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