# Vertex Angle of a Cone

• October 4th 2010, 07:02 PM
Latszer
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 2pi/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 di erent angle? )

So the work that I have is that I created two vectors that go from the origin along the y-axis. 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?

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