1.14
1 Vectors and Matrices
Definition: mathematical expressions possessing a
magnitude and a direction that add up accordingly to
the parallelogram law.
Examples: forces, displacements, velocities etc.
Graphical representation – an arrow with the
direction of the vector and the length corresponding
to the magnitude of the vector. Length of vector 𝑃̅ is
denoted as P or |𝑃̅|.
Three basic types of vectors:
Free vector is not attached to any point or a line in a
body.
Example: Displacement of the body moving without
rotation.
Sliding vector has unique line of application
Example: Force acting on a rigid body.
Fixed vector has unique point of application (and,
thereof, line of application).
Example: Force acting on a deformable body.
Note: in statics, forces are sliding vectors as there is
no deformation. Hence, they have unique
magnitude, direction, and line of action. However,
we can slide the force along its line of action.
1.15
 Vector properties
Negative vector
http://mathinsight.org/image/
vector_opposite
a +(-a)=0
Multiplication by a scalar
http://webphysics.iupui.edu
/JITTworkshop/152Basics/
vectors/vectors.html
1.16
 Vector addition
Graphical Addition
Triangle rule:
http://www.maths.usyd.edu.
au/u/MOW/vectors/vectors3/v-3-3.html
Parallelogram rule:
http://www.maths.usyd.edu.
au/u/MOW/vectors/vectors3/v-3-3.html
Note 1:
𝑃̅ + 𝑄̅ = 𝑄̅ + 𝑃̅
Note 2: The above methods do not general give line of
action
Note 3: Parallelogram rule can be used to resolve the
vector into components (not unique and not necessarily
orthogonal)
̅ , generally |𝑃̅| + |𝑄̅ | = |𝑊
̅|
Note 3: While 𝑃̅ + 𝑄̅ = 𝑊
1.17
To find numerical values of length (magnitude) and
angle (direction) of vector, use trigonometry.
Sine law
𝐴
sin(𝑎)
=
𝐵
sin(𝑏)
=
𝐶
,
sin(𝑐 )
good when two
angles and one side are known.
Cosine law 𝐶 = √𝐴2 +𝐵2 − 2𝐴𝐵𝑐𝑜𝑠 𝑐, good for
addition, when two vectors and angle between
them are known.
Multiple vector addition
http://www.drcruzan.com/MathVectors.html
1.18
Vector addition, Example 1:
Example 2
1.19
Example 3
1.20
Addition using rectangular components:
2D
http://www.maths.usyd.edu.au
/u/MOW/vectors/vectors-7/v7-2.html
ϴ
Cartesian unit vectors: |𝑖̅| = |𝑗̅| = 1
Any vector can be then represented in Cartesian
coordinates:
𝑄̅ = 𝑄𝑥 ∗ 𝑖̅ + 𝑄𝑦 ∗ 𝑗̅
Where: 𝑄𝑥 = |𝑄̅ | ∗ cos Θ; 𝑄𝑦 = |𝑄̅ | ∗ sin Θ = |𝑄̅ | ∗ cos(90 − Θ)
From here:
|𝑄̅ | = √𝑄𝑥2 + 𝑄𝑦2
and
Θ = tan−1
𝑄𝑦
⁄𝑄
𝑥
Adding two vectors:
𝑄̅ = 𝑄𝑥 ∗ 𝑖̅ + 𝑄𝑦 ∗ 𝑗̅
and 𝑃̅ = 𝑃𝑥 ∗ 𝑖̅ + 𝑃𝑦 ∗ 𝑗̅
𝑅̅ = 𝑄̅ + 𝑃̅ = 𝑅𝑥 ∗ 𝑖̅ + 𝑅𝑦 ∗ 𝑗̅ = 𝑄𝑥 ∗ 𝑖̅ + 𝑄𝑦 ∗ 𝑗̅ + 𝑃𝑥 ∗ 𝑖̅ + 𝑃𝑦 ∗ 𝑗̅
= (𝑄𝑥 + 𝑃𝑥 ) ∗ 𝑖̅ + (𝑄𝑦 + 𝑃𝑦 ) ∗ 𝑗̅
Similar for more than two forces.
1.21
Examples:
1.22
3D
Similar to 2D.
̅| = 𝟏
Unit vectors: |𝒊̅| = |𝒋̅| = |𝒌
𝑄̅ = 𝑄𝑥 𝑖̅ + 𝑄𝑦 𝑗̅ + 𝑄𝑧 𝑘̅
and |𝑄̅ | = √𝑄𝑥2 + 𝑄𝑦2 + 𝑄𝑧2
Directional Cosines:
If we have vector 𝑄̅ = 𝑄𝑥 𝑖̅ + 𝑄𝑦 𝑗̅ + 𝑄𝑧 𝑘̅, we
can introduce directional cosines 𝑐𝑜𝑠(𝛼) =
𝑄𝑦
𝑄𝑥
𝑄
, 𝑐𝑜𝑠(𝛽) = , 𝑐𝑜𝑠(𝛾) = 𝑧, where 𝛼, 𝛽, 𝛾𝑄
𝑄
𝑄
coordinate direction angles between vector
𝑄̅ and positive 𝑖̅, 𝑗̅, 𝑘̅ directions of coordinate
system.
Then 𝑄̅ = 𝑄𝑐𝑜𝑠(𝛼)𝑖̅ + 𝑄𝑐𝑜𝑠(𝛽)𝑗̅ + 𝑄𝑐𝑜𝑠(𝛾)𝑘̅
= 𝑄(𝑐𝑜𝑠(𝛼)𝑖̅ + 𝑐𝑜𝑠(𝛽)𝑗̅ + 𝑐𝑜𝑠(𝛾)𝑘̅) = 𝑄𝑢
̅̅̅̅,
𝑄
Where ̅̅̅̅
𝑢𝑄 is unit vector in the direction of 𝑄̅,
𝑢𝑄
̅̅̅̅=𝑐𝑜𝑠(𝛼)𝑖̅
+ 𝑐𝑜𝑠(𝛽)𝑗̅ + 𝑐𝑜𝑠(𝛾)𝑘̅ and 𝑐𝑜𝑠 2 (𝛼) + 𝑐𝑜𝑠 2 (𝛽) + 𝑐𝑜𝑠 2 (𝛾) = 1
1.23
 Matrices
Definition: A rectangular array of numbers arranged in
rows and columns.
Matrix A of an order mXn
http://en.wikipedia.org/wiki/Matrix_
%28mathematics%29
Each element of the matrix aij is the row i and column j
element of A. For example, a12 denotes the element in
the first row and second column.
Square Matrix: m=n
Note: A vector can be represented as a matrix of the
order 1X3, in the following way:
𝑄̅ = 𝑄𝑥 𝑖̅ + 𝑄𝑦 𝑗̅ + 𝑄𝑧 𝑘̅ = (𝑄𝑥 𝑄𝑦 𝑄𝑧 )
Submatrix is obtained from the matrix by deleting any
collection of rows and columns.
Example:
http://en.wikipedia.org/wiki/Matrix_
%28mathematics%29
Note: most of the matrix algebra is beyond the scope of this
course.
1.24
Determinant of a matrix:
Determinant can be calculated only for a square matrix.
Determinant of a matrix A is denoted det(A) or |A|.
For a 2X2 matrix, the determinant is defined as follows:
http://en.wikipedia.org/wiki/Determinant#2
.C2.A0.C3.97.C2.A02_matrices
For matrices of higher order, the process is inductive. The
determinant is calculated by going through either a row or a
column, multiplying the elements in it by the determinant of a
submatrix that is obtained by removing the row and the column of
the corresponding element for the overall matrix and alternating
the plus and minus signs. Of most interest to us is the
determinant of a 3X3 matrix:
http://en.wikipedia.org/wiki/Determinant#2
.C2.A0.C3.97.C2.A02_matrices
An alternative easy way to calculate the determinant of a 3X3
matrix is using the rule of Sarrus:
http://en.wikipedia.org/wiki/Determinant#2
.C2.A0.C3.97.C2.A02_matrices
𝑑𝑒𝑡 (𝐴) = 𝑎11 𝑎22 𝑎33 + 𝑎12 𝑎23 𝑎31
+ 𝑎13 𝑎21 𝑎32 − 𝑎13 𝑎22 𝑎31
− 𝑎11 𝑎23 𝑎32 − 𝑎12 𝑎21 𝑎33
Example of use: calculating the area of a parallelogram where
the first row is 1’s and the second and third are vectors of the
parallelogram’s sides.
1.25
 Vector operators
The Scalar (dot) product:
Definition: 𝐴̅ ∙ 𝐵̅ = |𝐴̅| ∙ |𝐵̅| ∙ 𝑐𝑜𝑠(𝜃), where 𝜃 is the angle
between vectors 𝐴̅ and 𝐵̅, with their tails joined.
As can be seen the dot product is a scalar (hence the
alternative name).
From definition it is obvious that for Cartesian unit vectors:
𝑖̅ ∙ 𝑖̅ = 𝑗̅ ∙ 𝑗̅ = 𝑘̅ ∙ 𝑘̅ = 1 ∙ 1 ∙ cos(0) = 1
𝑖̅ ∙ 𝑗̅ = 𝑖̅ ∙ 𝑘̅ = 𝑗̅ ∙ 𝑘̅ = 1 ∙ 1 ∙ cos(90°) = 0
Leading to:
𝐴̅ ∙ 𝐵̅ = (𝐴𝑥 𝑖̅ + 𝐴𝑦 𝑗̅ + 𝐴𝑧 𝑘̅ ) ∙ (𝐵𝑥 𝑖̅ + 𝐵𝑦 𝑗̅ + 𝐵𝑧 𝑘̅)
= 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧
Select computation method based on what is easier to
compute or given.
Dot product rules
Commutative: 𝐴̅ ∙ 𝐵̅ = 𝐵̅ ∙ 𝐴̅
Distributive: 𝐴̅ ∙ (𝐵̅ + 𝐶̅ ) = 𝐴̅ ∙ 𝐵̅ + 𝐴̅ ∙ 𝐶̅
Multiplication by a scalar: a(𝐴̅ ∙ 𝐵̅ ) = (𝑎𝐴̅) ∙ 𝐵̅ = 𝐴̅ ∙ (𝑎𝐵̅)
Example: 𝐴̅ = 2𝑖̅ − 3𝑗̅, 𝐵̅ = 6𝑖̅ + 4𝑗̅,
𝐴̅ ∙ 𝐵̅ =?.
𝐴̅ ∙ 𝐵̅ = (2𝑖̅ − 3𝑗̅) ∙ (6𝑖̅ + 4𝑗̅) = 2 ∙ 6 − 3 ∙ 4 = 0
In order to calculate the angle between these two vectors:
1.26
𝐴̅ ∙ 𝐵̅ = |𝐴̅| ∙ |𝐵̅| ∙ 𝑐𝑜𝑠(𝜃 ) = 0,
3D Example: 𝐴̅ = 𝑖̅ + 3𝑗̅ + 𝑘̅ ,
⟹ 𝛳 = 90°
𝐵̅ = 2𝑖̅ − 𝑗̅ + 𝑘̅
𝐴̅ ∙ 𝐵̅=?
𝐴̅ ∙ 𝐵̅ = 1 ∗ 2 − 3 ∗ 1 + 1 ∗ 1 = 0
Hence, 𝐴̅ and 𝐵̅ are perpendicular. You can always check it by
drawing them.
Applications of the dot product:
1) Angle between two vectors: 𝜃 = 𝑐𝑜𝑠 −1 (
𝐴̅∙𝐵̅
AB
)
You need to know components of 𝐴̅ and 𝐵̅.
2) Components of vector parallel and perpendicular to a line.
𝐹∥ = F ∙ cos(𝜃 ) = 𝐹̅ ∙ 𝑢
̅̅̅∥
𝐹̅∥ = 𝐹∥ ∙ 𝑢
̅̅̅∥ = (𝐹̅ ∙ ̅̅̅
𝑢∥ ) ∙ ̅̅̅
𝑢∥
̅̅̅
𝐹⊥ = 𝐹̅ − 𝐹̅∥ , 𝐹⊥ = √𝐹 2 − 𝐹∥2
Alternatively:
𝐹⊥ = 𝐹 ∙ 𝑠𝑖𝑛(𝜃 ),
𝜃 = 𝑐𝑜𝑠
−1
𝐹̅ ∙ ̅̅̅
𝑢∥
(
)
𝐹
̅̅̅
𝐹⊥ = 𝐹⊥ ∙ 𝑢
̅̅̅̅
̅̅̅̅
⊥ = 𝐹 ∙ 𝑠𝑖𝑛 (𝜃 ) ∙ 𝑢
⊥
= √𝐹 2 − 𝐹∥2 ∙ 𝑢
̅̅̅̅
⊥
1.27
Examples:
1.28
The vector (cross) product:
Definition: the vector product of two vectors 𝐴̅ and 𝐵̅ denoted as
𝐴̅ × 𝐵̅ is defined as vector 𝐶̅ which satisfies the following
conditions:
1. Direction: perpendicular to the plane
formed by 𝐴̅ and 𝐵̅
2. Magnitude: |𝐴̅ × 𝐵̅ | = |𝐴̅| ∗ |𝐵̅| ∗ 𝑠𝑖𝑛(𝜃),
where 𝜃 is an angle between |𝐴̅| and
|𝐵̅|. Hence, 𝐴̅ × 𝐴̅ = 0.
3. Sense: given by right hand rule (or
screwdriver rule).
Note 1: From the product definition
|𝑖̅ × 𝑖̅| = |𝑗̅ × 𝑗̅| = |𝑘̅ × 𝑘̅| = 1 ∙ 1 ∙ Sin(0) = 0
|𝑖̅ × 𝑗̅| = |𝑖̅ × 𝑘̅| = |𝑗̅ × 𝑘̅| = 1 ∙ 1 ∙ Sin(90°) = 1
Note 2: The vector product is not commutative 𝐴̅ × 𝐵̅ = −(𝐵̅ × 𝐴̅)
Note 3: The vector product is distributive
𝐴̅ × (𝐵̅1 + 𝐵̅2 ) = 𝐴̅ × 𝐵̅1 + 𝐴̅ × 𝐵̅2
How to calculate?
Example: using the definition
𝐴̅ = 2𝑖̅ − 𝑗̅,
𝐵̅ = 3𝑖̅ + 2𝑗̅,
𝐴̅ × 𝐵̅=?.
𝐴̅ × 𝐵̅ = (2𝑖̅ − 𝑗̅) × (3𝑖̅ + 2𝑗̅) = 2𝑖̅ × (3𝑖̅ + 2𝑗̅) − 𝑗̅ × (3𝑖̅ + 2𝑗̅)
= 2𝑖 × 3𝑖 + 2𝑖 × 2𝑗 − 𝑗 × 3𝑖 − 𝑗 × 2𝑗 = 4𝑘̅ + 3𝑘̅ = 7𝑘̅
Going by definition is cumbersome and complicated.
1.29
An alternative way to calculate:
It is easy to calculate both the magnitude and the direction of the
cross product by using a determinant of a matrix.
𝑖̅
𝑗̅
𝑘̅
𝐴̅ × 𝐵̅ = |𝐴𝑥 𝐴𝑦 𝐴𝑧 |
𝐵𝑥 𝐵𝑦 𝐵𝑧
= (𝐴𝑦 𝐵𝑧 − 𝐴𝑧 𝐵𝑦 ) ∗ 𝑖̅ + (𝐴𝑧 𝐵𝑥 − 𝐴𝑥 𝐵𝑧 ) ∗ 𝑗̅
+ (𝐴𝑥 𝐵𝑦 − 𝐴𝑦 𝐵𝑥 ) ∗ 𝑘̅
Mixed triple product: 𝐶̅ ∙ (𝐴̅ × 𝐵̅)
Interpretation: absolute value of the triple product is the volume
of the parallelepiped with vectors 𝐴̅, 𝐵̅ 𝑎𝑛𝑑 𝐶̅ for sides.
Mixed triple product is zero if the vectors forming it are coplanar.
Counterclockwise circular permutation:
𝐶̅ ∙ (𝐴̅ × 𝐵̅) = 𝐵̅ ∙ (𝐶̅ × 𝐴̅) = 𝐴̅ ∙ (𝐵̅ × 𝐶̅ )
= −𝐶̅ ∙ (𝐵̅ × 𝐴̅) = −𝐴̅ ∙ (𝐶̅ × 𝐵̅)
= −𝐵̅ ∙ (𝐴̅ × 𝐶̅ )
B
C
𝐶𝑥
𝐶̅ ∙ (𝐴̅ × 𝐵̅) = |𝐴𝑥
𝐵𝑥
Note: the triple product is a scalar.
𝐶𝑦
𝐴𝑦
𝐵𝑦
𝐶𝑧
𝐴𝑧 |
𝐵𝑧
A