Course Name : Physics – ICourse # PHY 107

Note-2: Vectors and their PropertiesGeometric and Algebraic PropertiesCartesian and Polar forms, Pythagorean Theorem

Abu Mohammad KhanDepartment of Mathematics and PhysicsNorth South Universityhttps://abukhan.weebly.com

Copyright: It is unlawful to distribute or use this lecture material without the written permission from

the author

Topics to be studied

◮ What is a ‘vector’?

◮ How to represent a vector?

◮ Unit vectors and their properties

◮ Addition/Subtraction rules

◮ Product rules for vectors:

1. Dot/Scalar product2. Cross/Vector product

◮ Polar form of product rules and the geometrical interpretations

◮ Triple Scalar Product : The Volume

◮ Examples

◮ Suggested Problems

Vectors: Definitions and Properties:

◮ Definition: A vector is an abstract quantity which has ‘Length’ and a ‘Sense ofdirection’. This is denoted by ~A and geometrically represented by a directedarrow, like,

Ab

a

from an arbitrary initial point ‘a’ to an arbitrary final point ‘b’. The length of thevector is the shortest distance from the initial to the final point.

◮ The number of parameters required to express a vector algebraically depends onthe dimensions of space. In simple terms, it is the number of coordinates, calledthe dimensions of space. In one dimension, there is only one coordinate: thex-axis. In three dimensions, there are three axes: x, y and z.

◮ How many parameters do we need to represent the length and direction?

Vector Addition Rules

◮ Triangular or Tip-to-Toe rule1: The diagram below show to add two vectors usingthis rule.

1Source: The diagram is taken from the Textbook

◮ Parallelogram rule2:

2Source: The diagram is taken from the Textbook

◮ Associative law of vector addition3:

3Source: The diagram is taken from the Textbook

◮ Vector Subtraction4: A negative is a vector whose direction has been reversed,and the length or magnitude remains same (Left diagram). Subtracting a vector isreally an addition of a negative vector (Right diagram).

4Source: These diagrams are taken from the Textbook

Vector components

◮ Algebraic form of a vector in two dimensions5: We will represent a vectoralgebraically in two ways: Polar form and Cartesian (or Component) forms.

5Source: These diagrams are taken from the Textbook

◮ The abstract vector ~a can be expressed algebraically in the following way:~a = (Magnitude, Direction) = (x-component, y-component)

(Polar form) (Cartesian form)= ( a, θa) = (ax , ay)

◮ Note that in the above equation, the pair of tuples, (a, θa) and (ax , ay), are equalin vectors sense, meaning that these two expressions represent the same object,but in two different ways.

◮ Here a 6= ax 6= ay 6= θa. All four are different parameters. But the question is, ifone pair is known, then how to find out the other pair?

◮ What is/are the mathematical relation(s) between these two tuples?

◮ The answer lies in the infamous ‘Pythagorean Theorem’.

Pythagorean Theorem and the Representation of a vector:

◮ Let’s choose two dimensions (i.e. two coordinate axes) for simplicity. A generalvector can be expressed as: ~A = (Ax , Ay ) = (A, θA).

◮ Geometrically the vector ~A is represented as in the following diagram:

)

A = |A|

(0,0)

A x

A x

A yA y

Aθ

y−axis

x−axis

(A x, A y

◮ From the above diagrams, it is clear that the length of the vector (hypotenuse),the x-component (base) and the y-component (height) form a right-angle triangle.

◮ The angle, θA, gives the direction of the vector, and is measured from the base tothe hypotenuse counter-clockwise (positive angle).

◮ By Pythagorean Theorem, we can easily write:

A2 = A2x + A2

y

θA = tan−1

(Ay

Ax

)

}

=⇒ This gives the conversion from theCartesian to the Polar form of a vector

Ax = A cos θAAy = A sin θA

}

=⇒ This gives the conversion from the Polar tothe Cartesian form of a vector

◮ Note that the Pythagorean theorem is defined only for the 1st quadrant.

◮ Since a vector can lie in any direction on a plane, we need to extend thePythagorean Theorem to all four quadrants. But the question is:

“How”?

◮ Consider the following two vectors given in Cartesian coordinates as

~A = (1, 1) −→ θA = tan−1

(1

1

)

= 45◦ .

~B = (−1, −1) −→ θB = tan−1

(−1

−1

)

= 45◦ !!!!

◮ Certainly these two vectors have the same directions ( as the algebra shows).

◮ But this is impossible since one vector is in the 1st quadrant, and the other is inthe 3rd quadrant.

◮ Thus these two vectors have different directions. Note that the length of thesetwo vectors are same by Pythagorean theorem which is

√2.

◮ The solution is:

~B = (−1, −1) −→ θB = tan−1

(−1

−1

)

= 180◦ + tan−1

(1

1

)

= 225◦ .

◮ Similar problem also occur between the vectors in the 2nd and 4th quadrant.

◮ The direction of a vector in a two dimensional plane is given by:

Direction, θA =

tan−1

(|Ay ||Ax |

)

(For a vector in the 1st quadrant)

180◦ − tan−1

(|Ay ||Ax |

)

(For a vector in the 2nd quadrant)

180◦ + tan−1

(|Ay ||Ax |

)

(For a vector in the 3rd quadrant)

360◦ − tan−1

(|Ay ||Ax |

)

(For a vector in the 4th quadrant)

◮ Note that the angle is positive by convention when measured counter-clockwise.

◮ In three dimension, we need two angles to fix the direction which is known as thespherical polar coordinates (MAT250), and hence in more than two dimensions,we will mainly use Cartesian coordinates (or the component form of a vector)

Examples:

Problem # 3.15:

The two vectors ~a and ~b in the adjacentfigure have equal magnitudes of 10.0m andthe angles are θ1 = 30◦ and θ2 = 105◦.Find the (a) x and (b) y-components oftheir vector sum ~r , (c) the magnitude of ~r ,and (d) the angle ~r makes with the positivedirection of the x-axis.

y

a

b

θ1

θ2

x

Solution: Problem # 3.15:

From the adjacent diagram: ~r = ~a + ~b, and the vectors inpolar form are:~a = (a, θa) = (a, θ1) = (10m , 30◦).~b = (b, θb) = (b, θ1 + θ2) = (10m , 135◦).Therefore, the components of ~r are:

(a) rx = ax + bx = a cos θa + b cos θb =(10 cos 30◦ + 10 cos 135◦)m = 1.60m. X

(b) ry = ay + by = a sin θa + b sin θb =(10 sin 30◦ + 10 sin 135◦)m = 12.1m. X

(c) r =√

r2x + r2y =√

(1.60)2 + (12.1)2 m = 12.2m. X

(d) θr = tan−1

(ry

rx

)

= tan−1

(12.1

1.60

)

= 82.3◦. X

y

b

θ1

θ2

aθr

r θ1

x

Unit Vectors:

◮ Definition: any vector whose length or magnitude is one is called a unit vector. InCartesian Coordinate system, the unit vector along the x-axis is denoted by i , andsimilarly j , k are the unit vectors along the y - and z-axes.

◮ In three dimensions, any vector is expressed as

~A = (Ax , Ay , Az) = Ax i + Ay j + Az k .

◮ The unit vectors are mutually perpendicular.

◮ For any two vectors, the addition/subtraction is given by

~A± ~B = (Ax , Ay , Az)± (Bx , By , Bz) = (Ax ±Bx)i +(Ay ±By)j +(Az ±Bz)k .

◮ The polar form of the sum can be obtained by Pythagorean Theorem.

Product Rules for Unit Vectors:

◮ There are two rules for product between two vectors:

1. Dot or Scalar Product: It is defined for the unit vectors as

i · i = j · j = k · k = 1︸ ︷︷ ︸

Definition of unit vector

and i · j = j · k = k · i = 0︸ ︷︷ ︸

Definition of orthogonality

2. Cross or Vector Product: It is defined for the unit vectors as

i × i = j × j = k × k = 0 , (Definition of parallel vectors)

i × j = k , j × k = i , k × i = j (Right-hand-rule)

j × i = −k, k × j = −i , i × k = −j (Note the change in sign)

◮ Geometrically the cross-product implies rotation. The right-hand-rule definespositive rotation which is counter-clockwise by convention.

◮ Hence, clockwise rotation is negative.

In Cartesian form, the vector products between any two vectors can easily be computedand easy to remember.

◮ Using the properties of the dot product between the unit vector, the dot productbetween any two vector is expressed as

~A · ~B = (Ax , Ay , Az) · (Bx , By , Bz) = AxBx + AyBy + AzBz → Scalar .

◮ Similarly, using the cross-product properties of unit vectors, the Cross productbetween any two vectors is now given by

~A× ~B = i(AyBz − AzBy

)− j

(AxBz − AzBx

)+ k

(AxBy − AyBx

),

=

∣∣∣∣∣∣

i j k

Ax Ay Az

Bx By Bz

∣∣∣∣∣∣

→ Vector .

Polar form of the Dot product (Geometrical Interpretation):

A

B

A cosθB

B cosθA

θθ

◮ The ‘Dot Product’ in polar form is given by

~A · ~B = |A|(|B | cos θ) = |B |(|A| cos θ) ≡ AB cos θ .

◮ So, it is really a scalar multiplication (multiplying a number by another number),also known as scaling. The result is a scalar.

Polar Form of the Cross Product (Geometrical Interpretation):

B

n

A

θ

090

θArea of the Parallelogram = AB sin

◮ The ‘Cross Product’ in polar form isdefined as

~A× ~B = |A||B | sin θ n = AB sin θ ,

= (Area of the Parallelogram) n .

◮ Here n is the rotation axis. It is also thedirection of the ‘Cross Product of ~A and ~B.

◮ Note that A surface or plane is a vectorquantity. The area is the magnitude of thesurface and the direction of the surface orplane is given by n which is known as thenormal unit vector.

Volume or Triple Scalar product:

A

B

Cz−axis

x−axis

y−axis

◮ The area of the bottom face= |~A× ~B| = AB sin 90◦ = AB .

◮ Therefore, the volume of the parallelepiped

= ~C · ~A× ~B ,

=

Cx Cy Cz

Ax Ay Az

Bx By Bz

→ Scalar Quantity

◮ Using the properties of Determinant, it isto show that:

~C · ~A× ~B = ~A · ~B × ~C = ~B · ~C × ~A .

Examples:

Problem # : Vectors ~C and ~D have magnitudes of 3 and 4 units respectively. Whatis the angle between the directions of ~C and ~D if the magnitude of the vector product| ~C × ~D| is zero? 12 units?Solution: By the definition of vector cross product, we get: | ~C × ~D| = CD sin θ. Usingthis and the given conditions, we easily can calculate:

1. 0 = (3)(4) sin θ =⇒ θ = 0◦, 180◦.That is, the vectors are either parallel or anti-parallel.

2. 12 = (3)(4) sin θ =⇒ θ = 90◦ or π/2 radian.That is, the vectors are perpendicular or orthogonal to each other.

Suggested Problems:

Chapter # 3: 3, 5, 8, 9, 11, 15, 16, 18, 22, 26, 27, 30, 33, 34, 36, 43, 44, 45, 47 and63.