| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Reflection and symmetry |
| Difficulty | Challenging +1.2 This is a multi-part vectors question requiring systematic application of standard techniques: finding intersection points (solving simultaneous equations), checking perpendicularity (dot product), verifying a point lies on a line (substitution), and reflection in a line (perpendicular projection). While part (d) requires careful multi-step work combining several concepts, each technique is standard C4 material with no novel insight required. The reflection component elevates it slightly above average difficulty. |
| Spec | 4.03c Matrix multiplication: properties (associative, not commutative)4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Setting up equations: i: \(-9 + 2\lambda = 3 + 3\mu\), j: \(\lambda = 1 - \mu\), k: \(10 - \lambda = 17 + 5\mu\) | M1 | Need any two of these correct equations seen anywhere in part (a) |
| \((1) - 2(2)\): \(-9 = 1 + 5\mu \Rightarrow \mu = -2\) | dM1 | Attempts to solve simultaneous equations to find one of either \(\lambda\) or \(\mu\) |
| \((2)\) gives: \(\lambda = 1 - -2 = 3\); both \(\lambda = 3\) & \(\mu = -2\) | A1 | Both \(\underline{\lambda = 3}\) & \(\underline{\mu = -2}\) |
| \(\mathbf{r} = \begin{pmatrix}-9\\0\\10\end{pmatrix} + 3\begin{pmatrix}2\\1\\-1\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}3\\1\\17\end{pmatrix} - 2\begin{pmatrix}3\\-1\\5\end{pmatrix}\) | ddM1 | Substitutes their value of either \(\lambda\) or \(\mu\) into line \(l_1\) or \(l_2\) respectively. Implied by any two correct components of \((-3, 3, 7)\) |
| Intersect at \(\mathbf{r} = \begin{pmatrix}-3\\3\\7\end{pmatrix}\) or \(\mathbf{r} = -3\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}\) or \((-3, 3, 7)\) | A1 | \(\begin{pmatrix}-3\\3\\7\end{pmatrix}\) or \(-3\mathbf{i}+3\mathbf{j}+7\mathbf{k}\) or \((-3,3,7)\) |
| Check k: \(\lambda=3\): LHS \(= 10 - \lambda = 10 - 3 = 7\); \(\mu=-2\): RHS \(= 17 + 5\mu = 17 - 10 = 7\) | B1 | Either check \(\lambda=3\), \(\mu=-2\) in third equation or check both give same coordinates on other line. Conclusion not needed. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{d}_1 = 2\mathbf{i}+\mathbf{j}-\mathbf{k}\), \(\mathbf{d}_2 = 3\mathbf{i}-\mathbf{j}+5\mathbf{k}\); \(\mathbf{d}_1 \bullet \mathbf{d}_2 = \begin{pmatrix}2\\1\\-1\end{pmatrix}\bullet\begin{pmatrix}3\\-1\\5\end{pmatrix} = (2\times3)+(1\times-1)+(-1\times5) = 0\) | M1 | Dot product calculation between the two direction vectors: \((2\times3)+(1\times-1)+(-1\times5)\) or \(6-1-5\) |
| Then \(l_1\) is perpendicular to \(l_2\) | A1 | Result \(= 0\) and appropriate conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equating i: \(-9 + 2\lambda = 5 \Rightarrow \lambda = 7\); \(\mathbf{r} = \begin{pmatrix}-9\\0\\10\end{pmatrix} + 7\begin{pmatrix}2\\1\\-1\end{pmatrix} = \begin{pmatrix}5\\7\\3\end{pmatrix}\) \((= \overrightarrow{OA}\), hence point A lies on \(l_1\)) | B1 | Substitutes candidate's \(\lambda = 7\) into line \(l_1\) and finds \(5\mathbf{i}+7\mathbf{j}+3\mathbf{k}\). Conclusion not needed. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(\overrightarrow{OX} = -3\mathbf{i}+3\mathbf{j}+7\mathbf{k}\) be point of intersection; \(\overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix}-3\\3\\7\end{pmatrix} - \begin{pmatrix}5\\7\\3\end{pmatrix} = \begin{pmatrix}-8\\-4\\4\end{pmatrix}\) | M1\(\checkmark\pm\) | Finding the difference between their \(\overrightarrow{OX}\) (can be implied) and \(\overrightarrow{OA}\): \(\overrightarrow{AX} = \pm\left(\begin{pmatrix}-3\\3\\7\end{pmatrix}-\begin{pmatrix}5\\7\\3\end{pmatrix}\right)\) |
| \(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OA} + 2\overrightarrow{AX} = \begin{pmatrix}5\\7\\3\end{pmatrix} + 2\begin{pmatrix}-8\\-4\\4\end{pmatrix}\) | dM1\(\checkmark\) | \(\begin{pmatrix}5\\7\\3\end{pmatrix} + 2\left(\text{their } \overrightarrow{AX}\right)\) |
| Hence \(\overrightarrow{OB} = \begin{pmatrix}-11\\-1\\11\end{pmatrix}\) or \(\overrightarrow{OB} = -11\mathbf{i}-\mathbf{j}+11\mathbf{k}\) or \((-11,-1,11)\) | A1 | \(\begin{pmatrix}-11\\-1\\11\end{pmatrix}\) or \(-11\mathbf{i}-\mathbf{j}+11\mathbf{k}\) or \((-11,-1,11)\) |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Setting up equations: **i**: $-9 + 2\lambda = 3 + 3\mu$, **j**: $\lambda = 1 - \mu$, **k**: $10 - \lambda = 17 + 5\mu$ | M1 | Need any two of these correct equations seen anywhere in part (a) |
| $(1) - 2(2)$: $-9 = 1 + 5\mu \Rightarrow \mu = -2$ | dM1 | Attempts to solve simultaneous equations to find one of either $\lambda$ or $\mu$ |
| $(2)$ gives: $\lambda = 1 - -2 = 3$; both $\lambda = 3$ & $\mu = -2$ | A1 | Both $\underline{\lambda = 3}$ & $\underline{\mu = -2}$ |
| $\mathbf{r} = \begin{pmatrix}-9\\0\\10\end{pmatrix} + 3\begin{pmatrix}2\\1\\-1\end{pmatrix}$ or $\mathbf{r} = \begin{pmatrix}3\\1\\17\end{pmatrix} - 2\begin{pmatrix}3\\-1\\5\end{pmatrix}$ | ddM1 | Substitutes their value of either $\lambda$ or $\mu$ into line $l_1$ or $l_2$ respectively. Implied by any two correct components of $(-3, 3, 7)$ |
| Intersect at $\mathbf{r} = \begin{pmatrix}-3\\3\\7\end{pmatrix}$ or $\mathbf{r} = -3\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}$ or $(-3, 3, 7)$ | A1 | $\begin{pmatrix}-3\\3\\7\end{pmatrix}$ or $-3\mathbf{i}+3\mathbf{j}+7\mathbf{k}$ or $(-3,3,7)$ |
| Check **k**: $\lambda=3$: LHS $= 10 - \lambda = 10 - 3 = 7$; $\mu=-2$: RHS $= 17 + 5\mu = 17 - 10 = 7$ | B1 | Either check $\lambda=3$, $\mu=-2$ in third equation or check both give same coordinates on other line. Conclusion not needed. |
**[6 marks]**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{d}_1 = 2\mathbf{i}+\mathbf{j}-\mathbf{k}$, $\mathbf{d}_2 = 3\mathbf{i}-\mathbf{j}+5\mathbf{k}$; $\mathbf{d}_1 \bullet \mathbf{d}_2 = \begin{pmatrix}2\\1\\-1\end{pmatrix}\bullet\begin{pmatrix}3\\-1\\5\end{pmatrix} = (2\times3)+(1\times-1)+(-1\times5) = 0$ | M1 | Dot product calculation between the two direction vectors: $(2\times3)+(1\times-1)+(-1\times5)$ or $6-1-5$ |
| Then $l_1$ is perpendicular to $l_2$ | A1 | Result $= 0$ and appropriate conclusion |
**[2 marks]**
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equating **i**: $-9 + 2\lambda = 5 \Rightarrow \lambda = 7$; $\mathbf{r} = \begin{pmatrix}-9\\0\\10\end{pmatrix} + 7\begin{pmatrix}2\\1\\-1\end{pmatrix} = \begin{pmatrix}5\\7\\3\end{pmatrix}$ $(= \overrightarrow{OA}$, hence point A lies on $l_1$) | B1 | Substitutes candidate's $\lambda = 7$ into line $l_1$ and finds $5\mathbf{i}+7\mathbf{j}+3\mathbf{k}$. Conclusion not needed. |
**[1 mark]**
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $\overrightarrow{OX} = -3\mathbf{i}+3\mathbf{j}+7\mathbf{k}$ be point of intersection; $\overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix}-3\\3\\7\end{pmatrix} - \begin{pmatrix}5\\7\\3\end{pmatrix} = \begin{pmatrix}-8\\-4\\4\end{pmatrix}$ | M1$\checkmark\pm$ | Finding the difference between their $\overrightarrow{OX}$ (can be implied) and $\overrightarrow{OA}$: $\overrightarrow{AX} = \pm\left(\begin{pmatrix}-3\\3\\7\end{pmatrix}-\begin{pmatrix}5\\7\\3\end{pmatrix}\right)$ |
| $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OA} + 2\overrightarrow{AX} = \begin{pmatrix}5\\7\\3\end{pmatrix} + 2\begin{pmatrix}-8\\-4\\4\end{pmatrix}$ | dM1$\checkmark$ | $\begin{pmatrix}5\\7\\3\end{pmatrix} + 2\left(\text{their } \overrightarrow{AX}\right)$ |
| Hence $\overrightarrow{OB} = \begin{pmatrix}-11\\-1\\11\end{pmatrix}$ or $\overrightarrow{OB} = -11\mathbf{i}-\mathbf{j}+11\mathbf{k}$ or $(-11,-1,11)$ | A1 | $\begin{pmatrix}-11\\-1\\11\end{pmatrix}$ or $-11\mathbf{i}-\mathbf{j}+11\mathbf{k}$ or $(-11,-1,11)$ |
**[3 marks] — Total: 12 marks**
---6. With respect to a fixed origin $O$, the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations
$$\begin{array} { l l }
l _ { 1 } : & \mathbf { r } = ( - 9 \mathbf { i } + 10 \mathbf { k } ) + \lambda ( 2 \mathbf { i } + \mathbf { j } - \mathbf { k } ) \\
l _ { 2 } : & \mathbf { r } = ( 3 \mathbf { i } + \mathbf { j } + 17 \mathbf { k } ) + \mu ( 3 \mathbf { i } - \mathbf { j } + 5 \mathbf { k } )
\end{array}$$
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 $A$ has position vector $5 \mathbf { i } + 7 \mathbf { j } + 3 \mathbf { k }$.
\item Show that $A$ 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 $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2008 Q6 [12]}}