Question 7 - A-Level Maths

Edexcel FP3 2009 June — Question 7 11 marks

Exam Board	Edexcel
Module	FP3 (Further Pure Mathematics 3)
Year	2009
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Lines & Planes
Type	Line intersection with line
Difficulty	Standard +0.3 This is a standard Further Maths vectors question covering routine techniques: equating line equations to find intersection (part a), finding a plane equation using cross product of direction vectors (part b), and applying the skew lines distance formula (part c). All three parts follow textbook methods with straightforward algebra, making it slightly easier than average even for FP3 level.
Spec	4.04b Plane equations: cartesian and vector forms 4.04e Line intersections: parallel, skew, or intersecting 4.04i Shortest distance: between a point and a line

7. The lines $l _ { 1 }$ and $l _ { 2 }$ have equations $$\mathbf { r } = \left( \begin{array} { r } 1 \\ - 1 \\ 2 \end{array} \right) + \lambda \left( \begin{array} { r } - 1 \\ 3 \\ 4 \end{array} \right) \text { and } \quad \mathbf { r } = \left( \begin{array} { r } \alpha \\ - 4 \\ 0 \end{array} \right) + \mu \left( \begin{array} { l } 0 \\ 3 \\ 2 \end{array} \right) .$$ If the lines $l _ { 1 }$ and $l _ { 2 }$ intersect, find

the value of $\alpha$,
an equation for the plane containing the lines $l _ { 1 }$ and $l _ { 2 }$, giving your answer in the form $a x + b y + c z + d = 0$, where $a , b , c$ and $d$ are constants. For other values of $\alpha$, the lines $l _ { 1 }$ and $l _ { 2 }$ do not intersect and are skew lines.
Given that $\alpha = 2$,
find the shortest distance between the lines $l _ { 1 }$ and $l _ { 2 }$.

Show mark scheme Show mark scheme source

Question 7:

Part (a):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
If the lines meet, $-1+3\lambda=-4+3\mu$ and $2+4\lambda=2\mu$	M1
Solve to give $\lambda=0$ ($\mu=1$ but this need not be seen)	M1 A1
Also $1-\lambda=\alpha$ and so $\alpha=1$	B1	(4)

Part (b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&3&4\\0&3&2\end{vmatrix} = -6\mathbf{i}+2\mathbf{j}-3\mathbf{k}$ is perpendicular to both lines and hence to the plane	M1 A1
The plane has equation $\mathbf{r\cdot n=a\cdot n}$, which is $-6x+2y-3z=-14$, i.e. $-6x+2y-3z+14=0$	M1, A1 o.a.e.	(4)

Alternative for (b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Use $(1,-1,2)$ and $(\alpha,-4,0)$ in equation $ax+by+cz+d=0$	M1
And third point so three equations, and attempt to solve	M1
Obtain $6x-2y+3z=$	A1
$(6x-2y+3z)-14=0$	A1 o.a.e.	(4)

Part (c):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$(\mathbf{a_1}-\mathbf{a_2})=\mathbf{i}-3\mathbf{j}-2\mathbf{k}$	M1
Use formula \(\frac{(\mathbf{a_1}-\mathbf{a_2})\bullet\mathbf{n}}{	\mathbf{n}	} = \frac{(\mathbf{i}-3\mathbf{j}-2\mathbf{k})\bullet(-6\mathbf{i}+2\mathbf{j}-3\mathbf{k})}{\sqrt{36+4+9}} = \left(\frac{-6}{7}\right)\)
Distance is $\frac{6}{7}$	A1	(3) [11]

# Question 7:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| If the lines meet, $-1+3\lambda=-4+3\mu$ and $2+4\lambda=2\mu$ | M1 | |
| Solve to give $\lambda=0$ ($\mu=1$ but this need not be seen) | M1 A1 | |
| Also $1-\lambda=\alpha$ and so $\alpha=1$ | B1 | (4) |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&3&4\\0&3&2\end{vmatrix} = -6\mathbf{i}+2\mathbf{j}-3\mathbf{k}$ is perpendicular to both lines and hence to the plane | M1 A1 | |
| The plane has equation $\mathbf{r\cdot n=a\cdot n}$, which is $-6x+2y-3z=-14$, i.e. $-6x+2y-3z+14=0$ | M1, A1 o.a.e. | (4) |

## Alternative for (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Use $(1,-1,2)$ and $(\alpha,-4,0)$ in equation $ax+by+cz+d=0$ | M1 | |
| And third point so three equations, and attempt to solve | M1 | |
| Obtain $6x-2y+3z=$ | A1 | |
| $(6x-2y+3z)-14=0$ | A1 o.a.e. | (4) |

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\mathbf{a_1}-\mathbf{a_2})=\mathbf{i}-3\mathbf{j}-2\mathbf{k}$ | M1 | |
| Use formula $\frac{(\mathbf{a_1}-\mathbf{a_2})\bullet\mathbf{n}}{|\mathbf{n}|} = \frac{(\mathbf{i}-3\mathbf{j}-2\mathbf{k})\bullet(-6\mathbf{i}+2\mathbf{j}-3\mathbf{k})}{\sqrt{36+4+9}} = \left(\frac{-6}{7}\right)$ | M1 | |
| Distance is $\frac{6}{7}$ | A1 | (3) **[11]** |

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7. The lines $l _ { 1 }$ and $l _ { 2 }$ have equations

$$\mathbf { r } = \left( \begin{array} { r } 
1 \\
- 1 \\
2
\end{array} \right) + \lambda \left( \begin{array} { r } 
- 1 \\
3 \\
4
\end{array} \right) \text { and } \quad \mathbf { r } = \left( \begin{array} { r } 
\alpha \\
- 4 \\
0
\end{array} \right) + \mu \left( \begin{array} { l } 
0 \\
3 \\
2
\end{array} \right) .$$

If the lines $l _ { 1 }$ and $l _ { 2 }$ intersect, find
\begin{enumerate}[label=(\alph*)]
\item the value of $\alpha$,
\item an equation for the plane containing the lines $l _ { 1 }$ and $l _ { 2 }$, giving your answer in the form $a x + b y + c z + d = 0$, where $a , b , c$ and $d$ are constants.

For other values of $\alpha$, the lines $l _ { 1 }$ and $l _ { 2 }$ do not intersect and are skew lines.\\
Given that $\alpha = 2$,
\item find the shortest distance between the lines $l _ { 1 }$ and $l _ { 2 }$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP3 2009 Q7 [11]}}

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