Vector Equation:

OP = OB + t(OA)
(x,y) = (-3,-2) + t(3,7)
 

Equations of Lines:

Find equations of the line that passes through (-2,1) and (1,3).

(a)  Direction Vector:  (3,2).  This is a component vector for the two points.
      Vector Equation:  (x,y) = (-2,1) + t(3,2)
              Choose a starting point which will be one of the two points above.   Add the direction vector times t to find
                other points on the line.

(b)  Parametric Equation:
                             x = -2 + 3t  and  y = 1 + 2t
      Add the vectors in the vector equation.

(c)  Cartesian Equation:
       y = 2/3 x + 7/3
      The slope can be obtained from the direction vector by writing y/x.
      Substitute in one of the given points to find b.

Lines in Space:

(x,y,z) = (x1, y1, z1) + t(a,b,c)
          where (x1, y1, z1) is a point in the plane and (a,b,c) is a direction vector

Parametric equations:
x = x1 + at
y = y1 + bt
z = z1 + ct
 

Symmetric Equations:

Solve for t in each part of the parametric equations to obtain x - x1  =  y - y1  = z – z1
                                                                                              a            b             c

Example:

Find parametric and symmetric equations of a line passing through (5,2,10) and parallel to v = 4i + 3k

Answer:
Vectors are parallel when they have the same direction vectors, so a good direction vector is <4, 0, 3>.
So the vector equation becomes  (x,y,z) = <5,2,10> + <4,0,3> t.

Parametric Equations:  x = 5 + 4t
                                   y = 2 + -2t
                                   z = 10 + 3t

Symmetric Equations:  x - 5   =  y - 2   =   z - 10
                                     4             -2            3
 
 

Standard Equation of a Plane in Space:

a(x – x1) + b(y – y1) + c(z – z1) = 0 where the plane contains (x1,y1,z1) and has a normal vector n = <a,b,c>
   (Normal means perpendicular)

Regroup:
ax - ax1 + by - by1 + cz - cz1 = 0
ax + by + cz = ax1 + by1 + cz1
  ax1 + by1 + cz1 is a constant so call it d
ax + by + cz = d
ax + by + cz + d = 0

Remember:  Cross products are always normal to the two vectors involved.

Example:
Find the equation of a plane containing A(4,-1,3), B(2,5,1), and C(-1,2,1).
Find a normal vector to the plane by first finding two component vectors contained in the plane.
AB = <-2,6,-2>   and   BC = <-3,-3,0>  Any two component vectors will work.
Find a normal to these vectors by finding their cross products.

|   i    j    k |
| -2   6  -2 |  =  (0 - 6) i - (0 - 6) j + (6 - (-18)) k  =  -6 i + 6 j + 24 k = <-6,6,24>
| -3  -3   0 |

This is <a,b,c>.  Choose a point in the plane for (x1,y1,z1) and write the equation.
-6x + 6y + 24y = (-6)(4) + 6(-1) + 24(3)
-6x + 6y + 24y = -24 - 6 + 72
-6x + 6y + 24y = 42
 

Perpendicular Planes:

Two distinct planes are parallel or intersect in a line.

The angle between the normal vectors to the planes is equal to the angle between the two planes.
        Use cos q =         u×v
                                 ||u|| ||v||

Example:
Find the angle between two planes given by
  3x + 2y –   z = 7
    x -  4y + 2z = 0
and find parametric equations for their line of intersection

Answer:
The normal vector to the first plane is <3, 2, -1> and the normal vector to the second plane is <1, -4, 2>.
 cos q =                     <3, 2, -1>×<1, -4, 2>                     =   3 - 8 - 2   =    -7     =   -1
                    ||\/(32 + 22 + (-1)2)|| ||\/(12 + (-4)2 + 22)||          \/14 \/21       7\/6         \/6

cos-1 (-1/ \/6 )  =  114.095o

The angle between the two planes is 114.095o

To find the equation of their intersection, find the line that is parallel to the cross product of their normal vectors..
  |  i    j   k |
  | 3   2  -1|  =  (4 - 4) i - (6 - (-1)) j + (-12 - 2) k = 0 i - 5 j - 14 k  =  <0,-5,-14 >
  | 1  -4   2|
This is the direction vector for the line of intersection.

A point shared by these two planes is (2,1,1).
So the equation is 0x - 5y - 14z = 0(2) - 5(1) - 14(1)
                                 -5y - 14z = -19
 

Sketching Planes in Space:

Sketch the traces and draw a triangle.  Then shade it.
 

Distance Between a Plane and a Point Q not on the Plane:

D = ||projuPQ|| = |PQ × n|
                                 ||n||
where P is a point in the plane and n is normal to the plane.

Example:
Find the distance between Q(1,2,3) and 2x – y + z = 4

n = <2,-1,1>
To find a point in the plane, let y = 0 and z = 0 to find x = 2.  P = (2,0,0)
Find vector PQ:  <1 - 2, 2 - 0, 3 - 0>  =  <-1, 2, 3>
Now, apply the formula:

D =  |PQ × n|  =  |<-1, 2, 3> × <2,-1,1>|  =  | -2 - 2 + 3|    =     1
          ||n||              \/22 + (-1)2 + 12)           \/(4 + 1 + 1)        \/6


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