Given three points P_{1}, P_{2}, P_{3}, we
can define two vectors in the plane determined by
the points as **V _{1}** = P

Suppose P_{1} = (1,1,1), P_{2} = (2,3,1), P_{3}
= (0,4,2). Then **V _{1} = **<1,2,0> and

i j k

1 2 0

-1 2 1

which is <2,-1,4>. (Check that the dot products of this with **V _{1}
**and

<2,-1,4> · <x - 1, y - 1, z - 1> = 0

or 2x - y + 4z = 5

(It's a good thing to substitute each of the three points into this equation to confirm that you did everything right.)

Note that the coefficients of x, y and z are 2, -1 and 4, which is
the normal vector we just found. A short cut for writing the equation
of a plane is to write n_{1}x + n_{2}y + n_{3}z
(in this case, 2x-y+4z) on the left side, and substitute any of the
three points to determine the value on the right side. So using P_{1}
for example we have

2x - y + 4z = 2(1) -1 + 4(1) = 2-1+4 = 5.

How would you find the intersection of a line with a plane? Substitute the parametric representations for x, y and z from the line into the equation of the plane and solve for the value of the parameter. This is what we'll have to do when we start shooting projectiles at objects!