Question 7 - A-Level Maths

CAIE Further Paper 1 2022 June — Question 7 18 marks

Exam Board	CAIE
Module	Further Paper 1 (Further Paper 1)
Year	2022
Session	June
Marks	18
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Cross Product & Distances
Type	Shortest distance between two skew lines
Difficulty	Challenging +1.2 This is a standard Further Maths question on skew lines and planes requiring the cross product formula for distance between skew lines, plane equations in different forms, and angle between planes. While it involves multiple parts and several techniques, each step follows well-established procedures without requiring novel insight. The algebraic manipulation is straightforward, making it moderately above average difficulty but routine for Further Maths students.
Spec	4.04a Line equations: 2D and 3D, cartesian and vector forms 4.04b Plane equations: cartesian and vector forms 4.04c Scalar product: calculate and use for angles 4.04h Shortest distances: between parallel lines and between skew lines 4.04i Shortest distance: between a point and a line

7 The position vectors of the points $A , B , C , D$ are $$7 \mathbf { i } + 4 \mathbf { j } - \mathbf { k } , \quad 11 \mathbf { i } + 3 \mathbf { j } , \quad 2 \mathbf { i } + 6 \mathbf { j } + 3 \mathbf { k } , \quad 2 \mathbf { i } + 7 \mathbf { j } + \lambda \mathbf { k }$$ respectively.

Given that the shortest distance between the line $A B$ and the line $C D$ is 3 , show that $\lambda ^ { 2 } - 5 \lambda + 4 = 0$.
Let $\Pi _ { 1 }$ be the plane $A B D$ when $\lambda = 1$.
Let $\Pi _ { 2 }$ be the plane $A B D$ when $\lambda = 4$.
1. Write down an equation of $\Pi _ { 1 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + \mathbf { s b } + \mathbf { t c }$.
2. Find an equation of $\Pi _ { 2 }$, giving your answer in the form $a x + b y + c z = d$.
Find the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$.
If you use the following page to complete the answer to any question, the question number must be clearly shown.

Show mark scheme Show mark scheme source

Question 7(a):

Answer	Marks	Guidance
$\overrightarrow{AB} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}$, $\overrightarrow{CD} = \mathbf{j} + (\lambda-3)\mathbf{k}$	B1	Finds direction vectors of the two lines
$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ 0 & 1 & \lambda-3 \end{vmatrix} = \begin{pmatrix} 2-\lambda \\ 12-4\lambda \\ 4 \end{pmatrix}$	M1 A1	Finds common perpendicular. May also be done by setting up simultaneous equations and solving them
\(\frac{1}{\sqrt{(\lambda-2)^2+(4\lambda-12)^2+16}}\left	\begin{pmatrix}-5\\2\\4\end{pmatrix}\cdot\begin{pmatrix}2-\lambda\\12-4\lambda\\4\end{pmatrix}\right	\)
\(\left	\frac{30-3\lambda}{\sqrt{17\lambda^2-100\lambda+164}}\right	= 3 \Rightarrow 9(\lambda-10)^2 = 9(17\lambda^2 - 100\lambda + 164)\)
$16\lambda^2 - 80\lambda + 64 = 0$ leading to $\lambda^2 - 5\lambda + 4 = 0$	A1	AG

Question 7(b)(i):

Answer	Marks	Guidance
$\mathbf{r} = 7\mathbf{i} + 4\mathbf{j} - \mathbf{k} + s(4\mathbf{i} - \mathbf{j} + \mathbf{k}) + t(-5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})$	M1 A1	OE. M1 for using a correct point and attempting to find relevant direction vectors

Question 7(b)(ii):

Answer	Marks	Guidance
$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ -5 & 3 & 5 \end{vmatrix} = \begin{pmatrix}-8\\-25\\7\end{pmatrix}$	M1 A1	Finds normal to the plane $\Pi_2$
$-8(7) - 25(4) + 7(-1)$ leading to $-8x - 25y + 7z = -163$	M1 A1	Substitutes point

Question 7(c):

Answer	Marks	Guidance
Answer	Marks	Guidance
$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ -5 & 3 & 2 \end{vmatrix} = \begin{pmatrix} -5 \\ -13 \\ 7 \end{pmatrix}$	M1 A1	Finds normal to the plane $\Pi_1$
$\begin{pmatrix} -5 \\ -13 \\ 7 \end{pmatrix} \cdot \begin{pmatrix} -8 \\ -25 \\ 7 \end{pmatrix} = \sqrt{243}\sqrt{738}\cos\theta$ leading to $\cos\theta = \dfrac{414}{\sqrt{243}\sqrt{738}}$	M1 A1	Uses dot product of normal vectors. $\cos = 0.9776176...$
$12.1°$	A1	Mark final answer. Accept $0.212^c$
5

## Question 7(a):

$\overrightarrow{AB} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}$, $\overrightarrow{CD} = \mathbf{j} + (\lambda-3)\mathbf{k}$ | **B1** | Finds direction vectors of the two lines

$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ 0 & 1 & \lambda-3 \end{vmatrix} = \begin{pmatrix} 2-\lambda \\ 12-4\lambda \\ 4 \end{pmatrix}$ | **M1 A1** | Finds common perpendicular. May also be done by setting up simultaneous equations and solving them

$\frac{1}{\sqrt{(\lambda-2)^2+(4\lambda-12)^2+16}}\left|\begin{pmatrix}-5\\2\\4\end{pmatrix}\cdot\begin{pmatrix}2-\lambda\\12-4\lambda\\4\end{pmatrix}\right|$ | **M1 A1** | Uses formula for perpendicular distance

$\left|\frac{30-3\lambda}{\sqrt{17\lambda^2-100\lambda+164}}\right| = 3 \Rightarrow 9(\lambda-10)^2 = 9(17\lambda^2 - 100\lambda + 164)$ | **M1** | Sets equal to 3 and forms quadratic in $\lambda$

$16\lambda^2 - 80\lambda + 64 = 0$ leading to $\lambda^2 - 5\lambda + 4 = 0$ | **A1** | AG

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## Question 7(b)(i):

$\mathbf{r} = 7\mathbf{i} + 4\mathbf{j} - \mathbf{k} + s(4\mathbf{i} - \mathbf{j} + \mathbf{k}) + t(-5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})$ | **M1 A1** | OE. M1 for using a correct point and attempting to find relevant direction vectors

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## Question 7(b)(ii):

$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ -5 & 3 & 5 \end{vmatrix} = \begin{pmatrix}-8\\-25\\7\end{pmatrix}$ | **M1 A1** | Finds normal to the plane $\Pi_2$

$-8(7) - 25(4) + 7(-1)$ leading to $-8x - 25y + 7z = -163$ | **M1 A1** | Substitutes point

## Question 7(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ -5 & 3 & 2 \end{vmatrix} = \begin{pmatrix} -5 \\ -13 \\ 7 \end{pmatrix}$ | M1 A1 | Finds normal to the plane $\Pi_1$ |
| $\begin{pmatrix} -5 \\ -13 \\ 7 \end{pmatrix} \cdot \begin{pmatrix} -8 \\ -25 \\ 7 \end{pmatrix} = \sqrt{243}\sqrt{738}\cos\theta$ leading to $\cos\theta = \dfrac{414}{\sqrt{243}\sqrt{738}}$ | M1 A1 | Uses dot product of normal vectors. $\cos = 0.9776176...$ |
| $12.1°$ | A1 | Mark final answer. Accept $0.212^c$ |
| | **5** | |

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7 The position vectors of the points $A , B , C , D$ are

$$7 \mathbf { i } + 4 \mathbf { j } - \mathbf { k } , \quad 11 \mathbf { i } + 3 \mathbf { j } , \quad 2 \mathbf { i } + 6 \mathbf { j } + 3 \mathbf { k } , \quad 2 \mathbf { i } + 7 \mathbf { j } + \lambda \mathbf { k }$$

respectively.
\begin{enumerate}[label=(\alph*)]
\item Given that the shortest distance between the line $A B$ and the line $C D$ is 3 , show that $\lambda ^ { 2 } - 5 \lambda + 4 = 0$.\\

Let $\Pi _ { 1 }$ be the plane $A B D$ when $\lambda = 1$.\\
Let $\Pi _ { 2 }$ be the plane $A B D$ when $\lambda = 4$.
\item \begin{enumerate}[label=(\roman*)]
\item Write down an equation of $\Pi _ { 1 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + \mathbf { s b } + \mathbf { t c }$.
\item Find an equation of $\Pi _ { 2 }$, giving your answer in the form $a x + b y + c z = d$.
\end{enumerate}\item Find the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$.\\

If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q7 [18]}}

This paper (7 questions)

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Q1 6 Q2 7 Q3 10 Q4 9 Q5 12 Q6 13 Q7 18

Answer	Marks	Guidance
\(\overrightarrow{AB} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}\), \(\overrightarrow{CD} = \mathbf{j} + (\lambda-3)\mathbf{k}\)	B1	Finds direction vectors of the two lines
\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ 0 & 1 & \lambda-3 \end{vmatrix} = \begin{pmatrix} 2-\lambda \\ 12-4\lambda \\ 4 \end{pmatrix}\)	M1 A1	Finds common perpendicular. May also be done by setting up simultaneous equations and solving them
\(\frac{1}{\sqrt{(\lambda-2)^2+(4\lambda-12)^2+16}}\left	\begin{pmatrix}-5\\2\\4\end{pmatrix}\cdot\begin{pmatrix}2-\lambda\\12-4\lambda\\4\end{pmatrix}\right	\)
\(\left	\frac{30-3\lambda}{\sqrt{17\lambda^2-100\lambda+164}}\right	= 3 \Rightarrow 9(\lambda-10)^2 = 9(17\lambda^2 - 100\lambda + 164)\)
\(16\lambda^2 - 80\lambda + 64 = 0\) leading to \(\lambda^2 - 5\lambda + 4 = 0\)	A1	AG

Answer	Marks	Guidance
\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ -5 & 3 & 5 \end{vmatrix} = \begin{pmatrix}-8\\-25\\7\end{pmatrix}\)	M1 A1	Finds normal to the plane \(\Pi_2\)
\(-8(7) - 25(4) + 7(-1)\) leading to \(-8x - 25y + 7z = -163\)	M1 A1	Substitutes point