Parameterising circle in 3D

[vintwc]

Homework Statement

Parameterise the circle whose diameter is the line joining (1,0,0) and (0,0,1) and which lies in the plane x+y+z=1

Homework Equations

1. Let N be a unit normal vector for the plane.
2. Let C be the circle center, and let R be the radius.
3. Let U be a unit vector from C toward a point on the circle.
4. Let V = N x U.
5. Let t be the parameter.
6. A point P is on the circle if...

P = C + R cos(t) U + R sin(t) V

The Attempt at a Solution

This is the first time I am solving this sort of question. By following the procedure, I have N=1/sqrt3(1,1,1), C=(1/2,0,1/2), U=1/sqrt2(1/2,0,-1/2) (from C to (1,0,0)), V=0.5/sqrt6(-1,2,-1) (vector product with N and U). R=1/sqrt2.

I'm not sure whether these values are right.

 

[ladybuggle]

hello warwick mathematician

 

[vintwc]

Can anyone check my results? I have got a feeling its somewhat wrong.

 

[ladybuggle]

edit: whoops, should probably check first. will post my solution below.

 

[ladybuggle]

Ok, so from question 1) on the same sheet, we have the formula for a circle, C in $$\mathbb{R}^{n}$$

$$C:=\{\underline{a}+\underline{v_{1}}\cos(t) + \underline{v_2}}\sin(t) | t \in [0,2\pi]\}$$
Where $$\underline{v_{1}}$$ and $$\underline{v_{2}}$$ are orthogonal and of equal length.

If we draw the graph (or get grapher to do it for us), it looks something like this:
http://img210.imageshack.us/img210/9686/screenshot20100124at220.png
the plane being $$x+y+z=1$$, and the circle being the one we're after (the points marked are $$(1,0,0)$$ and $$(0,0,1)$$

It's not difficult to work out that the centre of the circle is $$\left( \dfrac{1}{2},0,\dfrac{1}{2} \right)$$
Using a bit of trig (try sketching just the x and z axes and looking at the circle from side-on), we find that the radius of the circle is $$\dfrac{1}{\sqrt{2}}$$.

Now we need to find two vectors, $$\underline{v_{1}}$$ and $$\underline{v_{2}}$$, that are orthogonal, lie on the plane $$x+y+z=1$$ and have a length of $$\dfrac{1}{\sqrt{2}}$$.

We needn't worry about the length of the vectors at the moment, we can always multiply them by a scalar at the end to sort that out. The hardest part here is getting them to lie on the plane and be orthogonal to each other.

If we find the normal to the plane $$x+y+z=1$$, it is (fairly obviously) $$\left(1,1,1\right)$$, then any vector that lies on the plane is going to be orthogonal to it (ie $$\underline{v_{1}} \cdot \left(1,1,1\right) = 0, \underline{v_{2}} \cdot \left(1,1,1\right) = 0$$.

This is fairly straightforward: take $$\underline{v_{1}}=\left(0,-1,1\right)$$. This lies in the plane. We now need to find $$\underline{v_2}}$$ so that it is orthogonal to both $$\underline{v_{1}}, \left(1,1,1\right)$$. With a little bit of fiddling, an example would be $$\left(-2,1,1\right)$$.

Just one more little bit before we're done... we need to make sure the length of both vectors is $$\dfrac{1}{\sqrt{2}}$$. Finding the current lengths of them, dividing by them, then multiplying by $$\dfrac{1}{\sqrt{2}}$$, we get:
$$\underline{v_{1}}=\dfrac{1}{2}\left(0,-1,1\right), \underline{v_{2}}=\dfrac{1}{2\sqrt{3}}\left(-2,1,1\right)$$

So altogether, parameterised, we get:
$$C:=\{\left( \dfrac{1}{2},0,\dfrac{1}{2} \right)+\dfrac{1}{2}\left(0,-1,1\right)\cos(t) + \dfrac{1}{2\sqrt{3}}\left(-2,1,1\right)\sin(t) | t \in [0,2\pi]\}$$

I hope this helped - and hope the latex is ok, it's the first time I've used it.
Tom by the way - rather tall, blonde and I sit pretty close to the back ;)

 

[vintwc]

thanks mate. I'm actually a MMORSE student (hoping to transfer to Math next year) and maybe one of the difficulties that I'm facing in this question is that we have not done vectors & matrices during term 1. However, I understand everything in your solution except for the small bit where you get v1=(0,-1,1) which I will leave it for my supervisor to explain to me.