- Vector equation.
A plane passing through a point r0 with a normal vector n has the equation:
n·(r - r0) = 0
In our case we can take, say, r0 = {0, -4, 0} and n = {3, -1, 2}. So:
{3, -1, 2}·{x, y + 4, z} = 0 - Parametric equation.
We take the equation of a plane as an algebraic equation and solve it like a system of equations.
3x - y + 2z - 4 = 0
As the coefficient of y is -1, it's good to take is as the dependent variable.
y = 3x + 2z - 4
x and z are independent. So, we get the parametric equations:
x = x
y = 3x + 2z - 4
z = z
#Vectors-Linear Systems
7 messages · Page 1 of 1 (latest)
Sorry, there was a small mistake. I corrected it.
Oh. Then it's just the parametric equation, just written differently.
In that case:
{x, y, z} = {0, -4, 0} + {1, 3, 0}u + {0, 2, 1}v
Oh, they picked another point. And chose a more difficult vector for some reason. Oh well...
Note that there is an infinite amount of correct answers for a question like this.
You have a choice of infinitely many points r0, for example.
And an infinite amount of choices for the vectors that the plane sits on.
Well, yes, quite a lot of formulas in analytic geometry.