
Vertex Angle of a Cone
Here is the problem I have for my homework...
Find an equation for the cone in 3D space which has vertex angle 2pi/3. (Hint: What is the vertex angle for z = sqrt(x^2+x^2)? How could you change this equation to get a cone with dierent angle? )
So the work that I have is that I created two vectors that go from the origin along the yaxis. I got the two vectors to be A=i+k and B=i+k.
The angle between two vectors is A.B=A B cos(theta)
A.B = 1 and A B = sqrt(2)*sqrt(2) = 2
Therefore theta = arccos(1/2), which is 2pi/3.
First of all, is this correct? And if it is not, what might I be doing wrong, do I have the definition of a vertex angle correct and can I just create vectors like that?
Thanks for your help,
Tyler

The first thing I would do is draw a two dimensional version two lines meeting at the origin with angle $\displaystyle 2\pi/3$. That means that each line makes angle $\displaystyle \pi/3$ with the yaxis and so angle $\displaystyle \pi/2 \pi/3= \pi/6$ with the xaxis. One line has slope $\displaystyle tan(\pi/6)= \frac{2}{\sqrt{3}}= \frac{2\sqrt{3}}{3}$ and the other has slope $\displaystyle \frac{2}\sqrt{3}}{3}$. If you are in the xzplane, that would give equations $\displaystyle z= \frac{2\sqrt{3}}{3}x$ for one of the lines. Now rotate the whole thing about the z axis, letting "x" become "r" in cylindrical coordinates: $\displaystyle z= \frac{2\sqrt{3}}{3}r$.
Your two vectors, A= i+ k and B= i+ k make angle $\displaystyle \frac{\pi}{4}$ with both axes and angle between them of 90 degrees (which is correct for the question "What is the vertex angle for z = sqrt(x^2+x^2)?").
$\displaystyle A\cdot B= (1)(1)+ (1)(1)= 0$, not 1.