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# Vectors - PowerPoint PPT Presentation

Vectors. Definitions. Scalar – magnitude only Vector – magnitude and direction I am traveling at 65 mph – speed is a scalar. It has magnitude but no direction. I am traveling north at 65 mph – speed is a vector. It has both magnitude and direction. Graphical Representation.

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## PowerPoint Slideshow about ' Vectors' - toki

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### Vectors

Scalar – magnitude only

Vector – magnitude and direction

I am traveling at 65 mph – speed is a scalar. It has magnitude but no direction.

I am traveling north at 65 mph – speed is a vector. It has both magnitude and direction.

The vector V is denoted graphically by an arrow.

The length of the Arrow represents the magnitude of the vector.

The direction of the arrow represents the direction of the vector.

V

A vector may be represented by a letter with an arrow over it, e.g.

V

A vector may be represented by a letter in bold faced type, e.g.

V

For ease of typing or word processing, vectors will be represented by the bold faced type.

In two dimensions, a vector will have an x-component (parallel to the X-axis) and a y- component (parallel to the Y-axis).

In vector terms,

V = Vx+ Vy

Y

VyV

X

Vx

V = Vx+ Vy

Vectors may be added graphically by placing the tail of the second vector at the head of the first vector and then drawing a new vector from the origin to the head of the second vector.

B

C C = A + B

A

V = Vx+ Vyin 2 dimensions

V

Vy

Vx

V = Vx+ Vy+ Vz

Z

V

Y

VxVz

Vy

X

V = Vx+ Vy

the magnitude of the two component vectors is given by the relations

|Vx| = |V| cosϴ

|Vy| = |V| sin ϴ

The Pythagorean theorem then gives a relation between the magnitudes of the x and y components, i.e.

|V|2 = |Vx |2 + |Vy |2 in 2-dimensions

And

|V|2 = |Vx |2 + |Vy |2 + |Vz |2 in 3-d

Use of Unit Vectors magnitudes of the x and y components, i.e.i j k

It is convenient to define three unit vectors

iparallel to the X axis

j parallel to the Y axis

k parallel to the Z axis

And to express the components of the vector in terms of a scalar times the unit vector along that axis.

Vx = Vxiwhere Vx = |Vx|

Z magnitudes of the x and y components, i.e.

k

j Y

I

X

Dot or Scalar Product magnitudes of the x and y components, i.e.

The dot or scalar product of two vectors A · B

Is a scalar quantity.

A = Axi + Ayj + Azk

B = Bxi + Byj + Bzk

A · B = |A||Bcosϴ

A · B = AxBx + AyBy + AzBz

Example of Dot Product magnitudes of the x and y components, i.e.

Consider A = 2i + j – 3k

B = -i - 3j + k

A·B = (2)(-1) + (1)(-3) + (-3)(1) = -2-3-3 = -8

|A| = [22 + 12 + (-3)2]1/2 = [14]1/2 = 3.74

|B| = [(-1)2 + (-3)2 + (1)2]1/2 = [11]1/2 = 3.32

A·B = (3.74)(3.32) cosϴ = 12.41 cosϴ = - 8

cos ϴ = - 8/12.41 = - 0.645

ϴ = cos-1 (- 0.645) = 130.2⁰

Given a vector magnitudes of the x and y components, i.e.A = 2i + 3j – k, we can find a vector C that is normal to A by using the fact that the dot product A·C = 0 if A is normal to C.

C = Cxi + Cyj + Czk

A·C = 2Cx + 3Cy – Cz = 0

You now have three unknowns and only one equation.

A·C magnitudes of the x and y components, i.e.= 2Cx + 3Cy – Cz = 0

Let Cy = 1

2Cx + 3 – Cz = 0

Let Cx = 1

2 + 3 – Cz = 0

Cz = 5

So the vector C = i+ j +5k is normal to A.

The order of the vectors in the dot product does not affect the dot product itself, i.e.

A · B = B · A

Cross (Vector) Product the dot product itself, i.e.

The cross product of two vectors produces a third vector which is normal to the first two vectors, i.e.

C = A x B

So vector C is normal to both A and B.

Calculation of the dot product itself, i.e.C = A x B

If the vectors A and B are

A = Axi + Ayj + Azk

B = Bxi + Byj + Bzk

then

i j k

C = Ax AyAz

Bx By Bz

To evaluate the determinant, it is convenient to write the the dot product itself, i.e.iand j columns to the right and multiply along each of the diagonals 1, 2, and 3 and add them, then multiply along 4, 5, and 6 and subtract them.

1 2 3 4 5 6

i j k i j

C = Ax AyAzAx Ay

Bx By BzBxBy

C the dot product itself, i.e. = AyBzi+ AzBxj + AxByk- AzByi– AxBzj - AyBxk

Or

C = (AyBz – AzBy)i+ (AzBx– AxBz)j + (AxBy– AyBx)k

Example of cross product the dot product itself, i.e.

Calculate C = A x B where

A = i + 2j - 3k

B = 2i - 3j + k

i j k

C = 1 2 -3

2 -3 1

C the dot product itself, i.e.= (2)(1)i + (-3)(2)j + (1)(-3)k

- (-3)(-3)i– (1)(1)j – (2)(2)k

C = (2 – 9)i + (- 6 – 1)j + (- 3 -4)k

C = -7i -7j -7k

To check that we have not made any mistakes in calculating the cross product, we can calculate the dot product C·A which should be equal to 0 since vector C is normal to vectors A and B.

C·A = (-7)(1) + (-7)(2) + (-7)(-3) = -7 – 14 +21 = 0

So we have not made any mistakes in calculating C.

A x B = C

But

B x A = - C

B x A = - A x B

C = -7i – 7j – 7k

It has the same direction as the vector

C’ = - i – j – k

But a different magnitude.

|C| = 7 |C’|

Unit Vectors i.e.

To create a unit vector g in the same direction as G, simply divide the vector by its own magnitude, e.g.

g = G/|G|

If G = 2i + j – 3k

Then |G| = [22 + 12 + (-3)2]1/2 = [4 + 1 + 9]1/2

= [14]1/2 = 3.74

g = (2/3.74)i+ (1/3.74)j – (3/3.74)k

A · A = |A|2

i · i = 1 j · j = 1 k · k = 1

A x A = 0

i x i = 0 j x j = 0 k x k = 0

i x j = k j x k = i k x i = j

j x i = -k k x j = -ii x k = -j