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 ;)
