# Anmath14's question

### Anmath14's question

Given three points p=(2,1,3), q=(1,0,1) and r=(2,-1,1). Find the Cartesian equation of the plane containing those points.

Any plane (except one parallel to the xy-plane which would have every z component the same- and that doesn't happen here) can be written in the for z= Ax+ By+ C for constants A, B, and C.

Since the point (2, 1, 3) lies in the plane, x= 2, y= 1, z= 3 satisfy the equation: 3= 2A+ B+ C.
Since the point (1, 0, 1) lies in the plane, x= 1, y= 0, z= 1 satisfy the equation: 1= A+ C.
Since the point (2, -1, 1) lies in the plane, x= 2, y= -1, z= 1 satisfy the equation: 1= 2A- B+ C.
Solve those three equations for A, B, and C.

For example, I observe that the second equation does not involve y while the first and third equations involve B with opposite coefficients! Adding the first and second equations, 4= 4A+ 2C so we have 1= A+ C and 2= 2A+ C. Subtracting the first of those two equations from the second eliminates C: 1= A. Then C= 0 and 1= 2- B so B= 1.

The equation of the plane is z= x+ y.

Check: 3= 2+ 1, 1= 1+ 0, and 1= 2- 1.
HallsofIvy

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