Projection of space curves onto general planes

[Antiderivative]

So I've encountered many "what is the projection of the space curve C onto the xy-plane?" type of problems, but I recently came across a "what is the project of the space curve C onto this specific plane P?" type of question and wasn't sure how to proceed. The internet didn't yield me answers so I haven't made much headway. The problem and my attempt at a solution is outlined below:

Homework Statement

Compute the projection of the curve \vec{\mathbf{r}}(t) = \left\langle \mathrm{cos\:}t, \mathrm{sin\:}t, t \right\rangle onto the plane x + y + z = 0.

Homework Equations

I'm having trouble come up with an equation. I've tried drawing the relevant xy-, yz-, and xz-plane projections and seeing where the curves intersect, but I know that these intersection points do NOT necessarily correspond to the projection of the given curve onto the given plane.

The Attempt at a Solution

See reasoning above. I really don't know how to do this for a non-standard plane and so I'm completely lost as to how to make headway. I haven't been able to find relevant information on the internet either through a similar problem for some reason.

Can anybody help me out? If so, is there a way to do this for ANY plane P and ANY space curve \vec{\mathbf{r}}(t)? I feel like there should be yet Stewart's Multivariable Calculus yields nothing (at least the 5th edition doesn't) in this area. Thank you in advance.

 

[tiny-tim]

Hi Antiderivative!

Choose a basis for the plane.

Then use that basis, and the normal, as a new set of coordinates.

 

[Antiderivative]

Hi tiny-tim! Hmm... okay so the normal vector to the plane is N(1,1,1), and the plane goes through A(1/√2,0,-1/√2). Using that, I guess I can try to find an orthonormal basis. Clearly A•N = 0, so A and N are normal. The other vector can be found by finding N x A presumably.

I'm getting (-1/√2, √2, -1/√2). Since it's an orthonormal basis I'd have to normalize this to get (1/√6, √2/√3, -1/√6).

Okay so my basis vectors are (1/√2, 0, -1/√2) and (1/√6, √2/√3, -1/√6).

There's a way to convert the coordinate systems but I'm not sure I know how to do that. It involves a matrix of some kind presumably?

 

[Antiderivative]

Okay so I figured out that the three basis vectors in my new coordinate system would be

\left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right), \left( \frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}} \right), and \left( -\frac{1}{\sqrt{6}},\frac{\sqrt{2}}{\sqrt{3}},-\frac{1}{\sqrt{6}} \right).

Graphing these on Wolfram Alpha helped me see that they help to create the coordinate system where the new xy-plane is created by where the given plane exists.

Should I find the new \hat{x},\hat{y},\hat{z} in terms of the old x,y,z? I feel like that's so much work for a deceptively simple problem. Or is it actually necessary?

 

[Antiderivative]

Okay just for completion I want to post that I officially figured it out.

Using the change of basis, I ended up rewriting the old x,y,z in terms of the new ones, and deriving an equation for this "flattened" helix on the plane:

\left( \begin{array}{c}<br /> \hat{x} \\<br /> \hat{y} \\<br /> \hat{z} \end{array} \right) =<br /> \left( \begin{array}{c}<br /> \frac{2\mathrm{cos\:}t - \mathrm{sin\:}t - t}{3} \\<br /> \frac{2\mathrm{sin\:}t - \mathrm{cos\:}t - t}{3} \\<br /> \frac{2t - \mathrm{sin\:}t - \mathrm{cos\:}t}{3} \end{array} \right)

Graphing this in Mac's Grapher program or an equivalent software produces the attached diagram, which is what we're going for. Thanks tiny-tim for helping me visualize/understand the process!

 

[LCKurtz]

You can do it without changing basis vectors. Let's say you have a point $P_0$ on the plane and a unit normal $\hat n$ pointing to the side that $P_0$ is on. Let your curve be $\vec r(t)$. Then the projection curve is $\vec p(t)=\vec r(t) - (\vec r(t)-P_0\cdot \hat n)\hat n$.

I can give you more details later but have to run now.

 

[Antiderivative]

Yep I just used that method after the orthonormal one. In the end it's kind of the same thing because you'll end up getting v – (n•v/n•n) n, which is like the Gram-Schmidt process for creating an orthonormal basis. Analogous procedures! Geometric versus algebraic arguments I suppose.

 

[Charlie H]

You can do it without changing basis vectors. Let's say you have a point $P_0$ on the plane and a unit normal $\hat n$ pointing to the side that $P_0$ is on. Let your curve be $\vec r(t)$. Then the projection curve is $\vec p(t)=\vec r(t) - (\vec r(t)-P_0\cdot \hat n)\hat n$.

I can give you more details later but have to run now.

Could you please show me the details about your method please? I am stucking in this problem for a long time.thank you in advance!

 

[Charlie H]

Okay so I figured out that the three basis vectors in my new coordinate system would be

\left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right), \left( \frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}} \right), and \left( -\frac{1}{\sqrt{6}},\frac{\sqrt{2}}{\sqrt{3}},-\frac{1}{\sqrt{6}} \right).

Graphing these on Wolfram Alpha helped me see that they help to create the coordinate system where the new xy-plane is created by where the given plane exists.

Should I find the new \hat{x},\hat{y},\hat{z} in terms of the old x,y,z? I feel like that's so much work for a deceptively simple problem. Or is it actually necessary?

Could you please show me more details about your basis changing ? The result is right in normal ways, but I can't get that in your method. thank u in advance.