INTRODUCTION TO VECTORS - PART 1

Find the component form of a vector with initial point P at (1, 2) and terminal point Q at (4, 4). Name the vector v.

Given and P and Q , we find the component form of the vector to be

< 4 1, 4 2 >, which equals v = < 3, 2 >

Please note that we moved the vector PQ to standard position without changing its magnitude and direction.

Example 2:

Calculate the EXACT magnitude of v = < 3, 2 >.

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

Example 3:

Calculate the EXACT magnitude of v = < 0, 7 >.

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

Example 4:

Calculate the EXACT magnitude of v = < 4, 3 >.

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

Example 5:

Calculate the EXACT magnitude of v = < 8, 0 >.

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

Example 6:

Find the direction angle of vector .

Let's look at a picture of the vector and its direction angle. Remember that a direction angle is positive and measured from the positive x-axis to the vector in standard position as shown below!

From earlier work we know that , therefore, we find

and

Looking at the graph of the vector, we know that we are required to find a positive angle whose terminal side is in Quadrant IV.

We will use the Reference Angle of 60^o which is 60^o to find that the direction angle must equal 300^o.

Example 7:

Find the direction angle of vector .

Let's look at a picture of the vector and its direction angle.Remember that a direction angle is positive and measured from the positive x-axis to the vector in standard position as shown below!

Knowing that , we can find

and

Looking at the graph of the vector, we know that we are required to find a positive angle whose terminal side is in Quadrant IV. We will use the Reference Angle of 30^o which is 30^o to find that the direction angle must equal 210^o.

Example 8:

Let v = < 2, 5 > and w = < 3, 4 >.  Find v + w.

      v + w = < 2, 5 > + < 3, 4 >

                  = < 2 + 3, 5 + 4>

                  = < 1, 9>

Example 9:

Let v = < 2, 5 > and w = < 3, 4 >.  Find the following:

(a)  v

      v = (1)v

                = (1)< 2, 5 >

                = < (1)(2), (1)(5) >

                = <2, 5>

(b)  2v

      2v = 2< 2, 5 >

           = < 4, 10 >

(c)  w v

      w v = w + (1)v (Adding the negative of vector v!)

                = < 3, 4 > +  (1)< 2, 5 >

                = < 3, 4 > + < 2, 5 >

                = < 5, 1 >

(d)  v + 2w

      v + 2w = < 2, 5 > + 2< 3, 4 >

                  = < 4, 13 >