines $l _ { 1 }$ and $l _ { 2 }$
The point $A$ has position vector $- 5 \mathbf { i } - 3 \mathbf { j } + 16 \mathbf { k }$
\item Show that $A$ lies on $l _ { 1 }$
The point $C$ lies on the line $l _ { 1 }$ where $\overrightarrow { A B } = \overrightarrow { B C }$
\item Find the position vector of $C$.\\
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\hfill \mbox{\textit{Edexcel C34 2018 Q7 [13]}}