** INTRODUCTION TO VECTORS - PART 2**

Example 1:

Let

v= < 2, 5 > andw= < 3, 4 >, find the dot productv w.

v w= 2(3) + 5(4)= 6 + 20

= 14 which is a scalar!

Example 2:

Find a unit vector

uin the direction ofv= < 3, 2 >.Given the formula we need to find the magnitude of vector

v.Given ||

v|| = , we find ||v|| =

It follows that

Note: We just found a unit vector for the vector

vdrawn in Example 1. It lies on top of vectorvand has a "length" of 1. Incidentally, we found the magnitude of vectorvalready in Example 3.

Example 3:

Find a unit vector in the direction of

v= < 0, 7 >.Given the formula we need to find the magnitude of vector

v.Given ||

v|| = , we find ||v|| =It follows that

Example 4:

Write the following vectors in terms of the standard unit vectors

iandj:(a)

v= < 2, 5 >

Please note that a vector written in component form does not include the i and j unit vector and the horizontal and vertical components are separated by commas.

v= 2i+5j

Please note that a vector written in terms of the i and j unit vector is neither in angle brackets nor are the horizontal and vertical components separated by commas.(b)

w= < 5, 1 >

** w ** = 5**i **** j**

(c)

a=< 7, 0 >

a= 7i(d)

b= < 0, 10 >

b= 10j

(e)

u= < 6, 2 >

** u ** = 6**i** 2**j**

(f)

n= < 6, 2 >

** n **= 6**i** + 2**j**

Example 5:

Find the angle between

z= < 4, 3 > andw= < 3, 5 > and round to 1 decimal place.

Using the formula , lets find

||

z|| =||

w|| =and then the dot product

zw= 4(3) + 3(5) = 27Then

and

NOTE: During the process of solving for the angle always try to use given values without rounding first.

Example 6:

Given the following vectors, determine whether they are parallel, orthogonal or neither.

(a)

u= < 6, 4 > andv=< 2, 3 >

Parallel:Two nonzero vectors

uandvare parallel if there is some scalarcsuch thatu=cv. Therefore, can we factor a scalar out of vectoruso that the remaining vector is equal to vectorv?That is, < 6, 4 >

=2 < 3, 2 >. Since < 3, 2 > is not equal to < 2, 3 >, we can state that vectorsuandvare NOT parallel.

Orthogonal:Remember that two vectors

uandvare orthogonal if the dot product equals 0.Since

uv= 6(2) + 4(3) = 0, we can state that vectorsuandvare orthogonal.(b)

u= < 9, 15 > andv= < 3, 5 >

Parallel:Since < 9, 15 > = 3 < 3, 5 >, we can see that vector

uis a multiple of vectorv. This means that the vectors are parallel.(c)

u= < 1, 3 > andv= < 0, 5 >

Parallel:It should be obvious that there is no scalar that allows us to state that

u=cv. We can say with confidence that vectorsuandvare NOT parallel.

Orthogonal:Since

uv= 1(0) + 3(5) = 15, we can state that vectorsuandvare NOT orthogonal because the dot product does NOT equal 0.