Question

Show that the equation of the plane passing through the points (3,3,1), (3,21)and(8,6,3)is 4x+2y-13z=5

Answer 1

Since the plane includes the three points, given above, then each of these three points can be substituted in place of x, y, and z, in the equation ax+by+cz=1 to give three equations in three unknowns.  The unknowns, in this case, are the constants of the equation, a, b, and c.  The three equations are:

-1a +4b +3c = 12a +b +3c = 14a +2b +c = 1

Using Cramer's Rule, we solve the equations for a, b, and c by finding the determinant of the coefficient matrix, and of three special matrices obtained by replacing each of the three columns in turn by the constants {1,1,1}.  Here it is:

| -1 4 3| | 2 1 3| = 45 for the coefficient matrix| 4 2 1|       | 1 4 3| | 1 1 3| = 6  for the "a" column| 1 2 1|       | -1 1 3| | 2 1 3| = 6 for the "b" column| 4 1 1|       | -1 4 1| | 2 1 1| = 9 for the "c" column| 4 2 1| 

The values of a, b, and c are as follows:
a = 6/45 = 2/15
b = 6/45 = 2/15
c = 9/45 = 1/5

The equation that contains all three points is
(2/15)x + (2/15)y + (1/5)z = 1

Multiplying through by the common denominator, 15, we get
2x + 2y + 3z = 15

It's always a good idea to check the answers.  So I plugged the three points (-1,4,3), (2,1,3), and (4,2,1) into this equation, and verified the value of 2x+2y+3z is indeed 15 for each point.