find plane equation which pass by three points?

Question

Find the plane equation which pass by three points
p1=(-1,2,5)
p2=(2,3,.5)
p3=(4,1,.667)

please i need explanation on how to form a plane equation i have many equation but i can not use any of them, i feel confused .....

z-coordinate of third point is .667 not 2/3

Answer 1

I assume you mean that the plane passes "thru" the three points, not "by" them.

p1 = (-1, 2, 5)
p2 = (2, 3, .5)
p3 = (4, 1, .667)

Create the vectors u and v from the three points.

u = p1p2 = <2-(-1), 3-2, .5-5> = <3, 1, -4.5>
v = p1p3 = <4-(-1), 1-2, 5-.667> = <5, -1, -4.333>

Take the cross product to get the normal vector n to the plane.

u X v = <3, 1, -4.5> X <5, -1, -4.333>
u X v = -8.833i + 9.501j - 8k = <-8.833, 9.501, -8>

Use the normal vector n and one of the points, let's use p1, to write the equation of the plane.

-8.833(x - (-1)) + 9.501(y - 2) - 8(z - 5) = 0
-8.833x - 8.833 + 9.501y - 19.002 - 8z + 40 = 0
-8.833x + 9.501y - 8z + 12.165 = 0

Answer 2

Well there are lots of planes that "pass by" these points. But I assume you want the plane that passes THROUGH these three points. The general equation for a plane is
Ax + By + Cz = D

But assuming D is not zero, you can divide by D and get
(A/D)x + (B/D)y + (C/D)z = 1

If we re-index these coefficients as A, B, and C, then we have Ax + By + Cz = 1. Plugging in three points, we have:

-A + 2B + 5C = 1
2x + 3B + 5C = 1
4A + B + 0.667C = 1

You have 3 equations to solve for 3 unknowns. Once you have A, B, and C, you can write the equation of the plane. If by some chance you get "no real solution", then D must have been zero. So try again from the beginning with all the equations being equal to zero.

Was the z coordinate of the third point really 0.667 exactly, or was it 2/3?

Answer 3

you've been given 3 (non-collinear) factors because there is not any thanks to outline a airplane with in simple terms 2 factors - a line would properly be defined given 2 factors, yet an unlimited type of planes intersect at that line. commence with Ax + via + Cz + D = 0 - a widespread equation of a airplane. Plug in all the criteria. as an social gathering, because the airplane passes via (0,0,0) you know that D = 0. you'll come across the different constants via plugging in coordinates for the different factors and eliminating all yet between the three final constants. The very last consistent would not remember - the final equation would properly be elevated via a consistent without replacing the airplane being defined via it. A more advantageous systematic formulation makes use of determinants (see link less than).

Answer 4

if you mean linar eqaution just graph it and ;you would get the equation