The angle between two rays in 3D space

Question

This is a problem from Mathematics GRE Subject Test - #42.

In the xyz-space, what is the degree measure of the angle between the rays $z= (x>=0), y=0$ and $z= (y>=0), x=0$?

a)0; b)30; c)45; d)60; e)90

My Attempt at a Solution

Because the first set of rays are always along the line y=0, they must be spread out on solely the x-z plane, in the direction of the positive x-axis.

Similarly, the second set of rays would be on the y-z plane, and in the direction of the positive y axis. So I figured that because the rays are on perpendicular planes, they should have an angle of 90 degrees.

Sorry if this is drastically wrong, I'm at a loss where to proceed. I'm not totally even sure what topic to tag this under. Any help is much appreciated. Thanks

Answer 1

Well, this might not be kosher but:

o = (0, 0, 0) is the vertex of the two rays. Let a = (1,0,1) is in Ray 1. Let b = (0, 1,1) by in ray two. The distance between a and o is $\sqrt{2}$. Between b and o is $\sqrt{2}$ and between a and b is $\sqrt{2}$. So the three points form an equilateral triangle. So the angle is 60 degrees.

Answer 2

The rays are spanned by the vectors (1,0,1) and (0,1,1). The general answer to the angle between two vectors is give by the dot product formula:

$a\cdot b = \|a\|\, \|b\|\cos(\theta)$

In this case, $\|a\| = \|b\| = \sqrt{2}$ and $(1,0,1) \cdot (0,1,1)=1$. So $\cos(\theta)=1/2$ or $\theta=\arccos(1/2)=60^\circ$