| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | January |
| Marks | 12 |
| 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 line intersection problem requiring students to equate vector equations, form simultaneous equations from components, and solve. While it involves multiple steps, the method is routine and commonly practiced in C3/C4 courses, making it slightly easier than average. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}-2\\1\\4\end{pmatrix} \cdot \begin{pmatrix}q\\2\\1\end{pmatrix} = -2q+2+4 = 0 \Rightarrow q = 3\) | M1 A1* | Set scalar product of direction vectors \(= 0\); condone one sign error; given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equate \(y\) and \(z\) coordinates: \(-6+\lambda = -7+2\mu\) and \(-13+4\lambda = 4+\mu\) | M1 | Condone sign errors |
| Full method to find \(\lambda\) or \(\mu\) | dM1 | Dependent on previous M |
| \(\lambda = 5\) or \(\mu = 3\) | A1 | Either value |
| Substitute both values into \(x\) coordinate: \(14 - 2\times5 = p + 3\times3 \Rightarrow p = -5\) | ddM1 A1 | Dependent on both M marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}14\\-6\\-13\end{pmatrix} + 5\begin{pmatrix}-2\\1\\4\end{pmatrix} = \begin{pmatrix}4\\-1\\7\end{pmatrix}\) | M1, A1 | Or using \(l_2\) with \(\mu=3\); coordinates \((4,-1,7)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AX} = \begin{pmatrix}-2\\1\\4\end{pmatrix}\), use \(\overrightarrow{OB} = \overrightarrow{OA} \pm 2\overrightarrow{AX}\) | M1 | Either form; or use midpoint method |
| \(\overrightarrow{OB} = \begin{pmatrix}2\\0\\11\end{pmatrix}\) | A1 | |
| \(\overrightarrow{OB} = \begin{pmatrix}10\\-4\\-5\end{pmatrix}\) | A1 | Both position vectors required |
# Question 11(a) - Vectors:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}-2\\1\\4\end{pmatrix} \cdot \begin{pmatrix}q\\2\\1\end{pmatrix} = -2q+2+4 = 0 \Rightarrow q = 3$ | M1 A1* | Set scalar product of direction vectors $= 0$; condone one sign error; given answer |
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# Question 11(b) - Intersection:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equate $y$ and $z$ coordinates: $-6+\lambda = -7+2\mu$ and $-13+4\lambda = 4+\mu$ | M1 | Condone sign errors |
| Full method to find $\lambda$ or $\mu$ | dM1 | Dependent on previous M |
| $\lambda = 5$ or $\mu = 3$ | A1 | Either value |
| Substitute both values into $x$ coordinate: $14 - 2\times5 = p + 3\times3 \Rightarrow p = -5$ | ddM1 A1 | Dependent on both M marks |
---
# Question 11(c) - Point of Intersection:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}14\\-6\\-13\end{pmatrix} + 5\begin{pmatrix}-2\\1\\4\end{pmatrix} = \begin{pmatrix}4\\-1\\7\end{pmatrix}$ | M1, A1 | Or using $l_2$ with $\mu=3$; coordinates $(4,-1,7)$ |
---
# Question 11(d) - Position of B:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AX} = \begin{pmatrix}-2\\1\\4\end{pmatrix}$, use $\overrightarrow{OB} = \overrightarrow{OA} \pm 2\overrightarrow{AX}$ | M1 | Either form; or use midpoint method |
| $\overrightarrow{OB} = \begin{pmatrix}2\\0\\11\end{pmatrix}$ | A1 | |
| $\overrightarrow{OB} = \begin{pmatrix}10\\-4\\-5\end{pmatrix}$ | A1 | Both position vectors required |
---11. With respect to a fixed origin $O$ the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations
$$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r }
14 \\
- 6 \\
- 13
\end{array} \right) + \lambda \left( \begin{array} { r }
- 2 \\
1 \\
4
\end{array} \right) \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r }
p \\
- 7 \\
4
\end{array} \right) + \mu \left( \begin{array} { l }
q \\
2 \\
1
\end{array} \right)$$
where $\lambda$ and $\mu$ are scalar parameters and $p$ and $q$ are constants.
Given that $l _ { 1 }$ and $l _ { 2 }$ are perpendicular,
\begin{enumerate}[label=(\alph*)]
\item show that $q = 3$
Given further that $l _ { 1 }$ and $l _ { 2 }$ intersect at point $X$, find
\item the value of $p$,
\item the coordinates of $X$.
The point $A$ lies on $l _ { 1 }$ and has position vector $\left( \begin{array} { r } 6 \\ - 2 \\ 3 \end{array} \right)$\\
Given that point $B$ also lies on $l _ { 1 }$ and that $A B = 2 A X$
\item find the two possible position vectors of $B$.\\
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\section*{Question 11 continued}
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\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2015 Q11 [12]}}