| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Parallel and perpendicular lines |
| Difficulty | Standard +0.3 This is a standard C4 vectors question requiring students to work with vector equations of lines, likely testing parallel/perpendicular conditions using dot products or comparing direction vectors. While it involves 3D vectors (a step up from 2D), the techniques are routine and well-practiced at this level, making it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Setting up equations: i: \(11-2\lambda=-5+q\mu\) (1); j: \(2+\lambda=11+2\mu\) (2); k: \(17-4\lambda=p+2\mu\) (3) | M1 | Need to see equations (1) and (2). Condone one slip. Note \(q=-3\) |
| From (2): \(\lambda=9+2\mu\); substituting into (1): \(\mu=-2\) | dM1 | Attempts to solve (1) and (2) to find one of either \(\lambda\) or \(\mu\) |
| Any one of \(\lambda=5\) or \(\mu=-2\) | A1 | — |
| From (3): \(17-4(9+2\mu)=p+2\mu\), writing equation containing \(p\) and one of \(\lambda\) or \(\mu\) | A1 | Candidate writes correct equation containing \(p\) and one of \(\lambda\) or \(\mu\) already found |
| Substituting \(\mu=-2\): \(17-20=p-4 \Rightarrow p=1\) | ddM1 | Attempt to substitute into k component to give equation in \(p\) alone |
| \(p=1\) | A1 cso | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Two out of three coordinates implies first M1 | M1 | If no working shown, any two of three coordinates can imply M1 |
| Intersect at \(\mathbf{r}=\begin{pmatrix}1\\7\\-3\end{pmatrix}\) or \((1,7,-3)\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\overrightarrow{AX}=\overrightarrow{OX}-\overrightarrow{OA}=\begin{pmatrix}1\\7\\-3\end{pmatrix}-\begin{pmatrix}9\\3\\13\end{pmatrix}=\begin{pmatrix}-8\\4\\-16\end{pmatrix}\) | M1\(\checkmark\ \pm\) | Finding difference between their \(\overrightarrow{OX}\) (can be implied) and \(\overrightarrow{OA}\) |
| \(\overrightarrow{OB}=\overrightarrow{OX}+\overrightarrow{XB}=\overrightarrow{OX}+\overrightarrow{AX}=\begin{pmatrix}1\\7\\-3\end{pmatrix}+\begin{pmatrix}-8\\4\\-16\end{pmatrix}\) | dM1\(\checkmark\) | their \(\overrightarrow{OX}\) + their \(\overrightarrow{AX}\) |
| \(\overrightarrow{OB}=\begin{pmatrix}-7\\11\\-19\end{pmatrix}\) or \(-7\mathbf{i}+11\mathbf{j}-19\mathbf{k}\) or \((-7,11,-19)\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| At \(A\), \(\lambda=1\). At \(X\), \(\lambda=5\). Hence at \(B\), \(\lambda=5+(5-1)=9\) | M1\(\checkmark\) | \(\lambda_B=(\text{their }\lambda_X)+(\text{their }\lambda_X - \text{their }\lambda_A)\) or \(\lambda_B=2(\text{their }\lambda_X)-(\text{their }\lambda_A)\) |
| \(\overrightarrow{OB}=\begin{pmatrix}11\\2\\17\end{pmatrix}+9\begin{pmatrix}-2\\1\\-4\end{pmatrix}\) | dM1\(\checkmark\) | Substitutes their value of \(\lambda\) into line \(l_1\) |
| \(\overrightarrow{OB}=\begin{pmatrix}-7\\11\\-19\end{pmatrix}\) or \(-7\mathbf{i}+11\mathbf{j}-19\mathbf{k}\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\overrightarrow{OA}=9\mathbf{i}+3\mathbf{j}+13\mathbf{k}\), intersection \(\overrightarrow{OX}=\mathbf{i}+7\mathbf{j}-3\mathbf{k}\); applying \(\overrightarrow{AX}\) transformation to \(\overrightarrow{OX}\) | M1\(\checkmark\ \pm\) | Finding difference between their \(\overrightarrow{OX}\) and \(\overrightarrow{OA}\); \(\overrightarrow{AX}=\pm\left(\begin{pmatrix}1\\7\\-3\end{pmatrix}-\begin{pmatrix}9\\3\\13\end{pmatrix}\right)\) |
| Adding \(\overrightarrow{AX}\) to \(\overrightarrow{OX}\) | dM1\(\checkmark\) | their \(\overrightarrow{OX}\) + their \(\overrightarrow{AX}\) |
| \(\overrightarrow{OB}=\begin{pmatrix}-7\\11\\-19\end{pmatrix}\) or \(-7\mathbf{i}+11\mathbf{j}-19\mathbf{k}\) or \((-7,11,-19)\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(X\) is midpoint of \(AB\): \((1,7,-3)=\left(\frac{9+a}{2},\frac{3+b}{2},\frac{13+c}{2}\right)\) | M1\(\checkmark\) | Writing down any two of these equations correctly |
| \(a=2(1)-9=-7,\quad b=2(7)-3=11,\quad c=2(-3)-13=-19\) | dM1\(\checkmark\) | Attempt to find at least two of \(a\), \(b\), or \(c\) |
| \(\overrightarrow{OB}=\begin{pmatrix}-7\\11\\-19\end{pmatrix}\) or \(-7\mathbf{i}+11\mathbf{j}-19\mathbf{k}\) or \(a=-7, b=11, c=-19\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(\overrightarrow{OX} = \mathbf{i} + 7\mathbf{j} - 3\mathbf{k}\) be point of intersection | ||
| \(\overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix}1\\7\\-3\end{pmatrix} - \begin{pmatrix}9\\3\\13\end{pmatrix} = \begin{pmatrix}-8\\4\\-16\end{pmatrix}\) | M1 \(\checkmark\) \(\pm\) | Finding the difference between their \(\overrightarrow{OX}\) (can be implied) and \(\overrightarrow{OA}\); note \( |
| \( | \overrightarrow{AX} | = \sqrt{64+16+256} = \sqrt{336} = 4\sqrt{21}\) |
| \(\overrightarrow{BX} = \overrightarrow{OX} - \overrightarrow{OB} = \begin{pmatrix}1\\7\\-3\end{pmatrix} - \begin{pmatrix}11-2\lambda\\2+\lambda\\17-4\lambda\end{pmatrix} = \begin{pmatrix}-10+2\lambda\\5-\lambda\\-20+4\lambda\end{pmatrix}\) | ||
| \( | \overrightarrow{BX} | = |
| \(100-40\lambda+4\lambda^2+25-10\lambda+\lambda^2+400-160\lambda+16\lambda^2 = 336\) | ||
| \(21\lambda^2 - 210\lambda + 525 = 336\) | ||
| \(21\lambda^2 - 210\lambda + 189 = 0\) | ||
| \(\lambda^2 - 10\lambda + 9 = 0\) | ||
| \((\lambda-1)(\lambda-9) = 0\) | ||
| At \(A\), \(\lambda=1\) and at \(B\), \(\lambda=9\), so \(\overrightarrow{OB} = \begin{pmatrix}11-2(9)\\2+9\\17-4(9)\end{pmatrix}\) | ||
| \(\overrightarrow{OB} = \begin{pmatrix}-7\\11\\-19\end{pmatrix}\) or \(\overrightarrow{OB} = -7\mathbf{i}+11\mathbf{j}-19\mathbf{k}\) | A1 | \(\begin{pmatrix}-7\\11\\-19\end{pmatrix}\) or \(-7\mathbf{i}+11\mathbf{j}-19\mathbf{k}\) or \((-7,11,-19)\) |
## Question 4(b) — Aliter Way 2:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Setting up equations: **i**: $11-2\lambda=-5+q\mu$ (1); **j**: $2+\lambda=11+2\mu$ (2); **k**: $17-4\lambda=p+2\mu$ (3) | M1 | Need to see equations (1) and (2). Condone one slip. Note $q=-3$ |
| From (2): $\lambda=9+2\mu$; substituting into (1): $\mu=-2$ | dM1 | Attempts to solve (1) and (2) to find one of either $\lambda$ or $\mu$ |
| Any one of $\lambda=5$ or $\mu=-2$ | A1 | — |
| From (3): $17-4(9+2\mu)=p+2\mu$, writing equation containing $p$ and one of $\lambda$ or $\mu$ | A1 | Candidate writes correct equation containing $p$ and one of $\lambda$ or $\mu$ already found |
| Substituting $\mu=-2$: $17-20=p-4 \Rightarrow p=1$ | ddM1 | Attempt to substitute into **k** component to give equation in $p$ alone |
| $p=1$ | A1 cso | — |
---
## Question 4(c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Two out of three coordinates implies first M1 | M1 | If no working shown, any two of three coordinates can imply M1 |
| Intersect at $\mathbf{r}=\begin{pmatrix}1\\7\\-3\end{pmatrix}$ or $(1,7,-3)$ | A1 | — |
---
## Question 4(d) — Aliter Way 2:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AX}=\overrightarrow{OX}-\overrightarrow{OA}=\begin{pmatrix}1\\7\\-3\end{pmatrix}-\begin{pmatrix}9\\3\\13\end{pmatrix}=\begin{pmatrix}-8\\4\\-16\end{pmatrix}$ | M1$\checkmark\ \pm$ | Finding difference between their $\overrightarrow{OX}$ (can be implied) and $\overrightarrow{OA}$ |
| $\overrightarrow{OB}=\overrightarrow{OX}+\overrightarrow{XB}=\overrightarrow{OX}+\overrightarrow{AX}=\begin{pmatrix}1\\7\\-3\end{pmatrix}+\begin{pmatrix}-8\\4\\-16\end{pmatrix}$ | dM1$\checkmark$ | their $\overrightarrow{OX}$ + their $\overrightarrow{AX}$ |
| $\overrightarrow{OB}=\begin{pmatrix}-7\\11\\-19\end{pmatrix}$ or $-7\mathbf{i}+11\mathbf{j}-19\mathbf{k}$ or $(-7,11,-19)$ | A1 | — |
---
## Question 4(d) — Aliter Way 3:
| Working/Answer | Marks | Guidance |
|---|---|---|
| At $A$, $\lambda=1$. At $X$, $\lambda=5$. Hence at $B$, $\lambda=5+(5-1)=9$ | M1$\checkmark$ | $\lambda_B=(\text{their }\lambda_X)+(\text{their }\lambda_X - \text{their }\lambda_A)$ or $\lambda_B=2(\text{their }\lambda_X)-(\text{their }\lambda_A)$ |
| $\overrightarrow{OB}=\begin{pmatrix}11\\2\\17\end{pmatrix}+9\begin{pmatrix}-2\\1\\-4\end{pmatrix}$ | dM1$\checkmark$ | Substitutes their value of $\lambda$ into line $l_1$ |
| $\overrightarrow{OB}=\begin{pmatrix}-7\\11\\-19\end{pmatrix}$ or $-7\mathbf{i}+11\mathbf{j}-19\mathbf{k}$ | A1 | — |
---
## Question 4(d) — Aliter Way 4:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\overrightarrow{OA}=9\mathbf{i}+3\mathbf{j}+13\mathbf{k}$, intersection $\overrightarrow{OX}=\mathbf{i}+7\mathbf{j}-3\mathbf{k}$; applying $\overrightarrow{AX}$ transformation to $\overrightarrow{OX}$ | M1$\checkmark\ \pm$ | Finding difference between their $\overrightarrow{OX}$ and $\overrightarrow{OA}$; $\overrightarrow{AX}=\pm\left(\begin{pmatrix}1\\7\\-3\end{pmatrix}-\begin{pmatrix}9\\3\\13\end{pmatrix}\right)$ |
| Adding $\overrightarrow{AX}$ to $\overrightarrow{OX}$ | dM1$\checkmark$ | their $\overrightarrow{OX}$ + their $\overrightarrow{AX}$ |
| $\overrightarrow{OB}=\begin{pmatrix}-7\\11\\-19\end{pmatrix}$ or $-7\mathbf{i}+11\mathbf{j}-19\mathbf{k}$ or $(-7,11,-19)$ | A1 | — |
---
## Question 4(d) — Aliter Way 5:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $X$ is midpoint of $AB$: $(1,7,-3)=\left(\frac{9+a}{2},\frac{3+b}{2},\frac{13+c}{2}\right)$ | M1$\checkmark$ | Writing down any two of these equations correctly |
| $a=2(1)-9=-7,\quad b=2(7)-3=11,\quad c=2(-3)-13=-19$ | dM1$\checkmark$ | Attempt to find at least two of $a$, $b$, or $c$ |
| $\overrightarrow{OB}=\begin{pmatrix}-7\\11\\-19\end{pmatrix}$ or $-7\mathbf{i}+11\mathbf{j}-19\mathbf{k}$ or $a=-7, b=11, c=-19$ | A1 | — |
## Question 4(d) – Aliter Way 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $\overrightarrow{OX} = \mathbf{i} + 7\mathbf{j} - 3\mathbf{k}$ be point of intersection | | |
| $\overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix}1\\7\\-3\end{pmatrix} - \begin{pmatrix}9\\3\\13\end{pmatrix} = \begin{pmatrix}-8\\4\\-16\end{pmatrix}$ | M1 $\checkmark$ $\pm$ | Finding the difference between their $\overrightarrow{OX}$ (can be implied) and $\overrightarrow{OA}$; note $|\overrightarrow{AX}| = \sqrt{336}$ would imply M1 |
| $|\overrightarrow{AX}| = \sqrt{64+16+256} = \sqrt{336} = 4\sqrt{21}$ | | |
| $\overrightarrow{BX} = \overrightarrow{OX} - \overrightarrow{OB} = \begin{pmatrix}1\\7\\-3\end{pmatrix} - \begin{pmatrix}11-2\lambda\\2+\lambda\\17-4\lambda\end{pmatrix} = \begin{pmatrix}-10+2\lambda\\5-\lambda\\-20+4\lambda\end{pmatrix}$ | | |
| $|\overrightarrow{BX}| = |\overrightarrow{AX}| = \sqrt{336}$ gives $(-10+2\lambda)^2 + (5-\lambda)^2 + (-20+4\lambda)^2 = 336$ | dM1 $\checkmark$ | Writes distance equation $|\overrightarrow{BX}|^2 = 336$ where $\overrightarrow{BX} = \overrightarrow{OX} - \overrightarrow{OB}$ and $\overrightarrow{OB} = \begin{pmatrix}11-2\lambda\\2+\lambda\\17-4\lambda\end{pmatrix}$ |
| $100-40\lambda+4\lambda^2+25-10\lambda+\lambda^2+400-160\lambda+16\lambda^2 = 336$ | | |
| $21\lambda^2 - 210\lambda + 525 = 336$ | | |
| $21\lambda^2 - 210\lambda + 189 = 0$ | | |
| $\lambda^2 - 10\lambda + 9 = 0$ | | |
| $(\lambda-1)(\lambda-9) = 0$ | | |
| At $A$, $\lambda=1$ and at $B$, $\lambda=9$, so $\overrightarrow{OB} = \begin{pmatrix}11-2(9)\\2+9\\17-4(9)\end{pmatrix}$ | | |
| $\overrightarrow{OB} = \begin{pmatrix}-7\\11\\-19\end{pmatrix}$ or $\overrightarrow{OB} = -7\mathbf{i}+11\mathbf{j}-19\mathbf{k}$ | A1 | $\begin{pmatrix}-7\\11\\-19\end{pmatrix}$ or $-7\mathbf{i}+11\mathbf{j}-19\mathbf{k}$ or $(-7,11,-19)$ |
**Total: [3]**
---4. With respect to a fixed origin $O$ the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations
$$l _ { 1 } : \quad \mathbf { r } = \left( \begin{array} { c }
11 \\
2 \\
17
\end{array} \right) + \lambda \left( \begin{array} { c }
- 2 \\
1 \\
- 4
\end{array} \right) \quad l _ { 2 } : \quad \mathbf { r } = \left( \begin{array} { c }
- 5 \\
11 \\
p
\end{array} \right) + \mu \left( \begin{array} { l }
q \\
2 \\
2
\end{array} \right)$$
where $\lambda$ and $\mu$ are 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, find
\item the value of $p$,
\item the coordinates of the point of intersection.
The point $A$ lies on $l _ { 1 }$ and has position vector $\left( \begin{array} { c } 9 \\ 3 \\ 13 \end{array} \right)$. The point $C$ lies on $l _ { 2 }$.\\
Given that a circle, with centre $C$, cuts the line $l _ { 1 }$ at the points $A$ and $B$,
\item find the position vector of $B$.\\
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\section*{Question 4 continued}
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\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2009 Q4 [13]}}