| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Reflection and symmetry |
| Difficulty | Challenging +1.2 This is a multi-part 3D vectors question requiring standard techniques: finding intersection by equating components (part a), checking perpendicularity via dot product (part b), and reflection in a line (part c). While part (c) requires multiple steps (finding foot of perpendicular, then reflecting), all techniques are standard A-level procedures with no novel insight required. The computational work is moderate but straightforward, making this slightly above average difficulty. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}1\\5\\5\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\end{pmatrix} = \begin{pmatrix}0\\2\\12\end{pmatrix}+\mu\begin{pmatrix}3\\-1\\5\end{pmatrix} \Rightarrow 1+2\lambda=3\mu,\; 5+\lambda=2-\mu,\; 5-\lambda=12+5\mu\) (any two) | M1 | Writing down any two equations giving coordinates of intersection point; condone slips |
| Full method to find both \(\lambda\) and \(\mu\); \((2)+(3) \Rightarrow 10=14+4\mu \Rightarrow \mu=-1\) | dM1 | Dependent on previous M; must end with values for both |
| Both values correct: \(\mu=-1\), \(\lambda=-2\) | A1 | |
| Check values in 3rd equation: \(1+2(-2) = 3(-1)\) ✓ | B1 | Correct values of \(\lambda\) and \(\mu\) substituted into both sides of third equation with calculation/statement showing equal |
| Position vector of intersection: \(\begin{pmatrix}1\\5\\5\end{pmatrix}+-2\begin{pmatrix}2\\1\\-1\end{pmatrix} = \begin{pmatrix}-3\\3\\7\end{pmatrix}\) OR \(\begin{pmatrix}0\\2\\12\end{pmatrix}+(-1)\begin{pmatrix}3\\-1\\5\end{pmatrix} = \begin{pmatrix}-3\\3\\7\end{pmatrix}\) | ddM1, A1 | Substitutes their \(\lambda\) into \(l_1\) or \(\mu\) into \(l_2\); correct answer only, accept as vector or coordinate \((-3,3,7)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}2\\1\\-1\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\5\end{pmatrix} = 2\times3 + 1\times-1 + -1\times5 = 0\) | M1 | Clear attempt to find scalar product of direction vectors; must see attempt to multiply and add |
| Scalar product \(=0\), lines are perpendicular | A1 | Must be followed by reason and conclusion; accept "\(=0\), hence perpendicular" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix}-8\\-4\\4\end{pmatrix}\) | M1 | Attempt to find vector \(\overrightarrow{AX}\) where \(X\) is point of intersection using \(\overrightarrow{AX}=\overrightarrow{OX}-\overrightarrow{OA}\) |
| \(\overrightarrow{OB} = \overrightarrow{OX} + \overrightarrow{AX} = \begin{pmatrix}-3\\3\\7\end{pmatrix}+\begin{pmatrix}-8\\-4\\4\end{pmatrix} = \begin{pmatrix}-11\\-1\\11\end{pmatrix}\) | M1, A1 | Attempt to find coordinate/vector of \(B\) using \(\overrightarrow{OB}=\overrightarrow{OX}+\overrightarrow{AX}\) or \(\overrightarrow{OB}=\overrightarrow{OA}+2\times\overrightarrow{AX}\); correct answer only, do NOT accept coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to find \(\lambda\) on \(l_1\) at \(X\) (\(\lambda=-2\)) and \(A\) (\(\lambda=2\)) | M1 | |
| Uses difference between \(\lambda\) values to find \(\lambda\) (\(-6\)) at \(B\) | M1 | |
| cao | A1 |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}1\\5\\5\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\end{pmatrix} = \begin{pmatrix}0\\2\\12\end{pmatrix}+\mu\begin{pmatrix}3\\-1\\5\end{pmatrix} \Rightarrow 1+2\lambda=3\mu,\; 5+\lambda=2-\mu,\; 5-\lambda=12+5\mu$ (any two) | M1 | Writing down any two equations giving coordinates of intersection point; condone slips |
| Full method to find both $\lambda$ and $\mu$; $(2)+(3) \Rightarrow 10=14+4\mu \Rightarrow \mu=-1$ | dM1 | Dependent on previous M; must end with values for both |
| Both values correct: $\mu=-1$, $\lambda=-2$ | A1 | |
| Check values in 3rd equation: $1+2(-2) = 3(-1)$ ✓ | B1 | Correct values of $\lambda$ and $\mu$ substituted into both sides of third equation with calculation/statement showing equal |
| Position vector of intersection: $\begin{pmatrix}1\\5\\5\end{pmatrix}+-2\begin{pmatrix}2\\1\\-1\end{pmatrix} = \begin{pmatrix}-3\\3\\7\end{pmatrix}$ OR $\begin{pmatrix}0\\2\\12\end{pmatrix}+(-1)\begin{pmatrix}3\\-1\\5\end{pmatrix} = \begin{pmatrix}-3\\3\\7\end{pmatrix}$ | ddM1, A1 | Substitutes their $\lambda$ into $l_1$ or $\mu$ into $l_2$; correct answer only, accept as vector or coordinate $(-3,3,7)$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}2\\1\\-1\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\5\end{pmatrix} = 2\times3 + 1\times-1 + -1\times5 = 0$ | M1 | Clear attempt to find scalar product of direction vectors; must see attempt to multiply and add |
| Scalar product $=0$, lines are perpendicular | A1 | Must be followed by reason and conclusion; accept "$=0$, hence perpendicular" |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix}-8\\-4\\4\end{pmatrix}$ | M1 | Attempt to find vector $\overrightarrow{AX}$ where $X$ is point of intersection using $\overrightarrow{AX}=\overrightarrow{OX}-\overrightarrow{OA}$ |
| $\overrightarrow{OB} = \overrightarrow{OX} + \overrightarrow{AX} = \begin{pmatrix}-3\\3\\7\end{pmatrix}+\begin{pmatrix}-8\\-4\\4\end{pmatrix} = \begin{pmatrix}-11\\-1\\11\end{pmatrix}$ | M1, A1 | Attempt to find coordinate/vector of $B$ using $\overrightarrow{OB}=\overrightarrow{OX}+\overrightarrow{AX}$ or $\overrightarrow{OB}=\overrightarrow{OA}+2\times\overrightarrow{AX}$; correct answer only, do NOT accept coordinates |
### Part (c) Alternative:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to find $\lambda$ on $l_1$ at $X$ ($\lambda=-2$) and $A$ ($\lambda=2$) | M1 | |
| Uses difference between $\lambda$ values to find $\lambda$ ($-6$) at $B$ | M1 | |
| cao | A1 | |
---10. With respect to a fixed origin $O$, the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations
$$\begin{aligned}
& l _ { 1 } : \mathbf { r } = ( \mathbf { i } + 5 \mathbf { j } + 5 \mathbf { k } ) + \lambda ( 2 \mathbf { i } + \mathbf { j } - \mathbf { k } ) \\
& l _ { 2 } : \mathbf { r } = ( 2 \mathbf { j } + 12 \mathbf { k } ) + \mu ( 3 \mathbf { i } - \mathbf { j } + 5 \mathbf { k } )
\end{aligned}$$
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.
\item Show that $l _ { 1 }$ and $l _ { 2 }$ are perpendicular to each other.
The point $A$, with position vector $5 \mathbf { i } + 7 \mathbf { j } + 3 \mathbf { k }$, lies on $l _ { 1 }$\\
The point $B$ is the image of $A$ after reflection in the line $l _ { 2 }$
\item Find the position vector of $B$.\\
\includegraphics[max width=\textwidth, alt={}, center]{5b698944-41ac-4072-b5e1-c580b7752c39-35_133_163_2604_1786}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q10 [11]}}