from parametric to coordinate equation of a plane and reverse

Question

I've been having a few problems with converting from parametric to coordinate equation and from coordinate to parametric.I know it's easy for others,but i just can't seem to get it right. I.e: I have the following parametric equation of a plane,i need to find the coordinate equation:

$$ x=2 + 3\cdot u + 4\cdot v \\ y=4 + 0\cdot u - 1\cdot v \\ z=2 + 3\cdot u + 0\cdot v $$ I would really appreciate if someone could help me understand, and explain the reverse(coordinate $\to$ parametric). Thank you!

Answer 1

I think you have to substitute: $$u = \frac{z-2}{3}$$. Now $$v = 4 - y$$. Substituting: $$x = 2 + 3(\frac{z-2}{3}) + 4(4 - y)$$ and $$x = z - 4y +16$$

Answer 2

From parametric to Cartesian:
solve two of the equations (say the first two) with respect to $u, v$, so to express the two parameters in terms of two of the Cartesian variables (say $x, y$), then substitute this expressions into the last equation to obtain the Cartesian equation of the plane.

From Cartesian to parametric:
Set $x = u, y = v$ that already give two parametric equations, then substitute into the Cartesian equation of the plane to obtain the third parametric equation.

Note: the choice of which pair of equations to solve (in the first case) or which variables to set equal to the parameters (in the second case) depend on the possibility to perform the calculations (the determinant of the coefficients should be non null, etc.)

Answer 3

Vectorially, the parametric equation is

$$\vec P=\vec P_0+u\vec p+v\vec q.$$

Then a vector normal to the plane is $\vec p\times\vec q$ and it suffices to express that the plane contains $P_0$:

$$(\vec p\times\vec q)\cdot\vec P=(\vec p\times\vec q)\cdot\vec P_0,$$ which is of the form $ax+by+cz=d.$

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Conversely, consider the normal vector $\vec n=(a,b,c)$. We need two vectors parallel to the plane, i.e. perpendicular to $\vec n$. Take an arbitrary vector $\vec a$ not parallel to $\vec n$, and compute $\vec p=\vec a\times\vec n$ and $\vec q=(\vec a\times\vec n)\times\vec n$.

Now we need a point on the plane. A straight line orthogonal to the plane drawn from the origin will meet it, so

$$a(\lambda a)+b(\lambda b)+c(\lambda c)=d$$ gives the solution

$$\vec P_0=\frac d{a^2+b^2+c^2}(a,b,c).$$