1.  Use the distance formula: \/[(6 -(-3))² + (1 - 2)² + (2 - 5)²] = \/ (9² + (-1)² + (-3)²)
                                             = \/ (81 + 1 + 9) = \/ 91

 2.  Find two direction vectors using the given points:
      u = <2, 0, -2>       v = <-1,-2,6>
     I used the first and second points to get u and first and third to get v.
     The area of the triangle is ||(1/2)(u x v)||
                  |  i    j    k |
     u x v =  | 2   0   -2 |  =  -4i - 10j - 4k
                  |-1  -2   6  |

    ||(1/2)(u x v)|| = \/(16 + 100 + 16)  =  \/ 132   =  2\/33    =  \/33
                                         2                    2                 2

 3.  Use the distance from point to plane formula:  distance = |vector PQ × n|
                                                                                                   ||n||
      Q = (2, -1,4)   n = <3,5,-2>     We need to find a P in the plane.
      In 3x + 5y - 2z = 11, let x = -1 and z = -2; then y = 2.
      So P = (-1,2,-2).
      vector PQ = < 3, -3, 6 >

      distance = | < 3, -3, 6 > × <3,5,-2>|   =   |9 - 15 - 12|   =    18
                             \/(9 + 25 + 4)                        \/38               \/38

 4.  Use the three points to find two vectors.  I used the first two points and the first and last points.
      u = <-5,8,-1>   and   v = <1,2,0>
      Find their cross product for <a,b,c>.
                    |  i    j    k |
       u x v =  |-5   8   -1 |  =  2i - 1j - 18k
                    | 1   2    0  |
       So <2,-1,-18> are the coefficients in the equation.
       2x - y - 18z = d
       Plug in one of the original points.  I used the first one.
       2(2) - (-1) - 18(2) = d
       4 + 1 - 36 = d
       -31 = d
       So 2x - y - 18z = -31

 5.  Use  cos q =     u × v
                            ||u|| ||v||
       cos q =       |<3,-1,2> × <-1,4,5>|
                     \/(9 + 1 + 4) \/(1 + 16 + 25)

       cos q =         |3|
                     \/(14) \/(42)

       q = 82.893^o

 6.  ( 2 + 4  , 1 + (-1)  ,   5 + 3 )
           2             2               2
      =  (3, 0, 4)

 7.  The area of the parallelogram is ||u x v||.
                    |  i    j    k |
       u x v =  | 1   3    2 |  =  11i + 9j - 19k
                    | 5  -4    1 |

      ||u x v|| = \/(121 + 81 + 361) = \/563 = 23.728

 8.  Center = midpoint of diameter
      ( 2 + 4 , 1 + (-2) , 7 + 1 ) = (3, -1/2, 4)
           2           2            2
       Radius is the distance from the center to one end of the diameter.
       \/[(3 - 2)² + (-1/2 - 1)² + (4 - 7)²] = \/[1² + (-3/2)² + (-3)²] = \/(1 + 9/4 + 9) = \/49/4 = 7/2
       The equation is
       (x - 3)² + (y - (-1/2))² + (z - 4)² = 49/4
       (x - 3)² + (y + 1/2)² + (z - 4)² = 49/4

 9.  Subtract the initial point from the terminal point:
      < -2 - 3, 1 - 1, -8 - 6 > = < -5, 0, -14 >

10.  Use the component vector found in #9.  -5i + 0j - 14k = -5i - 14k

11.  Use the component vector found in #9.
        ||v|| = \/[(-5)² + 0² + 14²] = \/(25 + 196) = \/221 = 14.866

12.  Divide vector v found in #9 by its magnitude found in #11.
       <    -5   ,     0   ,   -14   >   or    <      -5       ,    0    ,    -14    >
          \/221   \/221   \/221                   14.866                14.866

13.  Change the signs on the answer in #12.
       <   5   ,     0   ,    14      >   or    <      5       ,    0    ,    14    >
          \/221   \/221   \/221                   14.866                14.866

14.  Complete the squares:
       3x² + 3y² + 3z² - 12x - 15y - 21z + 100 = 0.
       x² + y² + z² - 4x - 5y - 7z + 33 1/3 = 0.
       x² + y² + z² - 4x - 5y - 7z = -33 1/3
       x² - 4x + y² - 5y + z² - 7z = -33 1/3
       x² - 4x + 4 + y² - 5y + 25/4 + z² - 7z + 49/4 = -33 1/3 + 4 + 25/4 + 49/4
      (x - 2)² + (y - 5/2)² + (z - 7/2)² = 670/12
      Center = (2, 5/2, 7/2)
      Radius = \/(670/12) = 7.472
 
15.  component vector = terminal point - initial point
       < 4, 0, -2 > = < 1, 5, 2 > - < a, b, c >
       < 4, 0, -2 > = < 1 - a, 5 - b, 2 - c >
        4 = 1 - a       0 = 5 - b        -2 = 2 - c
        3 = -a         -5 = -b           - 4 = -c
        -3 = a          5 = b                4 = c
      initial point = (-3, 5, 4)

16.  u x v is a vector orthogonal to both u and v.
                    |  i    j    k |
       u x v =  |-1   2   -3 |  =  -5i - 7j - 3k
                    | 2  -1   -1 |

17.  The volume of a parallelepiped is found by taking the determinant of a matrix made up of vectors
       representing adjacent sides.  The absolute of the determinant is the volume.
      |  0   2   1 |
      |  2   0   4 |  =  0 | 0   4 |  -  2 | 2   4 |  +  1 | 2   0 |  =  0(0 - 8) - 2(-12 - 4) + 1(4 - 0)
      |  1   2  -6 |         | 2  -6 |        |1  -6 |         | 1   2 |
 
      = 0 - 2(-16) + 4 = 32 + 4 = 36

18.  (2,-3,1) can be the initial point
        x - 2 = y + 5 = z - 1 = t  will give the direction vector
          3          4          2

        x - 2  = t       x - 2 = 3t        x = 2 + 3t
          3

         y + 5 =  t     y + 5 = 4t       y = -5 + 4t
            4

         z - 1 = t.       z - 1 = 2t       z = 1 + 2t
           2

        The coefficients of t give the coordinates for the direction vector
 
        Parametric Equations for the line through the point (2, -3, 1) and parallel to the given line are
        x = 2 + 3t
        y = -3 + 4t
        z = 1 + 2t

19.  2x - 5y + 3z = 30
       Find the traces:  xy-trace is   2x - 5y = 30
                                xz-trace is   2x + 3z = 30
                                yz-trace is -5y + 3z = 30
      Find the intercepts for each trace:  x-intercept = (15, 0, 0)
                                                          y-intercept = (0, -6, 0)
                                                          z-intercept = (0, 0, 10)
     Graph each of these points and connect them (draw the traces).  Shade the triangle.

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