Mini-Lesson

Vectors

2 missions. Guess, learn, then check. Your place is saved.

Mission 1A plane heads north at 200 km/h in a 50 km/h east wind. Where to?
Mission 2A drone hovers at (3, 4, 12) m from its launch point. How far away?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

A plane heads north at 200 km/h in a 50 km/h east wind. Where to?

Make a guess: A plane heads north at 200 km/h in a 50 km/h east wind. Where to?

Vectors · two dimensions

What a vector is

A vector has size (magnitude) and direction; a scalar has size only. Velocity is a vector; speed is a scalar.
Write a vector as a column vector $(32)$ ( 3 ; 2 ) or with unit vectors $3 i + 2 j$ 3i + 2j, where $i$ i is one unit in the $x$ x-direction and $j$ j one unit in the $y$ y-direction.
Print uses bold $a$ a; by hand, underline it: $a$ a.
The movement from $A$ A to $B$ B is $A B$ AB→. In the diagram, $A B = 3 i + 2 j$ AB→ = 3i + 2j.

Vectors · two dimensions

Magnitude and direction

The magnitude of $a = x i + y j$ a = xi + yj is its length, by Pythagoras:

∣ a ∣ = x^{2} + y^{2}

Its direction comes from $tan θ = \frac{y}{x}$ tan θ = y/x. Sketch the vector first so you put the angle in the right quadrant.
Give the direction as an angle with $i$ i (or a bearing in context), saying which way it is measured.
Example: $3 i + 4 j$ 3i + 4j has magnitude 5 and makes $53. 1^{\circ}$ 53.1° with $i$ i.

Worked example Vectors · two dimensions

Magnitude and direction of a vector

Question

Find the magnitude and direction of

a = 5 i - 12 j

.

1
Magnitude by Pythagoras: $∣ a ∣ = 5^{2} + (- 12)^{2} = 169 = 13$ |a| = √(5² + (−12)²) = √169 = 13
2
Sketch: 5 right and 12 down, so $a$ a points below the positive $x$ x-axis
3
Angle below the axis: $tan^{- 1} \frac{12}{5} = 67. 4^{\circ}$ tan⁻¹12/5 = 67.4°
4
Measured anticlockwise from the positive $x$ x-axis: $36 0^{\circ} - 67. 4^{\circ} = 292. 6^{\circ}$ 360° − 67.4° = 292.6°

Answer

∣ a ∣ = 13

; direction

67. 4^{\circ}

below

i

(that is,

292. 6^{\circ}

anticlockwise from the positive

x

-axis)

⚠️ Watch out

a calculator gives

tan^{- 1} (- \frac{12}{5}) = - 67. 4^{\circ}

for any vector with this ratio, including

- 5 i + 12 j

. Only the sketch tells you the quadrant.

Calculate

?A force of 20 N acts at

3 0^{\circ}

above the direction of

i

. Write it as

a i + b j

and give

a

to 3 significant figures.

Vectors · two dimensions

Adding vectors and position vectors

Add or subtract component by component; a scalar multiplies every component. Geometrically, vectors add tip to tail (triangle law).
$a$ a and $b$ b are parallel if $b = k a$ b = ka for some scalar $k$ k.
The position vector of $A$ A is $O A = a$ OA→ = a, from the origin $O$ O.

A B = b - a

The distance $A B$ AB is the magnitude $∣ b - a ∣$ |b − a|.

Worked example Vectors · two dimensions

Distance and the fourth corner of a parallelogram

Question

A

,

B

and

C

have position vectors

2 i - j

,

7 i + 11 j

and

10 i + 4 j

. Find the distance

A B

, and the position vector of

D

such that

A BC D

is a parallelogram.

1
$A B = b - a = 5 i + 12 j$ AB→ = b − a = 5i + 12j, so $A B = 5^{2} + 1 2^{2} = 13$ AB = √(5² + 12²) = 13
2
In parallelogram $A BC D$ ABCD, $A D = BC$ AD→ = BC→
3
$BC = c - b = 3 i - 7 j$ BC→ = c − b = 3i − 7j
4
$d = a + BC = (2 i - j) + (3 i - 7 j)$ d = a + BC→ = (2i − j) + (3i − 7j)

Answer

A B = 13

and

d = 5 i - 8 j

⚠️ Watch out

the letters go round in order.

A D

equals

BC

, not

CB

.

Quick check

?Points

A

and

B

have position vectors

a = 3 i - 2 j

and

b = - i + 4 j

. Find

A B

.

Vectors · two dimensions

Unit vectors and vectors in mechanics

A unit vector has magnitude 1. The unit vector in the direction of $a$ a is $a^= \frac{a}{∣ a ∣}$ â = a/(|a|). For $6 i + 8 j$ 6i + 8j it is $0.6 i + 0.8 j$ 0.6i + 0.8j.
Forces, velocities and displacements are vectors. The resultant of several forces is their vector sum.
Two velocities combine the same way: a plane's velocity through the air plus the wind's velocity gives its velocity over the ground.
In a bearing problem, take $i$ i as east and $j$ j as north.

Calculate

?Two forces

F_{1} = (2 i + 3 j)

N and

F_{2} = (p i - 5 j)

N act on a particle. Their resultant is parallel to

3 i - j

. Find the value of

p

.

Mission check

?A plane heads north at 200 km/h in a 50 km/h wind blowing towards the east. How does it actually move?

Mission 2

A drone hovers at (3, 4, 12) m from its launch point. How far away?

Make a guess: A drone hovers at (3, 4, 12) m from its launch point. How far away?

Vectors · three dimensions

Vectors in three dimensions (A-level only)

Add a third axis, $z$ z, at right angles to the $x$ x- and $y$ y-axes. Its unit vector is $k$ k.
Write $x i + y j + z k$ xi + yj + zk or $x y z$ ( x ; y ; z ).
Pythagoras twice: the floor diagonal is $x^{2} + y^{2}$ √(x² + y²), then add the height:

∣ a ∣ = x^{2} + y^{2} + z^{2}

Example: $∣2 i + 3 j + 6 k ∣ = 4 + 9 + 36 = 7$ |2i + 3j + 6k| = √(4 + 9 + 36) = 7.
Position vectors, $A B = b - a$ AB→ = b − a and distance $∣ b - a ∣$ |b − a| work exactly as in 2-D.

Worked example Vectors · three dimensions

Distance and unit vector

Question

A

and

B

have coordinates

(2, 1, - 3)

and

(4, - 2, 3)

. Find the distance

A B

and a unit vector in the direction of

A B

.

1
End minus start: $A B = b - a = 2 i - 3 j + 6 k$ AB→ = b − a = 2i − 3j + 6k
2
Magnitude: $A B = 2^{2} + (- 3)^{2} + 6^{2} = 49 = 7$ AB = √(2² + (−3)² + 6²) = √49 = 7
3
Divide by the magnitude: $\frac{1}{7} (2 i - 3 j + 6 k)$ 1/7(2i − 3j + 6k)

Answer

A B = 7

; unit vector

\frac{2}{7} i - \frac{3}{7} j + \frac{6}{7} k

⚠️ Watch out

square each component including its sign:

(- 3)^{2} = 9

, never

- 9

.

Calculate

?Find the distance between the points

P (1, 5, - 2)

and

Q (4, - 1, 3)

. Give your answer to 3 significant figures.

Vectors · three dimensions

Parallel vectors and geometry in 3-D

$b$ b is parallel to $a$ a if $b = k a$ b = ka: every component is multiplied by the same $k$ k.
$A$ A, $B$ B and $C$ C are collinear if $A C = k A B$ AC→ = kAB→: parallel, and sharing the point $A$ A.
In parallelogram $A BC D$ ABCD the letters go round in order, so $A D = BC$ AD→ = BC→.
The midpoint of $A B$ AB has position vector $\frac{1}{2} (a + b)$ 1/2(a + b).

Worked example Vectors · three dimensions

Showing that three points are collinear

Question

Show that

A (1, 2, - 1)

,

B (3, 5, 1)

and

C (7, 11, 5)

lie on a straight line, and find the ratio

A B : BC

.

1
$A B = b - a = 2 i + 3 j + 2 k$ AB→ = b − a = 2i + 3j + 2k
2
$A C = c - a = 6 i + 9 j + 6 k = 3 A B$ AC→ = c − a = 6i + 9j + 6k = 3AB→
3
So $A C$ AC→ is parallel to $A B$ AB→, and both pass through $A$ A: the points are collinear
4
$BC = 4 i + 6 j + 4 k = 2 A B$ BC→ = 4i + 6j + 4k = 2AB→

Answer

collinear, with

A B : BC = 1 : 2

⚠️ Watch out

parallel alone is not enough. You must say the vectors share a point; parallel lines can be separate.

Quick check

?

A BC D

is a parallelogram with

A (1, 0, 2)

,

B (4, 3, - 1)

and

C (6, - 2, 0)

. Find the coordinates of

D

.

Vectors · three dimensions

Direction in 3-D: the angle with an axis

The angle $θ_{x}$ θ_x between $a = x i + y j + z k$ a = xi + yj + zk and the positive $x$ x-axis satisfies

cos θ_{x} = \frac{x}{∣ a ∣}

Likewise $cos θ_{y} = \frac{y}{∣ a ∣}$ cos θ_y = y/(|a|) and $cos θ_{z} = \frac{z}{∣ a ∣}$ cos θ_z = z/(|a|).
A negative component gives an obtuse angle: that is correct, not a slip.

Calculate

?Find the angle between the vector

4 i - 4 j + 7 k

and the positive

x

-axis. Give your answer in degrees to 1 decimal place.

Calculate

?Relative to a launch point

O

, with

k

vertically upwards and distances in metres, a drone flies in a straight line at a constant 2.5 m/s from

A (2, - 3, 5)

to

B (14, 1, 8)

. How long does the flight take?

s

Mission check

?A drone hovers at

3 i + 4 j + 12 k

m from its launch point. How far away is it in a straight line?

Recap

Big ideas

A 2-D vector is $x i + y j$ xi + yj or $(x y)$ ( x ; y ); its magnitude is $x^{2} + y^{2}$ √(x² + y²) and its direction comes from $tan θ = \frac{y}{x}$ tan θ = y/x with a sketch.
Add vectors component by component (tip to tail); $A B = b - a$ AB→ = b − a and the distance $A B = ∣ b - a ∣$ AB = |b − a|.
Forces and velocities are vectors: their resultant is the vector sum.
In 3-D, $a = x i + y j + z k$ a = xi + yj + zk and $∣ a ∣ = x^{2} + y^{2} + z^{2}$ |a| = √(x² + y² + z²).
$A B = b - a$ AB→ = b − a; distance is $∣ b - a ∣$ |b − a|; parallel means a scalar multiple, and collinear means parallel through a shared point.
The angle with the $x$ x-axis has $cos θ = \frac{x}{∣ a ∣}$ cos θ = x/(|a|).

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Vectors for AQA A-level Mathematics (7357). 🎉

Mission 1A plane heads north at 200 km/h in a 50 km/h east wind. Where to?
Mission 2A drone hovers at (3, 4, 12) m from its launch point. How far away?

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Specification points and sources for this lesson are listed at the bottom of the page.