Math Is Fun Forum

A.R.

Guest

Parametric Form to Standard Form of a Plane

I need to convert a plane from parametric form to standard form (ax + by + cz = d).

(x, y, z) = (1 + s + t, 3 − 2s + t, 1 − 3s − 3t)

So far I have,

x     1         1         1
y  = 3  + s -2  +  t  1
z     1        -3         3

To get normal vector,

n (dot) q = 0
[a,b,c] (dot) [1,-2,-3] = 0
a - 2b - 3c = 0

n (dot) r = 0
[a,b,c] (dot) [1,1,-3] = 0
a + b -3c = 0

So the two eqns are,

a + b -3c = 0
a - 2b - 3c = 0

Two of the first eqns + 2nd eqn gives,  3a - 9c = 0

a = 3, c = 1

b = 3c - a
   =3(1) - 3
   = 0

So my normal vector is n = [3, 0, 1]

Is it possible to have b = 0 in a normal vector or did I mess up somewhere?

I can continue from here, I'm just unsure if the n is correct with having a 0 in it.

A.R.

Guest

Re: Parametric Form to Standard Form of a Plane

Vector r IS suppose to be [1, 1, -3] Sorry about the mistake.