Thread: determinantial equation is an equation for the plane though three noncollin pts.

I was given this question for hw and I am pretty confused. It asks: Show that the determinantal equation

det[
x y z 1
x1 y1 z1 1
x2 y2 z2 1
x3 y3 z3 1 ]
=0

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.

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

Sorry I don't understand what you mean by this. If you could elaborate a little that would be great.

1) Since the variables, x, y, z appear only in the first row, expanding on the first row will give
$\displaystyle x\left|\begin{array}{ccc}y_1 & z_1 & 1 \\ y_2 & z_2 & 1\\ y_3 & z_3 & 1 \end{array}\right|- y\left|\begin{array}{ccc}x_1 & z_1 & 1 \\ x_2 & z_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|+ z\left|\begin{array}{ccc}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|- \left|\begin{array}{ccc}x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2\\ x_3 & y_3 & z_3\end{array}\right|= 0$
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 $\displaystyle x_1$, $\displaystyle x_2$, and $\displaystyle x_3$ you will get a determinant with two identical rows- such a determinant is necessarily 0 and so $\displaystyle (x_1, y_1 , z_1)$ is a point on that plane. The same is true for the other two points.

I understand this now. It was starring me right in the face. Thanks for your help.