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Vectors, Lines in Space

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					    UNIT 1
VECTORS, LINES
 and PLANES in
    SPACE
OBJECTIVES
 By the end of the unit, you must be
  able to:
  enumerate and apply properties of
   vectors in the plane and in space;
  perform and interpret vector
   operations;
  find the equations of a line and
   equation of a plane in space; and
  identify and sketch cylinders and
   quadric surfaces.
     1.1
VECTORS IN THE
    PLANE
NOTION

 VECTOR
   quantity that has both
   magnitude and direction
 SCALAR
   quantity that only has
   magnitude
Examples
   SCALARS      VECTORS
    speed        velocity
    length    displacement
     time     acceleration
  temperature     force
     mass     magnetic field
    density   electric field
    energy     momentum
 EXAM SCORE       LIFE
JUST FOR FUN


 LIFE IS NOT JUST
         MAGNITUDE
 ABOUT MAGNITUDE.
           DIRECTION
 IT NEEDS DIRECTION.
       -something I overheard
Geometric representation


  magnitude


                          terminal point




                     direction
  initial point
                      5
                      4
                      3
                      2
                      1

-5 -4 -3 -2 -1            1   2   3   4   5
                 -1
                                      
                 -2
                 -3       A or        A
                 -4
                 -5
Illustration 1.

   Consider a vector A with
   initial point at  2 ,  4  and
   terminal point at  5 ,6 .
  Determine its magnitude and
  direction.
Solution
                               5                5 ,6 
                               4

           magnitude
                   3
                               2
                               1

     -5 -4 -3 -2 -1                1   2   3   4   5
                          -1
                          -2
                          -3
                          -4
            2 ,  4    -5
Solution (continued)

 magnitude of A
      d 2 ,4  ,  5 ,6 

       5   2 2  6   4 2

       7 2  10 2
       149
Solution (continued)
                                5                5 ,6 
        10
tan                           4
         7                      3
              10
   Arc tan                   2
                      7         1                          10
        -5 -4 -3 -2 -1              1   2   3   4   5
                           -1
                           -2
                                        
                           -3
                           -4
             2 ,  4    -5
                                    7
Solution (continued)



   Hence, vector A has a
   magnitude of 149 and in
   the direction of Arc tan 10 .
                            7
Representations . . .
A vector has several representations
on the plane depending on the initial
and terminal point.

Position Representation
• initial point at the origin
• direction is measured from the
  positive x-axis in counter
  clockwise direction
Illustration 2. The following are
different representations of one
vector.

   Initial point   Terminal point

      3 , 2         1 ,1 
     5 ,1          1 ,  4 
       1 ,5           5 ,2 
                            5
                            4 ,5 
                             1
                            3
                            2
        3 , 2            1                   5 ,2 

   -5 -4 -3 -2 -1               1     2    3   4   5
                     -1
 5 ,1                   1 ,1 
                     -2
                     -3
                     -4                    4 , 3 
              1 ,  4 
                      -5
Vector in the plane



 A vector is an ordered pair of
 real numbers a , b . a and b
 are called components of the
 vector.
Representation


Position representation of a , b

    initial point:     0 ,0 
    terminal point:    a ,b 
MUST!!!

             x i , yi 
Initial point:
Terminal point:  xt , yt 

VECTOR COMPONENTS:

      xt  xi , yt  yi
Example 1.
Determine the components of
the vector with initial point at
 2 ,  4  and terminal point at
 5 ,6  .
Solution:
  xt  xi , yt  yi
     5    2  ,6    4 
     7 ,10
                      5                     5 ,6 
                      4
                      3
                      2
                      1
                                      7 ,10
-5 -4 -3 -2 -1            1   2   3    4   5
                 -1
                 -2
                 -3
                 -4
     2 ,  4 
               -5
Example 2.
Determine the components of
the vector with initial point at
 3 , 2  and terminal point at
1 ,1  .
Solution:
  xt  xi , yt  yi
     1   3  ,  1  2
     4 , 3
                      5

                      4
                      3
                      2       4 , 3
   3 , 2           1

-5 -4 -3 -2 -1            1   2   3   4   5
                 -1
                 -2
                      1 ,1 
                 -3
                 -4                4 , 3 
                 -5
Equality of vectors

Two vectors are equal if their
magnitudes and directions are
equal.
Vectors a , b and c , d are
equal if and only if a  c and
bd.
Consider nonzero vector A.
The magnitude of A,    A , is the
length of any of its
representations.

The direction angle of A,  A , is
the measure of the angle formed
by the vector with the positive x-
axis in the counterclockwise
direction.
Magnitude and direction

Consider vector     A  a ,b .
       A       a2  b2
                b
      tan  A 
                a
Also, A     A cos A , A sin A .
Example 3.
Determine the magnitude and
direction of vectors A   4 , 4
and B  1 ,  3 .
Solution:
  A   4 ,4
  A   4    2  4 2  32  4 2
           4
 tan A   4  1  A  Arc tan  1
Solution (continued)
                5
                        4
                                    A   4 ,4
                            A
                        3
                        2
                        1

  -5 -4 -3 -2 -1            1   2   3     4   5
                   -1
                                                            
                   -2                   Arc tan  1  
                   -3
                                                            4
                   -4
                   -5
Solution (continued)
                  4
        tan A   4  1
        A    Arc tan  1

            
                   
                4
            3
       A 
             4
Solution (continued)

 B  1 ,    3

 B     1   3 
         2        2    4 2

         3
tan B  1   3

  B  Arc tan 3 
Solution (continued)
B  1 ,    3                  2

 B  Arc tan 3             1
       
B  
       3
                -2   -1            1    2

                          -1           B
             5
 Also,  B     .         -2
              3
Example 4.
Determine the components of
the vector with a magnitude of
                            5
6 units in the direction of    .
                             3
Solution:
   A cos A , A sin A
          5       5
    6 cos 3 ,6 sin 3
Solution (continued)

  A cos A , A sin A
              5       5
        6 cos 3 ,6 sin 3
           1       3
        6   ,6    
           2  2 
        3 , 3 3
                                    5
Solution (continued)
                                    4

        5                          3
                                    2
         3
        vertical component
                                    1

 -5 -4 -3 -2 -1                         1   2   3   4   5
                               -1
                               -2               6 units
                               -3
                               -4                       3 , 3 3
                               -5


                             horizontal component
Operations on vectors
Consider vectors
   A a1 , a2      B  b1 , b2
 SUM: A  B  a1  b1 , a2  b2

NEGATIVE:  A   a1 ,  a2

DIFFERENCE:
        A  B  a1  b1 , a2  b2
Operations on vectors
SCALAR PRODUCT:
       cA ca1 , ca2
     where c is a constant
   If c  1 , cA stretches A .
   If 0  c  1 , cA shrinks A .

   If c  0 , cA reverses A .
Geometrically


    cA , c  0


                             cA , c  1
                            A
                 cA , 0  c  1
Geometrically

                B


                    A

                        AB
Geometrically
AB
                B
 A   B 


                    A
         B             AB
Example 5.
If A   2 , 4 and B  4 , 3 ,
evaluate the following:
                      1 A
   1. A  B       4. 2
   2. A  B       5. 2 A  3 B
  3. 2 A
Also, determine the respective
direction and magnitude.
Solutions
 A   2 ,4    B  4 ,3
 1. A  B   2  4 , 4  3
           2 ,7
      A  B  53
                       7
       A B  Arc tan
                       2
Solutions
 A   2 ,4    B  4 ,3
 2. A  B   2  4 , 4  3
            6 ,1
     AB       37
                      1
       AB  Arc tan    
                      6
Solutions
 A   2 ,4     B  4 ,3
 3. 2 A  2  2  , 2 4 
          4 ,8
    2 A  80  4 5  2 A
    2 A  Arc tan 2      A
Solutions
 A   2 ,4      B  4 ,3

 4. 1 A  1  2  , 1 4 
    2     2          2
              1 ,2
    1 A  5 1 A
    2              2
     1 A Arc tan 2      A
      2
Solutions
 A   2 ,4    B  4 ,3

5. 2 A  3 B  2  2 , 4  3 4 , 3
               4 , 8  12 , 9
               16 ,  1
   2 A  3 B  257
   2 A3 B  Arc tan 16  
                       1
Illustration                      15
 A   2 ,4                       12
                         2A
 B  4 ,3                         9
                                  6
                                  3

               -15 -12 -9 -6 -3        3   6   9 12 15
                                  -3
                                  -6

 2 A  3B                 3B     -9
                                  -12
   16 ,  1                     -15
Unit vector
       A unit vector has a
       magnitude of 1.
  i  1 , 0 : unit vector in the
             direction of positive
             x-axis
  j  0 ,1 : unit vector in the
             direction of positive
             y-axis
Unit vector


 Given A  a , b .

        A  ai  bj
     or A  a 1 , 0  b 0 ,1
Unit vector
 Given A  a , b .

 Unit vector in the direction of A:

                  a   b
       UA          ,
                  A   A
               cos A , sin A
Example 6.
Determine a unit vector in the
direction of 12 ,  5 .
Solution:
Let A  12 ,  5 .

 A  12 2   5 2
    144  25              12  5
                    UA       ,
    169  13              13 13
Illustration 3         10
                       8                   12  5
                       6
                            UA               ,
                                           13 13
                       4
                       2

   -10 -8   -6 -4 -2        2   4    6 8   10
                       -2
                       -4
                       -6
                                    A  12 ,  5
                       -8
                       -10
Example 7.
Determine a unit vector in the
direction of the vector with a
magnitude of 10 in the direction
   
of 6 .
Solution: Let B be the given vector.
  U B  cos B , sin B
                             3 1
       cos 6 , sin 6        2
                                ,2
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Description: A guide for learning vectors and lines.