Question about finding equations on different planes

[daon]

I couldn't articulate my question well enough in the allotted space

We all know how to graph lines in 3-space, but I really am just curious about finding the equation of say, a parabola, on a plane other than the xy yz, or xz axis given some information about the parabola.

Lets say a perfect parabola (like y=x^2) exists on a plane x + y + z = 10. And the vertex of the parabola is at (0,0,10). This question is very vague, I know, but I'm not sure what other information is needed here. What would the form of this equation look like?

Thanks.
Daon

 

[pka]

Recall that a parabola is by definition the set of points in a plane that are equally distant from a fixed point and a fixed line in that plane.
Any plane is a two dimensional subspace of \(\displaystyle R^3\).
For example the plane \(\displaystyle x+y+z=10\) is the set of points \(\displaystyle \( x , y, 10-x-y \)\).

To write the equation if a parabola in that plane you need a focus, a point in the plane, and a directrix, a line in the plane not containing the focus.
Now the vertex is also equally distant between these two.
So use the distance formulae.

 

[JakeD]

I couldn't articulate my question well enough in the allotted space

We all know how to graph lines in 3-space, but I really am just curious about finding the equation of say, a parabola, on a plane other than the xy yz, or xz axis given some information about the parabola.

Lets say a perfect parabola (like y=x^2) exists on a plane x + y + z = 10. And the vertex of the parabola is at (0,0,10). This question is very vague, I know, but I'm not sure what other information is needed here. What would the form of this equation look like?

Thanks.
Daon

Here is a way to generate using linear algebra the equations for parabolas that satisfy your conditions. There are an infinite number of such parabolas.

Take the coefficient vector of the equation defining the plane and call it \(\displaystyle \L \mathbf{m}_0 = (1,1,1) .\) Now find two vectors \(\displaystyle \L \bold m_1\) and \(\displaystyle \L \bold m_2\) with length 1 that are orthogonal to \(\displaystyle \L \bold m_0\) and orthogonal to each other. For example set \(\displaystyle \L \bold m_1 = (1,-1,0)/\sqrt{2}\) and \(\displaystyle \L \bold m_2 = (1,1,-2)/\sqrt{6} .\) Let \(\displaystyle \L \bold v = (0,0,10)\) be your vertex.

Then the points \(\displaystyle \L \bold m_1 t + \bold m_2 u + \bold v\) for any \(\displaystyle \L (t,u)\) will all lie on your given plane and the points \(\displaystyle \L \bold m_1 t + \bold m_2 t^2 + \bold v\) for any \(\displaystyle \L t\) will form a parabola with vertex \(\displaystyle \L v .\)

Written in terms of coordinates the points on the plane using the example vectors will be \(\displaystyle \L (t/\sqrt{2}+u/\sqrt{6},-t/\sqrt{2}+u/\sqrt{6},-2u/\sqrt{6}+10)\) and the points on the parabola will be \(\displaystyle \L (t/\sqrt{2}+t^2/\sqrt{6} ,-t/\sqrt{2} +t^2/\sqrt{6} ,-2t^2/\sqrt{6} +10) .\)

Since the vectors \(\displaystyle \L \bold m_1\) and \(\displaystyle \L \bold m_2\) are orthogonal and have length 1 (called orthonormal), distances will be preserved and the transformed parabola will have the same shape as the original. And since there are an infinite number of pairs of orthonormal vectors orthogonal to \(\displaystyle \L \bold m_0,\) there are an infinite number of parabolas satisfying your conditions.

This method gives parametric representations of the parabolas. I don't know (but I doubt) that these can be converted back to a single equation in \(\displaystyle \L (x,y,z).\) It is easy to convert them to two equations by say solving

\(\displaystyle \L z = - 2t^2 /\sqrt{6} + 10\)

for \(\displaystyle \L t\) and then substituting that into

\(\displaystyle \L x = t/\sqrt{2} + t^2 /\sqrt{6},\)
\(\displaystyle \L y = -t/\sqrt{2} + t^2 /\sqrt{6}.\)

Since it takes two equations (two planes) to determine a line in 3 dimensions, it seems generally true it would take two equations to determine a curve like a parabola, except perhaps in some special cases.

 

[daon]

Thank you for the informative replies, but can you answer a couple questions?

1) Why did you need two vectors orthagonal to the plane's normal vector? Although its obvious that these two vectors serve a purpose, its not obvious to me what they are doing.

2) Is it the choice of the unit vectors that govern the shape of your parabola? If so, how would I choose my vectors to generate a "basic parabola" like y = x^2 (so that if I were to conceptually manuver the plane over the graph of y=x^2 they'd be equivilant), and how would I choose my vectors for something like y=2x^2, etc.?

3) Does this method generally work for any 2 dimentional graph in 3 dimentions (like a circle for example)?

 

[JakeD]

Thank you for the informative replies, but can you answer a couple questions?

1) Why did you need two vectors orthagonal to the plane's normal vector? Although its obvious that these two vectors serve a purpose, its not obvious to me what they are doing.

The 2 orthogonal vectors are a basis for the 2-dimensional subspace that is the plane through the origin defined by \(\displaystyle \L x+y+z =0.\) So any point on that plane is a linear combination of these two vectors. The plane defined by \(\displaystyle \L x+y+z = 10\) is translated from the plane through the origin by adding the point \(\displaystyle \L (0,0,10).\)

2) Is it the choice of the unit vectors that govern the shape of your parabola? If so, how would I choose my vectors to generate a "basic parabola" like y = x^2 (so that if I were to conceptually manuver the plane over the graph of y=x^2 they'd be equivilant), and how would I choose my vectors for something like y=2x^2, etc.?

When the basis vectors are orthonormal, the shape of the parabola is preserved exactly. Choosing different orthonormal basis vectors just points the parabola in a different direction. To change the shape, set \(\displaystyle \L y = f(x)\) and use \(\displaystyle \L x \bold m_1 + f(x) \bold m_2 + v\) as your parametric representation of the parabola. Notice I changed the parameter to \(\displaystyle \L x\) from \(\displaystyle \L t\) for the next question.

3) Does this method generally work for any 2 dimensional graph in 3 dimentions (like a circle for example)?

Yes. If you solved the equation for the circle to get \(\displaystyle \L y = f(x)\) you could use the same parametric representation as above.

Another way to represent the circle would be as the set of points \(\displaystyle \L x \bold m_1 + y \bold m_2 + v\) for \(\displaystyle \L (x,y)\) satisfying the equation \(\displaystyle \L x^2 + y^2 = r^2.\) This is probably a better way to view it. The orthonormal vectors are just different basis vectors determining a different plane. Think of the equation of a 2-dimensional circle as applying to the \(\displaystyle \L xy\) plane which has standard basis \(\displaystyle \L \bold m_1 = (1,0,0)\) and \(\displaystyle \L \bold m_2 = (0,1,0)\) in \(\displaystyle \L \mathbf R^3 .\) The \(\displaystyle \L (x,y)\) in the equation are the coefficents of \(\displaystyle \L m_1\) and \(\displaystyle \L m_2 .\) Then all you're doing is switching to a different basis and then translating by adding \(\displaystyle \L \bold v.\)

 

[daon]

Thank you JakeD. It makes sense to me now .