What is 10.2: Magnitude and direction?

For v=(a,b), its magnitude is |v|=√(a^2+b^2) and a direction angle must be stated relative to a specified axis or bearing convention.

How many marks are 10.2 questions worth?

10.2 practice drills on this page are worth 2–5 marks each; marks in a real exam vary by question.

Which past exam papers has 10.2 appeared in?

We have not yet indexed a verified past-paper appearance for 10.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

10.2

Calculate the magnitude and direction of a vector and convert between component form and magnitude/direction form.

Draft — not yet indexed

Magnitude and direction

Worked answers and methods for 10.2 on Edexcel A-level Maths 9MA0.

Explanation

For $\mathbf{v}=(a,b)$ , its magnitude is $|\mathbf{v}|=\sqrt{a^2+b^2}$ and a direction angle must be stated relative to a specified axis or bearing convention.
A vector of magnitude $r$ at angle $\theta$ anticlockwise from the positive $x$ -axis has components $(r\cos\theta,r\sin\theta)$ ; use signs or a quadrant-aware angle calculation when reversing the process.
For example, magnitude $10$ at $30^\circ$ above the positive $x$ -axis gives $(10\cos30^\circ,10\sin30^\circ)=(5\sqrt3,5)$ .
The value from $\tan^{-1}(b/a)$ alone can select the wrong quadrant; a common error is to report an acute reference angle without checking the component signs.
A unit vector in the direction of non-zero $\mathbf a$ is $\mathbf a/|\mathbf a|$ ; divide every component by the vector's magnitude.

Worked example

A vector has magnitude $14$ and direction $120^\circ$ anticlockwise from the positive $x$ -axis. Write it in exact component form.

1.Use $(14\cos120^\circ,14\sin120^\circ)$ .
2.Since $\cos120^\circ=-\frac12$ and $\sin120^\circ=\frac{\sqrt3}{2}$ , the vector is $(-7,7\sqrt3)$ .

Answer: $(-7,7\sqrt3)$

Common mistakes

Don't normalise a vector by dividing by the sum of its components rather than by its magnitude.
Don't use sine for the horizontal component and cosine for the vertical component despite the stated reference axis.

Exam tip

Draw the direction from the positive x-axis, then use cosine horizontally and sine vertically with correct signs.

Worked practice

Tier 1 · Easy

Find the magnitude and direction of the vector

(3,4)

, giving the direction anticlockwise from the positive

x

-axis to

1

decimal place.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Magnitude $5$ ; direction $53.1^\circ$	3
Notes
The magnitude is $\sqrt{3^2+4^2}=5$ . Both components are positive, so the vector is in the first quadrant and $\theta=\tan^{-1}(4/3)=53.1^\circ$ .

(3 marks)

Tier 2 · Standard

Find the magnitude and direction of the vector

(-6,8)

, measuring the direction anticlockwise from the positive

x

-axis and giving the direction to

3

significant figures.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Magnitude $10$ Direction $127^\circ$ to $3$ significant figures	3
Notes
The magnitude is $\sqrt{(-6)^2+8^2}=\sqrt{100}=10$ . The vector lies in quadrant II. Its reference angle is $\arctan(8/6)=53.130\ldots^\circ$ , so its direction is $180^\circ-53.130\ldots^\circ=126.870\ldots^\circ$ .

(3 marks)

Tier 3 · Hard

A two-dimensional vector has magnitude

13

, and the cosine of the angle it makes with the positive

x

-axis is

\frac5{13}

. Find all possible component forms and explain the ambiguity.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$(5,12)$ or $(5,-12)$	4
Notes
The horizontal component is $13(5/13)=5$ . If the vertical component is $y$ , then $5^2+y^2=13^2$ , so $y^2=144$ and $y=\pm12$ . Cosine fixes the positive horizontal component but does not distinguish an angle above the axis from its reflection below the axis.

(4 marks)

Tier 1 · Easy

Find an exact unit vector in the direction of

(-8,6)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\left(-\dfrac45,\dfrac35\right)$	2
Notes
The magnitude of $(-8,6)$ is $\sqrt{64+36}=10$ . Dividing both components by $10$ gives the unit vector $(-8/10,6/10)=(-4/5,3/5)$ .

(2 marks)

Tier 2 · Standard

A displacement has magnitude

12

and bearing

150^\circ

, where bearings are measured clockwise from north. Taking east and north as the positive component directions, express the displacement in exact component form.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$(6,-6\sqrt3)$	3
Notes
For a bearing measured from north, the east component is $12\sin150^\circ=6$ and the north component is $12\cos150^\circ=-6\sqrt3$ . Hence the component form, in east-north order, is $(6,-6\sqrt3)$ .

(3 marks)

Tier 3 · Hard

The vector

\mathbf{v}=(k,k-7)

has magnitude

\sqrt{85}

. Its direction angle, measured anticlockwise from the positive

x

-axis, lies between

180^\circ

and

270^\circ

. Find

k

and the direction angle of

\mathbf{v}

, giving the angle to

1

decimal place.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$k=-2$ and the direction angle is $257.5^\circ$	5
Notes
$k^2+(k-7)^2=85$ gives $2k^2-14k-36=0$ , so $(k-9)(k+2)=0$ and $k=9$ or $k=-2$ . The stated third-quadrant direction selects $k=-2$ , giving $\mathbf{v}=(-2,-9)$ . Its reference angle is $\tan^{-1}(9/2)=77.471\ldots^\circ$ , so the direction is $180^\circ+77.471\ldots^\circ=257.5^\circ$ to $1$ decimal place.

(5 marks)

Tier 2 · Standard

A vector has magnitude

10

. Its horizontal component is twice its vertical component, and both components are negative. Find the exact component form and the direction angle anticlockwise from the positive

x

-axis, giving the angle to

1

decimal place.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	The vector is $(-4\sqrt5,-2\sqrt5)$ and its direction angle is $206.6^\circ$ .	4
Notes
Write the components as $(-2k,-k)$ with $k>0$ . The magnitude condition gives $\sqrt{4k^2+k^2}=k\sqrt5=10$ , so $k=2\sqrt5$ and the vector is $(-4\sqrt5,-2\sqrt5)$ . It lies in the third quadrant and has reference angle $\tan^{-1}(1/2)=26.565\ldots^\circ$ , so its direction is $180^\circ+26.565\ldots^\circ=206.6^\circ$ .

(4 marks)

Tier 3 · Hard

Vector

\mathbf{a}

has magnitude

8

and direction

30^\circ

anticlockwise from the positive

x

-axis. Vector

\mathbf{b}

has magnitude

6

and direction

150^\circ

anticlockwise from the positive

x

-axis. Find the exact component form of

\mathbf{a}+\mathbf{b}

, its exact magnitude and its direction angle to

1

decimal place.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$\mathbf{a}+\mathbf{b}=(\sqrt3,7)$ , with magnitude $2\sqrt{13}$ and direction $76.1^\circ$ .	5
Notes
$\mathbf{a}=(8\cos30^\circ,8\sin30^\circ)=(4\sqrt3,4)$ and $\mathbf{b}=(6\cos150^\circ,6\sin150^\circ)=(-3\sqrt3,3)$ . Hence $\mathbf{a}+\mathbf{b}=(\sqrt3,7)$ . Its magnitude is $\sqrt{3+49}=2\sqrt{13}$ . Both components are positive, so its direction is $\tan^{-1}(7/\sqrt3)=76.102\ldots^\circ=76.1^\circ$ to $1$ decimal place.

(5 marks)

Tier 3 · Hard

The vector

\mathbf{a}=(3,4)

. A vector

\mathbf{b}

has magnitude

5

, and

\mathbf{a}+\mathbf{b}

points along the positive

x

-axis. Find

\mathbf{b}

and the magnitude of the resultant. Give the direction angle of

\mathbf{b}

1

decimal place, measured anticlockwise from the positive

x

-axis.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\mathbf{b}=(3,-4)$ , $\|\mathbf{a}+\mathbf{b}\|=6$ , and the direction angle of $\mathbf{b}$ is $306.9^\circ$ .	5
Notes
For the resultant to be horizontal, the vertical component of $\mathbf{b}$ must be $-4$ . Write $\mathbf{b}=(x,-4)$ . Since $\|\mathbf{b}\|=5$ , $x^2+16=25$ , so $x=\pm3$ . If $x=-3$ , the resultant is the zero vector and does not point along the positive $x$ -axis. Therefore $\mathbf{b}=(3,-4)$ and $\mathbf{a}+\mathbf{b}=(6,0)$ , whose magnitude is $6$ . The direction of $\mathbf{b}$ is $360^\circ-\tan^{-1}(4/3)=306.869\ldots^\circ$ , giving $306.9^\circ$ .

(5 marks)

Q10

Tier 3 · Hard

10.

The vectors

\mathbf{a}=(\sqrt3,1)

and

\mathbf{b}=(-1,\sqrt3)

have direction angles measured anticlockwise from the positive

x

-axis. A vector

\mathbf{v}

has magnitude

8

and points along the internal bisector of the smaller angle between the directions of

\mathbf{a}

and

\mathbf{b}

. Find the direction angle of

\mathbf{v}

and its exact component form.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	The direction angle of $\mathbf{v}$ is $75^\circ$ , and $\mathbf{v}=(2\sqrt6-2\sqrt2,2\sqrt6+2\sqrt2)$ .	5
Notes
The direction angles of $\mathbf{a}$ and $\mathbf{b}$ are $30^\circ$ and $120^\circ$ , respectively. Their smaller angle is $90^\circ$ , so its internal bisector has direction angle $(30^\circ+120^\circ)/2=75^\circ$ . Hence $\mathbf{v}=(8\cos75^\circ,8\sin75^\circ)$ . Using $\cos75^\circ=(\sqrt6-\sqrt2)/4$ and $\sin75^\circ=(\sqrt6+\sqrt2)/4$ gives $\mathbf{v}=(2\sqrt6-2\sqrt2,2\sqrt6+2\sqrt2)$ .

(5 marks)

Verified exam appearances

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