Just verify that:
1. the equation is linear about x, y, z
2. plugin (xi,yi,zi) into the equation make the left side zero
I was given this question for hw and I am pretty confused. It asks: Show that the determinantal equation
x y z 1
x1 y1 z1 1
x2 y2 z2 1
x3 y3 z3 1 ]
is an equation for the plane though the three non collinear points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3).
Any help is really appreciated.
1) Since the variables, x, y, z appear only in the first row, expanding on the first row will give
or Ax- By+ Cz- D= 0 where A, B, C, and D are those numeric determinants. That's a linear equation in three variables and so is the equation of a plane.
If you replace x, y, z with , , and you will get a determinant with two identical rows- such a determinant is necessarily 0 and so is a point on that plane. The same is true for the other two points.