Unit Vectors & Direction Angles

What is a unit vector?

A vector with a magnitude of 1.

What are the basis unit vectors?

i j= =10 01, _&_ ,

What is a unit vector?

A vector with a magnitude of 1.

What are the basis unit vectors?

i j= =10 01, _&_ ,

i = 1 0,j = 01,

Any vector can be written in terms of i and j ….

A ebraicallya b a b a b ai bjlg :, , , , ,= + = + = +0 0 10 01

How about graphically?

Any vector can be written in terms of i and j ….

A ebraicallya b a b a b ai bjlg :, , , , ,= + = + = +0 0 10 01

i j

a,b( )

a, 0( )

0,b( )ai + bj is the Component Vector

component vectoru a b ai bj

_ :,= = +

Rules of addition, subtraction, and scalar multiplication apply to component vectors.

− + = − −2 3 2 6 4( )i j i j

2i ! 3 j( )! i + 5 j( ) = i !8 j

3i + 5 j( )+ (2i ! 7 j) = 5i ! 2 j

Remember the unit vector? It has a magnitude of 1.

Givenu is a vector where u

:_ _ _ , _ ≠ 0

Remember the magnitude of a vector? Magnitude = l u l

Consider vector

uu it has the same direction as u

but its magnitude is

_ :

_ _ _ _ _ _

_ _ _ _

1

1

→

The process of finding the unit vector in the direction of a given vector is called Normalizing the Vector.

recall

u a b

:

= +2 2 Get ready, here comes the definition of the normalized vector.

unit vector of u a b where u equals

uu a

a bb

a b

_ _ _ , , _ _ ...

,

= ≠

=+ +

0

12 2 2 2

Normalizing a vector:

Find the unit vector in the same direction as u = 4i – 3j

1. Find the magnitude of u.

u = + − = =4 3 25 52 2( )

2. Now use the normalized vector formula.

normalized vectoruu_ = 1

= − = −154 3 4

535

( )i j i j

We can find the direction angle of a vector using Trigonometry

a,b( )

a

b

θu

cos! = au

! = cos!1 au

"#$

%&'

or

Alsoba

....

tanθ =sin! = b

u

! = sin!1 bu

"#$

%&'

We can now write the vectors in trigonometric form

a,b( )

a

b

θu

cos , ......

cos

θ

θ

=

=

auso

a u

sin , ......

sin

θ

θ

=

=

buso

b uSo, u = ai + bj becomes ….

ai + bj = u cos!( )i + u sin!( ) j

Example:

Find the direction angle of u = 5i + 12j

5,12( )

θ

If we use tangent we won’t have to find the magnitude of u.

tan! = 125

! = tan!1 125

"#$

%&' ( 67.4

!

Finding the components of a vector, u u

direction angle

=

=

254

_ ,θ π

1. Relate the component form to the trigonometric form.

ai + bj = u cos!( )i + u sin!( ) j

u = u cos 5!4

!"#

$%& i + u sin 5!

4!"#

$%& j

u = 2 ! 22

"#$

%&'i + 2 ! 2

2"#$

%&'j u i j= − −2 2