Vectors_lecture Slide_print(6).pdf

VECTORS

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Subtopics  The

Geometric Approach  Vector spaces, linear independence and basis  Scalar(dot) product  Vector(cross) product

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Scalars and Vectors 

A scalar is a quantity which has magnitude (size) only e.g. length, temperature, angle.

A vector is a quantity which has both magnitude (size) and direction e.g. velocity, acceleration. Vector representation and notation   Vector can be represented as a ; AB ; a ; 

~

B

Terminal point A

magnitude(length) written as 𝒂 ; 𝐴𝐵 ; 𝑎

Initial point

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Sub topic 1 The Geometric Approach 4

Definition 

Two vectors are equal if they have the same length (magnitude) and the same direction.

5

The triangle rule for vector addition 



The sum of two vectors a and b is a new vector c that is obtained geometrically by adding vectors a and b simply head to tail. Law of parallelogram: b

a

a

a

a+b b

a-b b

b-a

b

a

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Example 

a)

b)

c)

During a voyage of the old Bass Strait ferry, the Abel Tasman, the rip at the entrance to Port Phillip Bay caused the navigator some concern. He needed to sail due north, but the current was flowing east at 20 km/h. In still water, the Abel Tasman had a maximum speed of 30 km/h. Calculate: the actual direction of travel if he had simply pointed the boat towards north, the course necessary to travel in the required direction, the ship’s velocity relative to Point Lonsdale.

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Example 

Solution:

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Example 

.

∴ The ship travels in a direction of 33.7ο east of north (N 33.7ο E).

9

Example .

∴ The ship must be pointed in the direction of 41.8ο west of north.

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Cartesian coordinates and components of a vector 

In three-dimensional space, the Cartesian coordinate system is formed by three perpendicular lines (coordinate axis) that intersect at one point (the origin). Usually O denotes the origin and the coordinate axes are denoted by (x, y, z), as shown in the figure

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Unit vector 

A unit vector has length unity(1), and is defined as v

vˆ 





v

A vector v is described as product of |v| and the direction of its unit vector ˆ i.e. v = |v| v Zero (null) vector The zero vector has no length and is a point. 12

Unit vector 

Let i, j and k be unit vectors in the directions of the x-, y- and z-axis of the Cartesian coordinate system, respectively. The tripplet of mutually perpendicular unit vectors (i, j, k) is called a basis of the Cartesian coordinate system.

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Position vector 

.

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Position vector 

.

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Basic operations - Example 

If a=3i+2j-5k, b=4i-j+3k and c=2i+3j-2k, calculate (i) a-b+c (ii) 2a+b-3c

Solution: (i) a= 3i+2j-5k -b=-4i+ j-3k c= 2i+3j-2k

(ii) 2a= 6i+4j-10k b= 4i - j+ 3k -3c=-6i -9j+ 6k

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Example 

ෝ in the direction of the Find the unit vector 𝒖 vector P1(1,1,3) to P2(-1,2,0).

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Example 

Consider the points A(5,-2,3) and B(2,1,-1). Find 𝐵𝐴 .



Solution:

𝐵𝐴 = 𝑂𝐴 − 𝑂𝐵 = 𝐵𝐴 =

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Direction Cosines 

In 3D-space,the angles between a non zero vector v and the vectors i, j, k are called the direction angles of v



The cosines of those angles are called the direction cosines of v.

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Direction Cosines 

   

Let , ,  be the direction angles of v = ai + bj+ ck The direction cosines of v are: cos α = a/|v|, cos β = b/|v| and cos γ = c/|v|

 a, b, c 

Note: 𝑐𝑜𝑠 2 α + 𝑐𝑜𝑠 2 β + 𝑐𝑜𝑠 2 γ = 1 20

Direction Cosines – Example 1 

Find the direction cosines of the vector v = 2i - 4j + 4k and approximate the direction angles to the nearest degree.



Solution:

v 

4  16  16  6

The direction cosines of v are: cos α = a/|v|, cos β = b/|v| and cos γ = c/|v|

cos  = 2/6 = 1/3, cos  = -4/6 = -2/3, cos  = 4/6 = 2/3   71°,   132°,   48°

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Direction Cosines – Example 2 

Find the unit vector and the direction cosines of the vector joining the two points A(4, 2, 2) and B(7, 16, 14).



Answer: ෢ = 𝐀𝐁

cos  =

, cos  =

, cos  =

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Sub topic 2 Vector spaces, linear independence and basis 23

Vector spaces 

Vector in Matrix form A 3-dimensional vector can be regarded as a 3X1 column matrix. 4 e.g. can be written as 2 3



Vector Space: In n-dimensions the vector space V can be regarded as the set of vectors in that space.

e.g. in 2 dimensions, n=2 and the vector space is the set of vectors in 2D (all the vectors ) in 3 dimensions, n=3 and the vector space is the set of vectors in 3D (all the vectors ) 24

Vector spaces 

A vector space over the real numbers is a set V containing elements known as vectors, u, v, w etc., along with rules for combining those vectors with each other using also elements from the real number set. The space is closed under application of the rules in use. That means that if we use the rules to combine vectors in the space, we just get other vectors in the space and not something new. For any u, v ∈ V and any scalar k (v + u) ∈ V and kv ∈ V For V to be a vector space, the means by which these rules act must satisfy a number of restrictions. For all u, v, w ∈ V , all k, l ∈ R we require the following eight rules.

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Rules over Vector space 

(over the real numbers).

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Vector spaces - Example 

Show that a set of solutions X to a homogeneous system of linear equations AX = 0 is a vector space. Solution: We need to show that that the set is closed under the operations of addition and multiplication by a scalar. Let 𝑋1 and 𝑋2 be any two solutions of AX = 0, then 𝑋1 + 𝑋2 and λ𝑋1 , where λ is a scalar, also satisfy the same homogeneous equation.

A(X1 + X2) = AX1 + AX2 = 0 + 0 = 0 and A(λX1) = λ AX1 = λ x 0 = 0 

Note: The set of solutions X of a non-homogeneous system

AX = B is not a vector

space, because

A(X1 + X2) = AX1 + AX2 = B + B = 2B ≠ B. 27

Linear combination of Vectors 

Definition:

Example : Given the vectors u1  i  j , u2  4 i  3 j , ~

~

~

~

5 u1  2 u2  5( i  j )  2(4 i  3 j )  3 i  11 j is a linear combination ~

~

~

~

~

~

~

~

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Linear independence 

Definition:

•

.

•

.

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Linear independence 

. Vectors 1 , 2 , 3 ,..., n V are said to be linearly independent if ~

~

~

~

k1 1  k2 2  ...  kn n  0 ~

~

~

is true only if k1 , k2 ,..., kn  0 

Example:

Are the vectors a  1,2,3 , b  2,4,6  linearly independent? ~

~

Answer : No, because b  2 a . ~

~

Hence, the linear combination 2 a  b  0  k1  2, k2  1 ~

~

 k1 a  k2 b  0, for k1  k2  0 ~

~

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Linear independence and Rank 

Methods to determine linear dependence or independence:

1)

Gauss elimination

2)

Determinants - If the determinant of the matrix = 0,the vectors are linearly dependent - If the determinant of the matrix ≠ 0,the vectors are linearly independent



Definition: The RANK of a matrix A is the number of totally non-zero rows in the final form of A after Gauss elimination. Note: If the rank is equal to the dimension of the vectors, the vectors are independent. 31

Linear independence - Examples Example 1 : Do the following vectors form a linearly independent set? 1 1 1  1  0, 2  3, 3  2       0 6 4 (a) Obtain your answer : (i) by Gaussian elimination (ii) by determinant (b) If the vectors are dependent, write down a combination of them that simplifies to the zero vector.

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Linear independence - Examples

33

Linear independence - Examples

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Linear independence - Examples 

.

Answer:

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Linear independence - Examples Using Gauss elimination method:

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Vector space spanned by vectors 

Definition: The set V of all vectors v of the form

is called the vector space spanned by the set of m vectors 𝑢𝑖 , 𝑖 = 1 … 𝑚. We denote V as

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Example 

Given vector v = -i+j+3k, determine if v ∊ Span(u, w), where u = 5i+j+k, w = i-2j-k.

Solution:

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Example

39

Basis 

Definition: The dimension of a vector space is defined to be the smallest number of linearly independent vectors needed to span the entire space.



Definition: If V is an n-dimensional vector space, then a basis for V is a set of n linearly independent vectors in V which span V.



Note: Any m vectors in an n dimensional vector space are necessarily linearly dependent if m > n. Dimensionality of V = Rank of matrix A whose columns are the given vectors 40

Basis – Example 

.



Solution:

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Sub topic 3 Scalar (dot) product of two vectors 42

Scalar (dot) product of two vectors 

If a and b are two vectors, with θ the angle between them, then the scalar (dot) product of a and b is written as a.b and is defined by

a



b 

If a and b are at right angles to each other, then the dot product of these two vectors are zero. 43

Scalar (dot) product of two vectors   



If 𝜃 = 90°, If 𝜃 = 0°, Note: (a) i.i = j.j = k.k = 1(i, j, k are unit vectors) (b) i.j = j.i = i.k = k.i = j.k = k.j = 0 .

44

Angle between two vectors 

and b are two non zero vectors and 𝜃 is the angle If a ~ ~ between them, then

Example 1: Given that a = i+2j+2k and b = -3i+6j+2k. Find i. ab ii. a=IaI iii. the angle between a and b 

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Dot product – Example 1

Solution: Given: a = i+2j+2k , b = -3i+6j+2k. i.

ab =

ii. a=IaI = iii. .

𝜃=

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Dot product – Example 2



Find v . u, |v| and |u|, given that vectors v = i + 3j – 5k, u = -2i + 3j – k.

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Dot product – Example 3 

Find the angle between a) u = 3i +6j - k and v = 2i + j + 4k. b) u = 3i +j + 2k and v = -2i + 5j - 3k.



Answer:

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Dot product in higher dimensions

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Work done by a force 

.



Work W done on an object by a force vector F acting in the direction of motion from point P to Q is :



Work W done on an object by a force vector F acting (not in the direction of motion) but makes an angle of θ with the displacement vector 𝑃𝑄 is :

𝑊 = 𝑭 𝑐𝑜𝑠θ 𝑃𝑄

W  F  PQ 50

𝑊 = 𝑭 𝑐𝑜𝑠θ 𝑃𝑄 = 𝑭. 𝑃𝑄 Unit of Work is Nm or Joule (or foot-pounds) 51

Example 1-Work Done 

A box is dragged along the floor by a rope that applies a force of 50 lb at an angle of 60° with the floor. How much work is done in moving the box 15 ft?



Solution:

52

Example2 - Work Done 



A force of F = 5i + 2j + 3k N acts on a particle which moves from A(1,7,11) to B(4,16,18) with the coordinates in metres. Find W, the work done by the force on the particle. Solution:

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Projection of a vector

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Projection of a vector

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Projection of a vector

or is called the vector component of v in the direction of w .

or is called the vector component of v perpendicular to the direction of w.

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Example 1 - Projection 

.



Solution:

(a)

𝑤 = 𝑤 = 42 + 42 = 32

(b)

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Example 2 - Projection 

(a)

(b)



Find the vector component of u = i + 2j - k in the direction of v = 2i - j - 5k, and the vector component of u perpendicular to the direction of v. Answer:

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Orthogonal and Orthonormal Basis 

Definition: Any n linearly independent vectors 𝑢1 , 𝑢2 , … , 𝑢𝑛 form an orthogonal basis in an n- dimensional vector space, if 𝒖𝑖 are mutually orthogonal, i.e. if for any 𝑖 ≠ 𝑘,

𝒖𝑖 . 𝒖𝑘 = 0 

Definition: If additionally to orthogonality, all basis vectors 𝒖𝑖 have unit length, i.e. if 𝑢𝑖 = 𝒖𝑖 = 1, i=1,2,…,n then the basis is called orthonormal.

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Example – Orthogonal Basis 

.

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Example – Orthogonal Basis

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The Gram-Schmidt process 

Definition: The Gram-Schmidt process is an algorithm for constructing an orthogonal(orthonormal) basis in ndimensional vector space from any set of n linearly independent vectors 𝒖𝑖 , i = 1,2,…,n.



Example:

62

Example – The Gram–Schmidt Process 

Solution:

63

Example – The Gram–Schmidt Process

∴

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Sub topic 4 Vector(cross) product 65

The vector (cross) product 

If a and b are two vectors, with θ the angle between them, then the vector (cross) product of a and b is written as and is defined by 

where n is a unit vector perpendicular to both a and b ,with ~ ~ ~ its direction given by the right-hand rule. a  a1 i  a2 j  a3 k ~



~

~

~

b  b1 i  b2 j  b3 k ~

~

~

~

Or,

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The vector (cross) product 

A unit vector perpendicular to both a and b is given by ~



~

.

Note: (a) i X i = j X j = k X k = 0 (c)

(b) i X j = k jXk=i kXi=j

j X i = -k k X j = -i i X k = -j 67

Cross product-Example 

.

Solution: 1.

2. Note:

Here

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Area of parallelogram and triangle using cross products

=

=

69

Scalar triple product

Scalar triple product can be found as

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Scalar triple product

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Example – Scalar triple product 

.



Solution:

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Vector triple product

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Example – Vector triple product 

.



Solution:

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Angular velocity(𝜔) -Cross product application

where the is the rate at which the angle 𝜃 is changing.

So,

𝑠

𝑣 = 𝑡; ω =

θ 𝑡

and 𝑣 = ω𝑟

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Angular velocity as a vector  Consider a rigid body(e.g. a wheel) rotating with an angular speed of 𝜔 radians/s about a fixed axis LM, which passes through a fixed point A.

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Angular velocity-Example 

A rigid body rotates with an angular velocity of 4 radians/sec about an axis in the direction of i +2j+4k and passing through the point A(3,5,8). is on the axis. Find the (linear)velocity of the point P(4,6,10) on the body. Solution: We know Unit vector

where = unit vector along i +2j+4k i +2j+4k i +2j+4k = 2 2 2= 21 1 +2 +4

∴

=4

i +2j+4k 21

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Angular velocity-Example

Hence,

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The Coriolis force

79

Moment of a force(torque M) The moment of force about a point is a measure of the tendency of the force to rotate an object about the point.

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Moment of a force(torque M)



The SI unit for torque is the newton meter(N.m).



The symbol for torque is 𝜏, the Greek letter tau. When it is called moment, it is commonly denoted by M.

81

Moment of a force(torque M)-Example 

A force of 4 newtons acts through P(2,3,5) in the direction of a = 4i+7j+9k. Find the moment of the force about A(6,10,3). Dimensions are in metres. Solution: We know

where

Unit vector

= unit vector along 4i +7j+9k =

∴

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Moment of a force(torque M)-Example

Hence,

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