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Let $x,y \in \mathbb{R}^3: x = (0,0,-1) , y = (0,1,0)$ . Find the Great Circle that contains $x,y$.

From theory, a Great Circle is the intersection of $S^2 = \{x \in \mathbb{R}^3: ||x|| = 1\}$ and the plane $P = ax+by+cz=0, |a|+|b|+|c|\neq0, a,b,c\in \mathbb{R}$.

To find $a,b,c$: Put $x,y$ in the equation of $P$ and that yields that $b=c=0$. That leaves me with $ax=0$ and since $a$ can't be $0$, $x$ must be zero. That means that the Great Circle has equation $y^2+z^2=1$.

That doesn't feel like it's the right answer. Can someone check my work? Thanks.

Alex Matt
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1 Answers1

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I you draw a picture, the great circle that contains the South Pole and the unit point on the $y$ axis is clearly a unit circle drawn in the $y0z$ plane. Thus the equation you have found for the circle is good, but as it is it is the equation of the cylinder with axis Ox and radius $1$. The equation(s) of this circle in 3D should be :

$$\begin{cases}y^2+z^2&=&1\\x&=&0\end{cases}\tag{1}$$

Check that taking $(x,y,z)=(0,0,-1)$ or $(0,1,0)$, both equations (1) are verified.

Jean Marie
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