Question 13 - A-Level Maths

AQA Further AS Paper 1 2019 June — Question 13 10 marks

Exam Board	AQA
Module	Further AS Paper 1 (Further AS Paper 1)
Year	2019
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Lines & Planes
Type	Angle between two lines
Difficulty	Standard +0.3 This is a standard Further Maths vectors question testing routine conversions between Cartesian and vector forms, perpendicularity conditions, and angle formulas. Part (a) is mechanical conversion, parts (b)(i-ii) apply basic dot product tests, and part (b)(iii) uses the standard angle formula with algebraic manipulation. While it requires multiple techniques and careful algebra, all steps follow textbook procedures with no novel insight required, making it slightly easier than average for Further Maths content.
Spec	4.04a Line equations: 2D and 3D, cartesian and vector forms 4.04c Scalar product: calculate and use for angles

Line $l_1$ has Cartesian equation $$x - 3 = \frac{2y + 2}{3} = 2 - z$$

Write the equation of line $l_1$ in the form $$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$$ where $\lambda$ is a parameter and $\mathbf{a}$ and $\mathbf{b}$ are vectors to be found. [2 marks]
Line $l_2$ passes through the points $P(3, 2, 0)$ and $Q(n, 5, n)$, where $n$ is a constant.
1. Show that the lines $l_1$ and $l_2$ are not perpendicular. [3 marks]
2. Explain briefly why lines $l_1$ and $l_2$ cannot be parallel. [2 marks]
3. Given that $\theta$ is the acute angle between lines $l_1$ and $l_2$, show that $$\cos \theta = \frac{p}{\sqrt{34n^2 + qn + 306}}$$ where $p$ and $q$ are constants to be found. [3 marks]

Show mark scheme Show mark scheme source

Question 13:

Answer	Marks
13(a)	Rewrites in the general Cartesian form, or as a vector in terms of just

one parameter.

𝑙𝑙1

Or finds the position vector of a point on the line.

Or finds a direction vector of

Or writes three equations expressing x, y and z in terms of the parameter.

Answer	Marks	Guidance
1	3.1a	M1

= =

1 1.5 −1

3 1

𝐫𝐫 =�−1�+𝜆𝜆�1.5�

𝑙𝑙

Writes in a correct vector form.

𝑙𝑙1

Accept in place of

𝑥𝑥

Answer	Marks	Guidance
NMS ca�n𝑦𝑦 s�core 2/2 𝐫𝐫	1.1b	A1

Answer	Marks
13(b)(i)	𝑧𝑧

Calculates the scalar product of their direction vectors of lines and

Must not use a position vector as a direction vector.

Answer	Marks	Guidance
𝑙𝑙1 𝑙𝑙2	3.1a	M1

1 𝑛𝑛−3

�1.5�.� 3 �= 𝑛𝑛−3+4.5−𝑛𝑛 = 1.5

−1 𝑛𝑛

The scalar product is not zero

lines and are not perpendicular

∴ 𝑙𝑙1 𝑙𝑙2

Calculates a correct scalar product of a vector parallel to with a

1 �1.5�

vector parallel to −1

Answer	Marks	Guidance
𝑛𝑛−3	1.1b	A1

� 3 �

Explains that, as the scalar product is not zero, then the lines are not

𝑛𝑛

perpendicular.

Follow through their direction vectors if their scalar product is non-zero.

Must follow M1.

Answer	Marks	Guidance
NMS scores 0/3	3.2a	E1F
Q	Marking instructions	AO

Answer	Marks
13(b)(ii)	Explains that two vectors are parallel if one is a multiple

of the other. and need not be referred to explicitly.

Possibly implied by seen where and are their

𝑙𝑙1 𝑙𝑙2

direction vectors.

Answer	Marks	Guidance
𝒄𝒄 = 𝑘𝑘𝒅𝒅 𝒄𝒄 𝒅𝒅	2.4	E1F

2 𝑛𝑛−3

� th3en � � 3 an � d

−2 𝑛𝑛

𝑛𝑛−3 = 2 𝑛𝑛 = −2

but cannot be both and

and cannot be parallel

𝑛𝑛 5 −2

Demonstrates that, as n varies, the two direction vectors

Answer	Marks	Guidance
can never be a multiple of each other.	3.1a	B1

Answer	Marks
13(b)(iii)	Uses the scalar product to form an equation in and

𝑛𝑛

Follow through their direction vectors for lines and

Answer	Marks	Guidance
cos𝜃𝜃	1.1a	M1

2 𝑛𝑛−3

2 2 2 2 2 2

� 3 �.� 3 �= �2 +3 +2 ×�(𝑛𝑛−3) +3 +𝑛𝑛 ×cos𝜃𝜃

−2 𝑛𝑛

2 2(𝑛𝑛−3)+9−2𝑛𝑛 = √17×�2𝑛𝑛 −6𝑛𝑛+18×cos𝜃𝜃

3 𝑙𝑙1 𝑙𝑙2

Forms a correct equation in and

Or gives a correct expression for

𝑛𝑛 cos𝜃𝜃

Accept a correct equation, or expression, for the

cos𝜃𝜃

Answer	Marks	Guidance
supplementary angle.	1.1b	A1

Writes

3

Answer	Marks	Guidance
2	2.1	R1

cos𝜃𝜃 = √34𝑛𝑛 −102𝑛𝑛+306

Answer	Marks	Guidance
Total	10	cos𝜃𝜃 = 2

√34𝑛𝑛 −102𝑛𝑛+306

Answer	Marks	Guidance
Q	Marking instructions	AO

Question 13:
--- 13(a) ---
13(a) | Rewrites in the general Cartesian form, or as a vector in terms of just
one parameter.
𝑙𝑙1
Or finds the position vector of a point on the line.
Or finds a direction vector of
Or writes three equations expressing x, y and z in terms of the parameter.
1 | 3.1a | M1 | 𝑥𝑥−3 𝑦𝑦+1 𝑧𝑧−2
= =
1 1.5 −1
3 1
𝐫𝐫 =�−1�+𝜆𝜆�1.5�
𝑙𝑙
Writes in a correct vector form.
𝑙𝑙1
Accept in place of
𝑥𝑥
NMS ca�n𝑦𝑦 s�core 2/2 𝐫𝐫 | 1.1b | A1
--- 13(b)(i) ---
13(b)(i) | 𝑧𝑧
Calculates the scalar product of their direction vectors of lines and
Must not use a position vector as a direction vector.
𝑙𝑙1 𝑙𝑙2 | 3.1a | M1 | 2 −1
1 𝑛𝑛−3
�1.5�.� 3 �= 𝑛𝑛−3+4.5−𝑛𝑛 = 1.5
−1 𝑛𝑛
The scalar product is not zero
lines and are not perpendicular
∴ 𝑙𝑙1 𝑙𝑙2
Calculates a correct scalar product of a vector parallel to with a
1
�1.5�
vector parallel to −1
𝑛𝑛−3 | 1.1b | A1
� 3 �
Explains that, as the scalar product is not zero, then the lines are not
𝑛𝑛
perpendicular.
Follow through their direction vectors if their scalar product is non-zero.
Must follow M1.
NMS scores 0/3 | 3.2a | E1F
Q | Marking instructions | AO | Marks | Typical solution
--- 13(b)(ii) ---
13(b)(ii) | Explains that two vectors are parallel if one is a multiple
of the other. and need not be referred to explicitly.
Possibly implied by seen where and are their
𝑙𝑙1 𝑙𝑙2
direction vectors.
𝒄𝒄 = 𝑘𝑘𝒅𝒅 𝒄𝒄 𝒅𝒅 | 2.4 | E1F | If and are parallel
2 𝑛𝑛−3
� th3en � � 3 an � d
−2 𝑛𝑛
𝑛𝑛−3 = 2 𝑛𝑛 = −2
but cannot be both and
and cannot be parallel
𝑛𝑛 5 −2
Demonstrates that, as n varies, the two direction vectors
can never be a multiple of each other. | 3.1a | B1
--- 13(b)(iii) ---
13(b)(iii) | Uses the scalar product to form an equation in and
𝑛𝑛
Follow through their direction vectors for lines and
cos𝜃𝜃 | 1.1a | M1 | ∴ 𝑙𝑙1 𝑙𝑙2
2 𝑛𝑛−3
2 2 2 2 2 2
� 3 �.� 3 �= �2 +3 +2 ×�(𝑛𝑛−3) +3 +𝑛𝑛 ×cos𝜃𝜃
−2 𝑛𝑛
2
2(𝑛𝑛−3)+9−2𝑛𝑛 = √17×�2𝑛𝑛 −6𝑛𝑛+18×cos𝜃𝜃
3
𝑙𝑙1 𝑙𝑙2
Forms a correct equation in and
Or gives a correct expression for
𝑛𝑛 cos𝜃𝜃
Accept a correct equation, or expression, for the
cos𝜃𝜃
supplementary angle. | 1.1b | A1
Writes
3
2 | 2.1 | R1
cos𝜃𝜃 = √34𝑛𝑛 −102𝑛𝑛+306
Total | 10 | cos𝜃𝜃 = 2
√34𝑛𝑛 −102𝑛𝑛+306
Q | Marking instructions | AO | Marks | Typical solution

Show LaTeX source

Line $l_1$ has Cartesian equation
$$x - 3 = \frac{2y + 2}{3} = 2 - z$$

\begin{enumerate}[label=(\alph*)]
\item Write the equation of line $l_1$ in the form
$$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$$
where $\lambda$ is a parameter and $\mathbf{a}$ and $\mathbf{b}$ are vectors to be found. [2 marks]

\item Line $l_2$ passes through the points $P(3, 2, 0)$ and $Q(n, 5, n)$, where $n$ is a constant.

\begin{enumerate}[label=(\roman*)]
\item Show that the lines $l_1$ and $l_2$ are not perpendicular. [3 marks]

\item Explain briefly why lines $l_1$ and $l_2$ cannot be parallel. [2 marks]

\item Given that $\theta$ is the acute angle between lines $l_1$ and $l_2$, show that
$$\cos \theta = \frac{p}{\sqrt{34n^2 + qn + 306}}$$
where $p$ and $q$ are constants to be found. [3 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2019 Q13 [10]}}

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Q1 1 Q2 1 Q3 1 Q4 2 Q5 8 Q6 5 Q7 5 Q8 7 Q9 7 Q10 6 Q11 8 Q12 12 Q13 10 Q14 7