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Let $\textbf{x}$ and $\textbf{y}$ be two points on the sphere. Show that the normal to the plane determined by the great circle through $\textbf{x}$ and $\textbf{y}$ intersects the sphere at the points $\pm \textbf{z}$, where

$$\bf z = \frac{x \times y}{||x \times y||}$$

Now obviously, if we let the center of the sphere lie at the origin, $\bf x \times y$ will be the normal vector to the plane that crosses the sphere and make that great circle.

Since that plane divides the sphere into two equal parts, the points were the normal intersects the sphere will be opposites of one another, thus $\pm \textbf{z}$.

I don't really get why we need to adjust that point by $\bf ||x \times y||$? Could somebody explain?

Also, the problem itself is easier, as I am not asked to find the points, just to show that they are valid. But I don't really know how could I do it?

I could reason, that if we place that sphere at the origin, the equation of the plane that cuts the sphere and makes the great circle is described by:

$${\bf w}\cdot({\bf x\times y})=0$$

But what next?

Thanks!

Sarunas
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1 Answers1

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I don't really get why we need to adjust that point by $\bf ||x\times y||$? Could somebody explain?

The magnitude of the cross product is $||{\bf x\times y}|| = ||{\bf x}|| \cdot ||{\bf y}|| \cdot \sin (\angle {\bf xOy})$. We're assuming that the sphere has unit radius, so $||{\bf x}|| = ||{\bf y}|| = 1$, and we want the points where the normal intersects the sphere. Therefore unless $\bf x \perp y$ we need to normalise.

I'm afraid I can't help with the second half, because I'm not sure what definitions you're working from. Obviously you don't already have a definition or theorem to the effect that the equation of a plane is ${\bf P . N} = 0$ where $\bf N$ is its normal, or you would have used it already.

Peter Taylor
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