How many planes pass through #3# distinct points in #3# dimensional space?

1 Answer
Jul 12, 2015

If the points are colinear then an infinite number of planes can be made to pass through them. If three distinct points are non-colinear then exactly one plane passes through them.

Explanation:

Let us look at the general case.

The general equation of a plane is #ax+by+cz = d# for constants #a#, #b#, #c#, #d#.

For a given plane, this equation is unique only up to a constant factor. That is, the same plane is also described by the equation:

#kax+kby+kcz=kd# for any #k != 0#

Let the three points be #(x_1, y_1, z_1)#, #(x_2, y_2, z_2)# and #(x_3, y_3, z_3)#.

Then we have a system of three linear equations:

#ax_1 + by_1 + cz_1 = d#

#ax_2 + by_2 + cz_2 = d#

#ax_3 + by_3 + cz_3 = d#

In the general case, this is a little painful to deal with using substitution and/or elimination, so permit me to show how to do it using matrix arithmetic.

Let #M = ((x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3))#

Then our three equations become:

#M((a), (b), (c)) = ((d), (d), (d))#

If we can find #M^(-1)# then we get:

#((a), (b), (c)) = M^(-1)M((a), (b), (c)) = M^(-1)((d), (d), (d))#

Let:

#X_1 = det((y_2, z_2), (y_3, z_3)) = y_2z_3 - z_2y_3#
#Y_1 = z_2x_3 - z_3x_2#
#Z_1 = x_2y_3 - y_2x_3#
#X_2 = y_3z_1 - z_3y_1#
#Y_2 = z_3x_1 - x_3z_1#
#Z_2 = x_3y_1 - y_3x_1#
#X_3 = y_1z_2 - z_1y_2#
#Y_3 = z_1x_2 - x_1z_2#
#Z_3 = x_1y_2 - y_1x_2#

The determinant of #M# is given by the formula:

#det(M) = x_1X_1 + y_1Y_1 + z_1Z_1#

If #det(M) != 0# then #M^-1# is given by:

#M^(-1) = 1/det(M)((X_1, X_2, X_3), (Y_1, Y_2, Y_3), (Z_1, Z_2, Z_3))#

How can this go wrong?

If the points are not distinct, then two of the rows of #M# will be identical and #det(M) = 0#

If the points are colinear, then #det(M) = 0#

If the plane also passes through #(0, 0, 0)# then #det(M) = 0#

This last case is the one where #d = 0#, so #M^(-1)((d),(d),(d))# would be fairly useless anyway.