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Math Help - Vertex Angle of a Cone

  1. #1
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    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 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?

    Thanks for your help,

    Tyler
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  2. #2
    MHF Contributor

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    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.
    Last edited by HallsofIvy; October 5th 2010 at 05:52 AM.
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