finding the equation of a plane.?

Question

ok...say that a plane contains the points (1,2,3),(-4,2,-1), and (5,-3,0). how do i find the equation of this plane? how do i find a vector perpendicular to this plane?

Answer 1

Let the equation be ax + by + cz = d and sub in all three points. Solve the three resulting equations for a, b, c, d. Actually, three equations in four unknowns will only allow you to solve for three of them in terms of the other one, e.g. you might get
a = ...*d; b = ...*d etc. Choose any value you like for d, e.g. if one answer was a = (2/7)*d you might choose d = 7 to avoid fractions.

To find a vector perpendicular to the plane, I have a hunch it might be (a, b, c) but anyway here's how I work it out:

Find two vectors in the plane and work out their vector product:

The join of (1,2,3) and (5,-3,0) is (5-1, -3-2, -3) = (4, -5, -3)
Join (1,2,3) (-4,2,-1) gives (-5,0,-4)

The vector product is (-5*(-4) - (-3)*0, (-3)*(-5)-(-4)*4, 4*0 - (-5)*(-5))
This gives
(20, 31,-25)


I hope by now you've done the first bit for yourself, but here's my effort at it:

a + 2b +3c = d
-4a + 2b - c = d
5a -3b = d

Subtract second from first:
5a + 4c = 0
Hence c = -5a/4

Subtract first from third:
4a - 5b - 3c = 0

Sub c = -5a/4:
4a - 5b +15a/4 = 0
Hence
b = 31a/20

Now choose a = 20 so that b = 31 and c = -25

Then third equation gives
d = 5*20 - 3*31 = 7
and the equation is
20x + 31y - 25z = 7

PS Jim's method for finding the perpendicular vector will give the correct result only if the plane passes through the origin.

Answer 2

In some ways, it's the same question.

Take the second one first.  If the three points are A, B and C, then AB and AC are two vectors parallel to the plane.  The cross product of AB and AC will be a vector perpendicular to the plane.  Call this vector P.

Once you have a perpendicular vector, the plane is the set of points whose vector from any (and all) of the points A, B and C is perpendicular to P.  If the vector is called x, the P.x = 0.  A little bit of algebra and you're done.

Answer 3

you may both use a in problem-free words algebraic attitude to bathe up it as 3 equations in 3 unknowns or use a geometrical attitude. I choose the latter. we've the three given factors A(3,-a million,2); B(8,2,4): and C(-a million, -2, -3). 2 directional vectors interior the wanted airplane are: AB = = <8 - 3, 2 - -a million, 4 - 2> = <5, 3, 2> AC = = <8 - -a million, 2 - -2, 4 - -3> = <9, 4, 7> the traditional vector n, of the airplane is orthogonal to both directional vectors. Take the bypass product. n = AB X AC = <5, 3, 2> X <9, 4, 7> = With a level interior the airplane and the traditional vector to the airplane we may be able to write the equation of the airplane. enable's choose factor B(8, 2, 4). undergo in ideas, the traditional vector is orthogonal to any vector that lies interior the airplane. And the dot manufactured from orthogonal vectors is 0. outline R(x,y,z) to be an arbitrary factor interior the airplane. Then vector BR lies interior the airplane. n • BR = 0 n • = 0 • = 0 thirteen(x - 8) - 17(y - 2) - 7(z - 4) = 0 13x - 104 - 17y + 34 - 7z + 28 = 0 13x - 17y - 7z - 40 2 = 0

Answer 4

I think the technique described here is easiest:

http://local.wasp.uwa.edu.au/~pbourke/geometry/planeeq/

and the easiest way to find a normal vector is to pick one of the three points, subtract its components from the other two points (i.e. creating vectors between them), and finding the cross-product of the two vectors.

(Thanks, Hy, I fixed it.)