Question 9 - A-Level Maths

CAIE FP1 2012 November — Question 9 12 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2012
Session	November
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Lines & Planes
Type	Plane containing line and point/vector
Difficulty	Standard +0.3 This is a standard Further Maths vectors question with three routine parts: showing a line is parallel to a plane (checking direction vector perpendicular to normal), finding intersection of line and plane (substitution and solving), and perpendicular distance (standard formula application). All techniques are textbook exercises requiring no novel insight, though the multi-part nature and FM context place it slightly above average A-level difficulty.
Spec	4.04a Line equations: 2D and 3D, cartesian and vector forms 4.04b Plane equations: cartesian and vector forms 4.04f Line-plane intersection: find point 4.04h Shortest distances: between parallel lines and between skew lines

9 The plane $\Pi$ has equation $$\mathbf { r } = 2 \mathbf { i } + 3 \mathbf { j } - \mathbf { k } + \lambda ( \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } ) + \mu ( 3 \mathbf { i } + \mathbf { j } - 2 \mathbf { k } )$$ The line $l$, which does not lie in $\Pi$, has equation $$\mathbf { r } = 3 \mathbf { i } + 6 \mathbf { j } + 12 \mathbf { k } + t ( 8 \mathbf { i } + 5 \mathbf { j } - 8 \mathbf { k } )$$ Show that $l$ is parallel to $\Pi$. Find the position vector of the point at which the line with equation $\mathbf { r } = 5 \mathbf { i } - 4 \mathbf { j } + 7 \mathbf { k } + s ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } )$ meets $\Pi$. Find the perpendicular distance from the point with position vector $9 \mathbf { i } + 11 \mathbf { j } + 2 \mathbf { k }$ to $l$.

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Question 9:

Part 1 — Plane normal/parallel (4 marks):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\mathbf{n} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-2&2\\3&1&-2\end{vmatrix} = 2\mathbf{i}+8\mathbf{j}+7\mathbf{k}$	M1A1	Finds vector normal to $\Pi$
$\begin{pmatrix}3+8t\\6+5t\\12-8t\end{pmatrix}\cdot\begin{pmatrix}2\\8\\7\end{pmatrix}=138$ or $\begin{pmatrix}8\\5\\-8\end{pmatrix}\cdot\begin{pmatrix}2\\8\\7\end{pmatrix}=0$	A1	Dot product with general point on $l_1$
Independent of $t \Rightarrow$ parallel	A1	4

Part 2 — Cartesian equation and intersection (4 marks):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\Pi: 2x+8y+7z=21$	B1	Cartesian equation of $\Pi$
Sub. $x=5+2s$, $y=-4-s$, $z=7+s$	M1	Substitutes general point of $l_2$
$\Rightarrow s=-2$	A1	Finds value of parameter
Line meets $\Pi$ at point with p.v. $\mathbf{i}-2\mathbf{j}+5\mathbf{k}$	A1	4

Part 3 — Distance (4 marks):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$AB = \sqrt{6^2+5^2+10^2}=\sqrt{161}$	B1	Finds distance from point to known point on $l$
$BC = \frac{1}{\sqrt{(6^2\cdot5^2+8^2)(6\mathbf{i}+5\mathbf{j}-10\mathbf{k})(8\mathbf{i}+5\mathbf{j}-8\mathbf{k})}} = \frac{153}{\sqrt{153}}=\sqrt{153}$	M1A1	Finds distance along $l$ to foot of perpendicular
$AC=\sqrt{161-153}=\sqrt{8} \text{ or } 2\sqrt{2} \quad (=2.83)$	A1	4

Alternative Method B:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\overrightarrow{BA}=6\mathbf{i}+5\mathbf{j}-10\mathbf{j}$	B1	Finds vector $\overrightarrow{BA}$
$\frac{1}{\sqrt{8^2+5^2+8}}\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\6&5&-10\\8&5&-8\end{vmatrix} = \sqrt{\frac{1224}{153}}=\sqrt{8} \text{ or } 2\sqrt{2}$	M1A1A1	4

Alternative Method C:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\overrightarrow{AC}=\begin{pmatrix}8t-6\\5t-5\\10-8t\end{pmatrix}$	B1	Finds vector $\overrightarrow{AC}$
$\begin{pmatrix}8t-6\\5t-5\\10-8t\end{pmatrix}\cdot\begin{pmatrix}8\\5\\-8\end{pmatrix}=0 \Rightarrow t=1$	M1A1	Uses $\overrightarrow{AC}$ perpendicular to $l$
$AC=\sqrt{2^2+0^2+2^2}=\sqrt{8}$	A1	4

# Question 9:

## Part 1 — Plane normal/parallel (4 marks):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{n} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-2&2\\3&1&-2\end{vmatrix} = 2\mathbf{i}+8\mathbf{j}+7\mathbf{k}$ | M1A1 | Finds vector normal to $\Pi$ |
| $\begin{pmatrix}3+8t\\6+5t\\12-8t\end{pmatrix}\cdot\begin{pmatrix}2\\8\\7\end{pmatrix}=138$ or $\begin{pmatrix}8\\5\\-8\end{pmatrix}\cdot\begin{pmatrix}2\\8\\7\end{pmatrix}=0$ | A1 | Dot product with general point on $l_1$ |
| Independent of $t \Rightarrow$ parallel | A1 | 4 | Deduces result |

## Part 2 — Cartesian equation and intersection (4 marks):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\Pi: 2x+8y+7z=21$ | B1 | Cartesian equation of $\Pi$ |
| Sub. $x=5+2s$, $y=-4-s$, $z=7+s$ | M1 | Substitutes general point of $l_2$ |
| $\Rightarrow s=-2$ | A1 | Finds value of parameter |
| Line meets $\Pi$ at point with p.v. $\mathbf{i}-2\mathbf{j}+5\mathbf{k}$ | A1 | 4 | Finds p.v. of intersection |

## Part 3 — Distance (4 marks):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $AB = \sqrt{6^2+5^2+10^2}=\sqrt{161}$ | B1 | Finds distance from point to known point on $l$ |
| $BC = \frac{1}{\sqrt{(6^2\cdot5^2+8^2)(6\mathbf{i}+5\mathbf{j}-10\mathbf{k})(8\mathbf{i}+5\mathbf{j}-8\mathbf{k})}} = \frac{153}{\sqrt{153}}=\sqrt{153}$ | M1A1 | Finds distance along $l$ to foot of perpendicular |
| $AC=\sqrt{161-153}=\sqrt{8} \text{ or } 2\sqrt{2} \quad (=2.83)$ | A1 | 4 | **[12]** |

### Alternative Method B:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{BA}=6\mathbf{i}+5\mathbf{j}-10\mathbf{j}$ | B1 | Finds vector $\overrightarrow{BA}$ |
| $\frac{1}{\sqrt{8^2+5^2+8}}\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\6&5&-10\\8&5&-8\end{vmatrix} = \sqrt{\frac{1224}{153}}=\sqrt{8} \text{ or } 2\sqrt{2}$ | M1A1A1 | 4 | Cross product method |

### Alternative Method C:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AC}=\begin{pmatrix}8t-6\\5t-5\\10-8t\end{pmatrix}$ | B1 | Finds vector $\overrightarrow{AC}$ |
| $\begin{pmatrix}8t-6\\5t-5\\10-8t\end{pmatrix}\cdot\begin{pmatrix}8\\5\\-8\end{pmatrix}=0 \Rightarrow t=1$ | M1A1 | Uses $\overrightarrow{AC}$ perpendicular to $l$ |
| $AC=\sqrt{2^2+0^2+2^2}=\sqrt{8}$ | A1 | 4 | **[12]** |

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9 The plane $\Pi$ has equation

$$\mathbf { r } = 2 \mathbf { i } + 3 \mathbf { j } - \mathbf { k } + \lambda ( \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } ) + \mu ( 3 \mathbf { i } + \mathbf { j } - 2 \mathbf { k } )$$

The line $l$, which does not lie in $\Pi$, has equation

$$\mathbf { r } = 3 \mathbf { i } + 6 \mathbf { j } + 12 \mathbf { k } + t ( 8 \mathbf { i } + 5 \mathbf { j } - 8 \mathbf { k } )$$

Show that $l$ is parallel to $\Pi$.

Find the position vector of the point at which the line with equation $\mathbf { r } = 5 \mathbf { i } - 4 \mathbf { j } + 7 \mathbf { k } + s ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } )$ meets $\Pi$.

Find the perpendicular distance from the point with position vector $9 \mathbf { i } + 11 \mathbf { j } + 2 \mathbf { k }$ to $l$.

\hfill \mbox{\textit{CAIE FP1 2012 Q9 [12]}}

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