Introduction

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# Introduction - PowerPoint PPT Presentation

Introduction. Definitions of Vectors and Scalars Physical quantities can be classified under two main headings -- Vectors and Scalars. A vector quantity is any quantity that has both magnitude (size) and direction. E.g., velocity, acceleration, force, momentum.

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Introduction

Definitions of Vectors and Scalars

Physical quantities can be classified under two main headings -- Vectors and Scalars.

A vector quantity is any quantity that has both magnitude (size) and direction.

E.g., velocity, acceleration, force, momentum.

A scalar quantity is any quantity that has magnitude only, while direction is not taken into account.

E.g., speed, pressure, temperature, energy.

Vectors

Vectors are represented by lines with arrow. The length of the line indicates the magnitude of the vector, and the direction of the line indicates the vector\'s direction. An arrow is used to denote the sense of the vector, i.e. for a horizontal vector, say, whether it acts from left to right or vice versa. The arrow is positioned at the end of the vector and its position is called the nose of the vector.

A vector of 20kN acting at an angle of 45º to the horizontal may be depicted by:

oa = 20kN at 45º to the horizontal

a

20kN

45º

o

Vectors

To distinguish between vector and scalar quantities, different conventions are used. The one these notes will adopt is to denote vector quantities in bold print. (Note in these presentations they will also be in blue while the scalar will be in plain text and red – the colour will not help with the copied notes).

Thus oa represents the vector quantity but oa is the magnitude of vector oa. Also the convention is that positive angles will be measured in an anticlockwise direction from the horizontal right facing line and negative angles clockwise from this line.

Thus 90º is a line vertically upwards and -90º is a line vertically downwards

Vectors

F1

Let us say we want to add two vectors together, say F1 at angle θ1 and F2 at angle θ2 as shown.

θ1

θ2

F2

F1

a

The resultant can be obtained by drawing oa to represent F1and then drawing ar to represent F2. The resultant of F1and F2is given by or. This is called the nose-to-tail method of vector addition.

F2

r

o

Vectors

Alternatively, by drawing lines parallel to F1 and F2 from the noses of F2 and F1 respectively, and letting the intersection of these lines be R, gives OR as the magnitude and direction of the resultant of adding F1 and F2. This is called the parallelogram method of vector addition.

F1

R

O

F2

These two methods are graphical and rely on the accuracy of the drawing of the lines. There is a purely mathematical method for adding these vectors and it is shown on the next slide.

Vectors

A vector can be resolved into two component parts such that the two new vectors are equal to the original vector. The two components are normally a horizontal component and a vertical component.

a

F sinθ

F

Consider the vector F.

θ

F cosθ

o

If we need to sum a number of vectors then if each is resolved into two directions then the resultant vertical and horizontal components can then be conventionally summed as they are in the same direction.

Vectors

Vectors F1and F2 are to be summed.

F2

The two horizontal vectors sum to give:

H = F1 cosθ1 + F2 cosθ2

The two vertical vectors sum to give:

V = F1 sinθ1 + F2 sinθ2

F2 sinθ2

F1

F1 sinθ1

θ2

θ1

F1 cosθ1

F2 cosθ2

Once we have the two resultant vectors V and H we can determine the single resultant by using Pythagoras and trig: Magnitude of the resultant

Angle to the horizontal is given by

Note

Resolving the vectors may result in vertical and horizontal components which are either up or down (for the vertical) or to the left or right (for the horizontal). To the right and up are taken as the positive direction and to the left and down as negative. If the angle is measured anticlockwise from the axis to the right then modern calculators will automatically generate the correct sign.

The angle that would be used is 120º

V = 10 sin 120º = 8.66

H = 10 cos 120º = -5

10N

60º

Vectors

Vector Subtraction

It must be remembered that the subtraction F1 – F2 can be thought of as F1+ (-F2).

So how do we find the negative of a vector?

In the diagram F is represented by oa.

The vector –oa can be obtained by drawing a vector from o in the opposite sense to have the same magnitude, shown as ob. ob = -oa

a

F

o

-F

Note

-F is the same as F but with an angle increase of 180º

b

Vectors

b

s

For two vectors acting at a point the resulting vector addition is os = oa + ob

o

a

b

d

If we now want ob + (-oa) or ob – oa then we will have od = ob - oa

o

-a

Comparing od with the first diagram, it is the same as the line ab. Therefore when we complete the parallelogram the two diagonals give us the sum and the difference vectors.

Vectors

Example

1. Vector F1 has magnitude 8, with direction θ=30°;

vector F2has magnitude 12, with direction θ=60°.

Use (1) vector diagram, (2) vector resolution

to get the resultant force F=F1+F2

2. Vector F1 has magnitude 6, with direction θ=120°;

vector F2has magnitude 10, with direction θ=-30°.

Use (1) vector diagram, (2) vector resolution

to get the resultant force F=F1+F2

3. Vector F1 has magnitude 9, with direction θ=240°;

vector F2has magnitude 18, with direction θ=-60°.

Use (1) vector diagram, (2) vector resolution

to get the resultant force F=F1+F2

When a vector x of magnitude x and direction θº is divided by the magnitude of the vector the result is a vector of unit length at an angle θº. The unit vector of a velocity 10 m/s at 50º is

In general the unit vector for oa is: oa/|oa|

oa being a vector with both magnitude and direction and

|oa| being the magnitude of the vector only.

Vectors

One method of completely specifying the direction of a vector in space relative to some reference point is to use three unit vectors mutually at right angles to each other.

z

This is called a unit triad.

k

y

j

i

o

x

The next slide shows how this is used to specify a three dimensional vector.

Vectors

In the diagram below one way of getting from o to r is to move x units in the i direction, to a point a, y units in the j direction to get to b and z units in the k direction to get to r.

The vector or is specified as: xi + yj + zk

r

k

z

α

j

o

x

β

i

a

b

y

Vectors

Example

1. A spatial vector r has magnitude of 10 at direction α=60° and β=30°.

Re-write the vector in the form of xi + yj + zk.

2. A spatial vector r has magnitude of 8 at direction α=120° and β=-30°.

Re-write the vector in the form of xi + yj + zk.

3. A spatial vector r has magnitude of 12 at direction α=240° and β=-60°.

Re-write the vector in the form of xi + yj + zk.

Vectors

The Scalar Product of Two Vectors

When a vector oa is multiplied by a scalar quantity k, the magnitude of the resultant vector will be k times the magnitude of oa and its direction will remain the same. Thus 2 x 5N at 20º results in a vector 10N at 20º.

One of the products of two vector quantities is called the scalar or dot product of the two vectors and is defined as the product of their magnitudes multiplied by the cosine of the angle between them The scalar product of oa and ob is shown as oa • ob. For vectors oa = oa at θ1º, and ob = ob at θ2º, where θ2> θ1, the scalar product is:

oa • ob = oa ob cos(θ2 - θ1)

Vectors

For the vectors v1 and v2 shown, the scalar product is

v1• v2 = v1 v2 cosθ

The cumulative law of algebra, a x b = b x a, applies to the scalar product.

v2

θ

v1

If v1 is oa and v2 is ob then this is shown below:

b

v2

By geometry it can be seen that the projection of ob on oa is v2 cosθ. But we know that:

v1 • v2 = v1 v2 cosθ = v1 (v2 cosθ)

a

θ

o

v1

v2 cosθ

v1 • v2 = v1 times the projection of v2 on v1

The scalar product or dot product between two vectors P and Q is defined as

• Scalar products:
• are commutative,
• are distributive,
• are not associative,
• Scalar products with Cartesian unit components,
Scalar Product of Two Vectors

Angle between two vectors:

• Projection of a vector on a given axis:
• For an axis defined by a unit vector:
Scalar Product of Two Vectors: Applications

Mixed triple product of three vectors,

• The six mixed triple products formed from S, P, and Q have equal magnitudes but not the same sign,
• Evaluating the mixed triple product,
Mixed Triple Product of Three Vectors

Similarly:

b

v2

By geometry it can be seen that the projection of oa on ob is v1 cosθ. But we know that:

v1 • v2 = v1 v2 cosθ = v2 (v1 cosθ)

v1 cosθ

a

θ

o

v1

v1 • v2 = v2 times the projection of v1 on v2

This shows that the scalar product of two vectors is the product of the magnitude of one vector and the magnitude of the projection of the other vector on it.

The angle between the two vectors can be expressed in terms of the vector constants as follows:

Vectors

Three dimensional space:

Let a = a1i + a2j + a3k and b = b1i + b2j + b3k

a•b = (a1i + a2j + a3k) • (b1i + b2j + b3k)

a•b = a1b1i•i + a1b2i•j + a1b3i•k + a2b1j•i + a2b2j•j + a2b3j•k + a3b1k•i + a3b2k•j + a3b3k•k

The unit vectors i, j and k have length 1 and are at 90º to each other and so any unit vector when scalar product combined with itself will give:

i•i = 1 x 1 x cos 0º = 1

Whilst any unit vector when scalar product combined with a different one will give: i•j = 1 x 1 x cos 90º = 0

Therefore a•b = a1b1+ a2b2 + a3b3

Vectors

Three dimensional space:

From the diagram the length of OP in terms of the side lengths can be determined as follows:

OP2 = OB2 + BP2 and OB2 = OA2 + AB2

Thus OP2 = OA2 + AB2 + BP2

OP2 = a2 + b2 + c2

P

c

O

a

A

B

b

For our two vectors:

Using,

Vectors

Example

1. For a = 2i - 3j + 4k, b = 5i + 2j + 6k,

find and .

2. For a = -5i + 3j -6 k, b = 2i - 2j + 3k,

find and .

3. For a = 7i - j + 3k, b =i + 3j - 4k,

find and .

4. For a = 2i + 3j + 5k, b = 4i + 2j - 3k,

find and .

Vectors

The Vector Product of Two Vectors

The second product of two vectors is called the vector product or cross product and is defined in terms of its modulus and the magnitudes of the two vectors and the sine of the angle between them. The vector product of vectors oa and ob is written as oa x ob and is defined by:

|oa x ob| = oa ob sinθ, where θ is the angle between the two vectors. The direction of oa x ob is perpendicular to both oa and ob as shown:

b

θ

a

oa x ob

b

ob x oa

θ

a

Vectors

The direction is obtained by considering that a right handed screw is screwed along oa x ob with its head at the origin and if the direction of oa x ob is correct, the head should rotate from oa to ob (left hand diagram – previous slide). If the vector product is reversed then the direction of ob x oa is reversed (right hand diagram).

This oa x ob ≠ ob x oa. The magnitudes are the same (oa ob sinθ) but their directions are 180º displaced i.e.

oa x ob = - ob x oa

Vectors

Three dimensional space:

Once again let a = a1i + a2j + a3k and b = b1i + b2j + b3k

axb = (a1i + a2j + a3k) x (b1i + b2j + b3k)

axb = a1b1ixi + a1b2ixj + a1b3ixk + a2b1jxi + a2b2jxj + a2b3jxk + a3b1kxi + a3b2kxj + a3b3kxk

The unit vectors i, j and k have length 1 and are at 90º to each other and so any unit vector when vector product combined with itself will give:

ixi = 1 x 1 x sin 0º = 0

Whilst any unit vector when vector product combined with a different one will give: ixj = 1 x 1 x sin 90º = 1

Vectors

k

j

i

Three dimensional space:

The direction will be the same as the thirds unit vector:

i x j = k j x i = -k

j x k = i k x j = -i

k x i = j i x k = -j

Therefore

axb = a1b2k - a1b3j - a2b1k + a2b3i + a3b1j - a3b2i

axb = (a2b3 - a3b2)i + (a3b1 - a1b3)j + (a1b2 - a2b1)k

The magnitude of the vector product of two vectors can be found by expressing it in scalar product form and then using the relationship a•b = a1b1+ a2b2 + a3b3

Vectors

k

j

i

Express the cross-product of two vectors in a determinant form

Vectors

Three dimensional space:

Squaring both sides of the vector product equation gives:

|a x b| = a b sinθ so (|a x b|)2 = a2 b2 sin2θ

sin2 θ + cos2 θ = 1 so sin2θ = 1 - cos2θ

So(|a x b|)2 = a2 b2 (1 - cos2θ)

(|a x b|)2 = a2 b2 - a2 b2 cos2θ

But we know that

a•b = a b cosθ therefore a•a= a2 cos0 = a2

And square this then multiply by a2b2

Vectors

Three dimensional space:

Using (|a x b|)2 = a2 b2 - a2 b2 cos2θ

And substituting in gives us…

(|a x b|)2 = (a • a)(b • b) – (a •b)2

Example for the vectors:

a = i + 4j – 2k and b = 2i – j + 3k

Determine a x b and |a x b|

Vectors

Example

1. For a = 2i - 3j + 4k, b = 5i + 2j + 6k,

find and .

2. For a = -5i + 3j -6 k, b = 2i - 2j + 3k,

find and .

3. For a = 7i - j + 3k, b =i + 3j - 4k,

find and .

4. For a = 2i + 3j + 5k, b = 4i + 2j - 3k,

find and .

Vectors