How do I find the equation of a plane having co ordinate points P1)10,20,5 P2)15,25,7 and P3) 27,9,9?

Question

equation of a plane: ax+bY+cZ+d=0. I need to be able to find abc and d for any give plane.

Thanks

P1-3 are xy and z co ordinates. I believe a simultanious equation is required but not sure where to start.

Answer 1

Suppose the equation is
ax + by + cz +d = 0
Then you know
10a + 20b + 5c + d =0
15a + 25b + 7c + d =0
27a + 9b +9c +d = 0
.

Answer 2

How do I find the equation of a plane having coordinate points P1(10,20,5); P2(15,25,7) and P3(27,9,9)?

First create two directional vectors of the plane from the three points.

P1P2 = = <15-10, 25-20, 7-5> = <5, 5, 2>
P1P3 = = <27-10, 9-20, 9-5> = <17, -11, 4>

The normal vector n, to the plane will be orthogonal to both of them. Take the cross product.

n = P1 X P2 = <5, 5, 2> X <17, -11, 4> = <42, 14, -140>

Any non-zero multiple of n is also a normal vector to the plane. Divide by 14.

n = <3, 1, -10>

With the normal vector n, and a point on the plane we can write the equation of the plane.
Let's choose P1(10, 20, 5).

3(x - 10) + 1(y - 20) - 10(z - 5) = 0
3x - 30 + y - 20 - 10z + 50 = 0
3x + y - 10z = 0

Answer 3

P1P2 = <5, 5, 2>
P1P3 = <17, -11, 4>

P1P2 X P1P3 = 44i +14j + 140k = <44,14,140>

ax + by + cz + d = 0

44(x - 10) + 14(y - 20) + 140(z - 140) = 0

Answer 4

Use the determinant form of the plane equation:

|x y z 1|
|x1 y1 z1 1|
|x2 y2 z2 1| =0
|x3 y3 z3 1|

Signed minors are coefficents of x,y, z and D (signed minor of 1)
x(+),y(-),z(+),D(-);