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

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

Vector Representation

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

VA vector may be represented by a letter in bold

faced type, e.g. VFor ease of typing or word processing, vectors

will be represented by the bold faced type.

Components of a Vector

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

Graphically the vector is broken into its components as follows:

Y Vy V

X Vx

V = Vx + Vy

Vector Addition

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

The components of a vector add up to form the vector itself, i.e.

V = Vx + Vy in 2 dimensions

V Vy

Vx

Or in three dimensions

V = Vx + Vy + Vz

Components in 3-d

Z

V Y Vx Vz

Vy

X

When we resolve a vector into its components, e.g.

V = Vx + Vy

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

|Vx | = |V| cos ϴ

| 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 i j k

It is convenient to define three unit vectors i parallel to the X axisj parallel to the Y axisk 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 = Vxi where Vx = | Vx |

Z

k j Y I X

Dot or Scalar Product

The dot or scalar product of two vectors A · BIs a scalar quantity.

A = Axi + Ayj + Azk

B = Bxi + Byj + Bzk

A · B = |A||B cos ϴ

A · B = AxBx + AyBy + AzBz

Example of Dot Product

Consider A = 2i + j – 3k B = -i - 3j + kA·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 = 3.32A·B = (3.74)(3.32) cos ϴ = 12.41 cos ϴ = - 8cos ϴ = - 8/12.41 = - 0.645ϴ = cos-1 (- 0.645) = 130.2⁰

Given a vector 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.

• How many equations do you need to solve for three unknowns?

• I can solve for three unknowns with only this one equation!

A·C = 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 cross product of two vectors produces a third vector which is normal to the first two vectors, i.e.

C = A x BSo vector C is normal to both A and B.

Calculation of 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 Ay Az

Bx By Bz

To evaluate the determinant, it is convenient to write the i and 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 Ay Az Ax Ay

Bx By Bz Bx By

C = AyBzi + AzBxj + AxByk - AzByi – AxBzj - AyBxk

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

Example of cross product

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 = (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)kC = -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 = 0So we have not made any mistakes in calculating

C.

The cross product is different if the order is reversed, i.e.

A x B = CBut

B x A = - CB x A = - A x B

When we look at the vector C = -7i – 7j – 7k

It has the same direction as the vectorC’ = - i – j – k

But a different magnitude.|C| = 7 |C’|

Unit Vectors

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

= [14]1/2 = 3.74g = (2/3.74)i + (1/3.74)j – (3/3.74)k

Useful Information

A · A = |A|2

i · i = 1 j · j = 1 k · k = 1

A x A = 0i x i = 0 j x j = 0 k x k = 0i x j = k j x k = i k x i = jj x i = -k k x j = -i i x k = -j