Academic Skills Advice
Vectors Summary
A vector includes magnitude (size) and direction.
Types of vectors:
can “slide” along the line of action.
not restricted, defined by magnitude & direction but can be anywhere.
̅̅̅̅ = 4𝑖̂ − 5𝑗̂,
one end is fixed and it usually starts at the origin (e.g. 𝑂𝐵
means start at the origin and go along 4 and down 5).
The vector with a length of 1 (sometimes called the normalised vector).
𝑖̂ is a unit vector in the 𝑥-direction
𝑗̂ is a unit vector in the 𝑦-direction
𝑘̂ is a unit vector in the 𝑘-direction
Line vector:
Free vector:
Position vector:
Unit vector (𝑛̂):
Addition of Vectors:
When adding vectors you can draw them as a chain with the 2nd vector starting where the 1st
one ended.
e.g.
𝐵
if ⃗⃗⃗⃗⃗
𝐴𝐵 is the vector
𝐴
𝐵
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗
Adding them gives: ⃗⃗⃗⃗⃗
𝐴𝐵 + 𝐵𝐶
𝐴𝐶
⃗⃗⃗⃗⃗ is the vector
and 𝐵𝐶
𝐵
𝐴
𝐶
𝐶
The resultant vector
When you are given the components you can just add the 𝑖̂′𝑠 and add the 𝑗̂′𝑠:
e.g. add the vectors: 𝒗 = 𝟐𝒊̂ + 𝟒𝒋̂ and 𝒖 = 𝟓𝒊̂ + 𝟐𝒋̂
𝑣 + 𝑢 = 7𝑖̂ + 6𝑗̂
Magnitude of Vectors:
To find the magnitude of vector 𝑎 (𝑑𝑒𝑛𝑜𝑡𝑒𝑑 |𝒂|) you would use Pythagoras.
̂ , find |𝒂|
e.g. if 𝒂 = 𝟔𝒊̂ − 𝟑𝒋̂ + 𝟐𝒌
|𝑎| = √62 + (−3)2 + 22 = √49 = 7
Finding the unit Vector:
̂ ) has a length of 1. So to find the unit vector (i.e. to make the length 1) we
The unit vector (𝒏
𝒏
̂=
need to divide each component of the vector by the original length, i.e. 𝒏
|𝒏|
̂ ) of the vector 𝒃 = 𝟐𝒊̂ − 𝟑𝒋̂ + 𝒛̂
e.g. find the unit vector (𝒃
|𝑏| = √22 + (−3)2 + 12 = √14
(2,−3,1)
𝑏
𝑏̂ = |𝑏| =
√14
Therefore:
𝑏̂ =
2
√14
𝑖̂ −
3
√14
𝑗̂ +
1
√14
𝑘̂
(and |𝑏̂| = 1)
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Direction cosines:
The direction cosines are the cosines of the angles between the vector and each of the 3
axes.
For example:
𝑙 = 𝑐𝑜𝑠(𝛼) where 𝛼 is the angle between the vector and the 𝑥-axis,
𝑚 = 𝑐𝑜𝑠(𝛽) where 𝛽 is the angle between the vector and the 𝑦-axis
and
𝑛 = 𝑐𝑜𝑠(𝛾) where 𝛾 is the angle between the vector and the 𝑧-axis.
i.e. If 𝑣 = 𝑎𝑖̂ + 𝑏𝑗̂ + 𝑐𝑘̂
The direction cosines (𝑙, 𝑚 𝑎𝑛𝑑 𝑛) are:
𝑎
𝑏
𝑙 = |𝑣|
𝑚 = |𝑣|
𝑐
𝑛 = |𝑣|
(Remember that |𝑣| can be found by using Pythagoras: |𝑣| = √𝑎2 + 𝑏 2 + 𝑐 2)
e.g.
̂
let 𝒂 = 𝟑𝒊̂ − 𝟐𝒋̂ + 𝟔𝒌
|𝑎| = √32 + (−2)2 + 62 = 7
𝟑
the direction cosines (𝑙, 𝑚 & 𝑛) are:
𝒍 = 𝟕,
𝒎=
−𝟐
,
𝟕
𝟔
𝒏=𝟕
3
(We have found that 𝑐𝑜𝑠(𝛼) = 7 and so the angle between the vector and the 𝑥-axis is:
3
𝛼 = 𝑐𝑜𝑠 −1 (7) = 64.60 . We can find the other angles in the same way)
Resolving Vectors:
We can use basic trigonometry to resolve a vector into its 𝑥 and 𝑦 components. Remember
that if you go:
through the angle you use cos
away from the angle you use sin
(because we are considering hypotenuse and adjacent)
(because we are considering hypotenuse and opposite)
e.g. Vector 𝑎 has an angle of 40o and a length of 8.
𝑦
𝑥 component:
𝑎𝑥 = 8𝑐𝑜𝑠40 = 6.13
8
𝑎𝑦
40
𝑎𝑥
𝑥
𝑦 component:
𝑎𝑦 = 8𝑠𝑖𝑛40 = 5.14
Vector Multiplication:
There are 2 types of vector multiplication – the dot (scalar) product and the cross (vector)
product. See the following pages for explanations.
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Dot (Scalar) Product:
The dot product of the vectors 𝑎 and 𝑏 is written as
𝒂. 𝒃
(a dot b)
The result of the dot product is a scalar: i.e. 𝒗𝒆𝒄𝒕𝒐𝒓. 𝒗𝒆𝒄𝒕𝒐𝒓 = 𝒔𝒄𝒂𝒍𝒂𝒓
For two vectors 𝑎 and 𝑏 the dot product is calculated by multiplying the coefficients of 𝒊
PLUS multiplying the coefficients of 𝒋 PLUS multiplying the coefficients of 𝒌 (you
could think of it as: (𝑖 × 𝑖) + (𝑗 × 𝑗) + (𝑘 × 𝑘))
e.g.
̂
𝒂 = 𝟐𝒊̂ + 𝟑𝒋̂ + 𝟓𝒌
̂
𝒃 = 𝟒𝒊̂ + 𝒋̂ + 𝟔𝒌
(𝑖′𝑠)
(𝑗′𝑠)
(𝑘′𝑠)
𝑎. 𝑏 = (2 × 4) + (3 × 1) + (5 × 6) = 𝟒𝟏
We have found that the dot product of the vectors 𝑎 and 𝑏 = 41
Finding the angle Between 2 vectors:
The dot product can also be used to find the angle between two vectors using the following
equation:
𝒂. 𝒃 = |𝒂||𝒃|𝒄𝒐𝒔𝜽
We already know that 𝑎. 𝑏 = 41 (above) and we can work out the right hand side as follows:
|𝑎||𝑏| = √22 + 32 + 52 × √42 + 12 + 62 = √38√53
Now we can use the equation to find the angle between the 2 vectors:
𝒂. 𝒃 = |𝒂||𝒃|𝒄𝒐𝒔𝜽
41 = √38√53 𝑐𝑜𝑠𝜃
41
𝜃 = 𝑐𝑜𝑠 −1 (
)
𝜃 = 24𝑜
Parallel or perpendicular?
If 𝒂. 𝒃 = 𝟎
then 𝑐𝑜𝑠𝜃 = 0
If 𝒂. 𝒃 = 𝒂𝒃 then 𝑐𝑜𝑠𝜃 = 1
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∴ 𝜃 = 90𝑜
∴ 𝜃 = 0𝑜
√38√53
perpendicular
parallel
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Cross (Vector) Product:
The cross product of the vectors 𝑎 and 𝑏 is written as
𝒂×𝒃
(a cross b)
The result of the cross product is a vector: i.e. 𝒗𝒆𝒄𝒕𝒐𝒓 × 𝒗𝒆𝒄𝒕𝒐𝒓 = 𝒗𝒆𝒄𝒕𝒐𝒓
For two vectors 𝑎 = 𝑎1 𝑖̂ + 𝑎2 𝑗̂ + 𝑎3 𝑘̂ and 𝑏 = 𝑏1 𝑖̂ + 𝑏2 𝑗̂ + 𝑏3 𝑘̂ the cross product is calculated
as follows:
̂
𝒂 × 𝒃 = (𝒂𝟐 𝒃𝟑 − 𝒂𝟑 𝒃𝟐 )𝒊̂ − (𝒂𝟏 𝒃𝟑 − 𝒂𝟑 𝒃𝟏 )𝒋̂ + (𝒂𝟏 𝒃𝟐 − 𝒂𝟐 𝒃𝟏 )𝒌
This looks like a complicated formula to remember but there is an easy way to do the cross
product using a matrix.
Set up a matrix with 𝑎 above 𝑏.
𝑖 𝑗 𝑘
𝑎 × 𝑏 = |2 4 3 |
1 5 −2
̂
e.g. 𝒂 = 𝟐𝒊̂ + 𝟒𝒋̂ + 𝟑𝒌
̂
𝒃 = 𝒊̂ + 𝟓𝒋̂ − 𝟐𝒌
To find the 𝑖 component, cover up everything in the 𝑖 row & column, and find the determinant
of what’s left (the minor). Do the same for 𝑗 and 𝑘 and apply the alternate signs (see matrix
summary sheet for more help). If you’re not sure about (or just don’t like) matrices – see the
final page for an easy way to remember, and write out, the formula.
4 3
2
𝑎 × 𝑏 = +|
| 𝑖̂ − |
5 −2
1
2 4 ̂
3
| 𝑗̂ + |
|𝑘
1 5
−2
= +((4 × −2) − (3 × 5))𝑖̂ − ((2 × −2) − (3 × 1))𝑗̂ + ((2 × 5) − (4 × 1))𝑘̂
̂
∴ 𝒂 × 𝒃 = −𝟐𝟑𝒊̂ + 𝟕𝒋̂ + 𝟔𝒌
(You have found the cross product)
Finding the angle Between 2 vectors:
The cross product can also be used to find the angle between two vectors using the
following equation:
|𝒂 × 𝒃| = |𝒂||𝒃|𝒔𝒊𝒏𝜽
So far we have found that 𝑎 × 𝑏 = −23𝑖̂ + 7𝑗̂ + 6𝑘̂
Use this to find the modulus (to complete the left hand side):
|𝑎 × 𝑏| = √(−23)2 + 72 + 62 = √614
Right hand side:
(𝟐𝟒. 𝟖)
|𝑎||𝑏| = √22 + 42 + 32 × √12 + 52 + (−2)2 = √29√30
Now we can find the angle between 𝑎 and 𝑏 using the cross product equation:
|𝒂 × 𝒃| = |𝒂||𝒃|𝒔𝒊𝒏𝜽
√614 = √29√30 𝑠𝑖𝑛𝜃
𝜃 = 𝑠𝑖𝑛−1 (
𝑜
√614
√29√30
)
𝜃 = 57.1
Parallel or perpendicular?
If 𝒂 × 𝒃 = 𝟎 then 𝑠𝑖𝑛𝜃 = 0
If 𝒂 × 𝒃 = 𝒂𝒃 then 𝑠𝑖𝑛𝜃 = 1
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∴ 𝜃 = 0𝑜
∴ 𝜃 = 90𝑜
parallel
perpendicular
4
Right Hand Rule:
The answer to the cross product (𝑎 × 𝑏) is a vector that acts perpendicular to both of the
original vectors (𝑎 𝑎𝑛𝑑 𝑏) and has magnitude |𝑎||𝑏|𝑠𝑖𝑛𝜃.
Useful information:
𝒃 × 𝒂 = − (𝒂 × 𝒃)
𝒂 × 𝒂 =𝟎
𝒃 × 𝒃 =𝟎
For RH rule:
̂×𝒌
̂=𝟎
𝒊̂ × 𝒊̂ = 𝒋̂ × 𝒋̂ = 𝒌
̂
̂
𝒊̂ × 𝒋̂ = 𝒌
𝒋̂ × 𝒊̂ = −𝒌
̂ = 𝒊̂
̂ × 𝒋̂ = −𝒊̂
𝒋̂ × 𝒌
𝒌
̂ × 𝒊̂ = 𝒋̂
̂ = −𝒋̂
𝒌
𝒊̂ × 𝒌
You could remember the RH rule as follows:
If the 2 numbers you are multiplying are:
In alphabetical order (going round in a circle: 𝑖, 𝑗, 𝑘, 𝑖, 𝑗, 𝑘 𝑒𝑡𝑐), answer is:
Not in alphabetical order, answer is:
-other letter.
other letter.
The reason it is called the Right Hand Rule isn’t covered here – but you can always search
for it if you’re interested.
Another way to find the cross product formula:
If you don’t like using Matrices the following is a way to remember the formula:
𝑎 × 𝑏 = (𝑎2 𝑏3 − 𝑎3 𝑏2 )𝑖̂ − (𝑎1 𝑏3 − 𝑎3 𝑏1 )𝑗̂ + (𝑎1 𝑏2 − 𝑎2 𝑏1 )𝑘̂
Step 1:
Write the formula out but without the subscript numbers:
𝑎 × 𝑏 = (𝑎 𝑏 − 𝑎 𝑏)𝑖̂ − (𝑎 𝑏 − 𝑎 𝑏)𝑗̂ + (𝑎 𝑏 − 𝑎 𝑏)𝑘̂
(Remember the alternate signs on the brackets)
Now we just need to fill in the subscript numbers with the 𝑎′𝑠 and 𝑏′𝑠
Step 2:
Decide on the numbers that go with the 𝒂’𝒔
Bracket 1, miss out 1, so write 2 then 3
Bracket 2, miss out 2, so write 1 then 3
Bracket 3, miss out 3, so write 1 then 2
The numbers with the 𝑎’𝑠:
Step 3:
𝑎 × 𝑏 = (𝑎𝟐 𝑏 − 𝑎𝟑 𝑏)𝑖̂ − (𝑎𝟏 𝑏 − 𝑎𝟑 𝑏)𝑗̂ + (𝑎𝟏 𝑏 − 𝑎𝟐 𝑏)𝑘̂
Decide on the numbers that go with the 𝒃’𝒔
In each bracket just use the opposite numbers from what you used for the 𝑎’𝑠
The numbers with the 𝑏’𝑠:
𝑎 × 𝑏 = (𝑎2 𝑏𝟑 − 𝑎3 𝑏𝟐 )𝑖̂ − (𝑎1 𝑏𝟑 − 𝑎3 𝑏𝟏 )𝑗̂ + (𝑎1 𝑏𝟐 − 𝑎2 𝑏𝟏 )𝑘̂
We have now written the complete formula:
𝑎 × 𝑏 = (𝑎2 𝑏3 − 𝑎3 𝑏2 )𝑖̂ − (𝑎1 𝑏3 − 𝑎3 𝑏1 )𝑗̂ + (𝑎1 𝑏2 − 𝑎2 𝑏1 )𝑘̂
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