Academic Skills Advice
Vectors Summary
A vector includes magnitude (size) and direction.
Types of vectors:
Line vector:
Free vector:
Position vector:
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).
Unit vector (𝑛̂):
Addition of Vectors:
When adding vectors you should draw them as a chain with the 2nd vector starting where the
1st one ended.
e.g.
𝐵
⃗⃗⃗⃗⃗ is the vector
if 𝐴𝐵
𝐴
𝐵
⃗⃗⃗⃗⃗ is the vector
and 𝐵𝐶
𝐶
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗
Then adding them gives: ⃗⃗⃗⃗⃗
𝐴𝐵 + 𝐵𝐶
𝐴𝐶
𝐵
𝐴
𝐶
The resultant vector
When you are given the components you can just add the 𝑖̂′𝑠 and add the 𝑗̂′𝑠:
e.g.
𝒛𝟏 = 𝟐𝒊̂ + 𝟒𝒋̂ and
add the vectors:
𝒛𝟐 = 𝟓𝒊̂ + 𝟐𝒋̂
𝑧1 + 𝑧2 = 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
̂=
need to divide each component of the vector by the original length, i.e. 𝒏
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 −
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3
√14
𝑗̂ +
1
√14
𝑘̂
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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.
e.g. If 𝑣 = 𝑎𝑖̂ + 𝑏𝑗̂ + 𝑐𝑘̂
The direction cosines (𝑙, 𝑚 𝑎𝑛𝑑 𝑛) are:
𝑎
𝑏
𝑐
𝑙 = |𝑣| 𝑚 = |𝑣|
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
the direction cosines (𝑙, 𝑚 & 𝑛) are:
𝟑
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:
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
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𝑥
𝑦 component:
𝑎𝑦 = 8𝑠𝑖𝑛40 = 5.14
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Dot (Scalar) Product:
The result of the dot product is a scalar: i.e. vector.vector = scalar
𝒂. 𝒃 = 𝒂𝒃𝒄𝒐𝒔𝜽
The dot product can be used to find the angle between 2 vectors.
e.g.
̂
𝒂 = 𝟐𝒊̂ + 𝟑𝒋̂ + 𝟓𝒌
̂
𝒃 = 𝟒𝒊̂ + 𝒋̂ + 𝟔𝒌
(𝑖′𝑠)
(𝑗′𝑠)
(𝑘′𝑠)
Left hand side (𝑎. 𝑏):
𝑎. 𝑏 = (2 × 4) + (3 × 1) + (5 × 6) = 41
Right hand side (𝑎𝑏):
𝑎𝑏 = |𝑎||𝑏| = √22 + 32 + 52 × √42 + 12 + 62 = √38√53
If necessary we can find the angle between 𝑎 and 𝑏 using the dot product equation:
𝒂. 𝒃 = 𝒂𝒃𝒄𝒐𝒔𝜽
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
© H Jackson 2014/15 / Academic Skills
∴ 𝜃 = 90𝑜
∴ 𝜃 = 0𝑜
√38√53
perpendicular
parallel
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Cross (Vector) Product:
The result of the cross product is a vector: i.e. vector x vector = vector
|𝒂 × 𝒃| = 𝒂𝒃𝒔𝒊𝒏𝜽
Left hand side (|𝑎 × 𝑏|):
If 𝑎 = 𝑎1 𝑖̂ + 𝑎2 𝑗̂ + 𝑎3 𝑘̂
then
and 𝑏 = 𝑏1 𝑖̂ + 𝑏2 𝑗̂ + 𝑏3 𝑘̂
𝑎 × 𝑏 = (𝑎2 𝑏3 − 𝑎3 𝑏2 )𝑖̂ − (𝑎1 𝑏3 − 𝑎3 𝑏1 )𝑗̂ + (𝑎1 𝑏2 − 𝑎2 𝑏1 )𝑘̂
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
𝑖 𝑗 𝑘
with 𝑎 above 𝑏.
̂
e.g. 𝒂 = 𝟐𝒊̂ + 𝟒𝒋̂ + 𝟑𝒌
𝑎 × 𝑏 = |2 4 3 |
1 5 −2
̂
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
To finish the left hand side, remember to find the magnitude:
|𝑎 × 𝑏| = √(−23)2 + 72 + 62 = √614
(𝟐𝟒. 𝟖)
Right hand side (𝑎𝑏):
(This is the same as the dot product)
𝑎𝑏 = |𝑎||𝑏| = √22 + 42 + 32 × √12 + 52 + (−2)2 = √29√30
If necessary we can find the angle between 𝑎 and 𝑏 using the cross product equation:
|𝒂 × 𝒃| = 𝒂𝒃𝒔𝒊𝒏𝜽
√614 = √29√30 𝑠𝑖𝑛𝜃
𝜃 = 57.1𝑜
Parallel or perpendicular?
If 𝒂. 𝒃 = 𝟎
then 𝑠𝑖𝑛𝜃 = 0
If 𝒂. 𝒃 = 𝒂𝒃 then 𝑠𝑖𝑛𝜃 = 1
© H Jackson 2014/15 / Academic Skills
∴ 𝜃 = 0𝑜
∴ 𝜃 = 90𝑜
parallel
perpendicular
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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
Right Hand Rule:
The product vector (𝑎 × 𝑏) of 2 vectors (𝑎 𝑎𝑛𝑑 𝑏) acts perpendicular to both vectors and has
magnitude 𝑎𝑏𝑠𝑖𝑛𝜃. (Also 𝑏 × 𝑎 = − (𝑎 × 𝑏)).
For RH rule: 𝑖̂ × 𝑖̂ = 𝑗̂ × 𝑗̂ = 𝑘̂ × 𝑘̂ = 0
𝑖̂ × 𝑗̂ = 𝑘̂
𝑗̂ × 𝑖̂ = −𝑘̂
𝑗̂ × 𝑘̂ = 𝑖̂
𝑘̂ × 𝑗̂ = −𝑖̂
𝑘̂ × 𝑖̂ = 𝑗̂
𝑖̂ × 𝑘̂ = −𝑗̂
You could remember the above 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.
© H Jackson 2014/15 / Academic Skills
other letter.
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