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can anyone help me with this problem:

Is it possible to construct three vectors (a,b,c) in 3D, such that angle between a and b is 30 degrees, between a and c is 150 degrees, and between b and c is 30 degrees?

If not, prove it.

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Fix $a$ and $c$ so that the angle between $a$ and $c$ is $5\pi/6.$ All possible vectors which have angle $\pi/6$ with $a$ form a cone about $a$ with cone angle $\pi/3$. Likewise, all possible vectors which have angle $\pi/6$ with $c$ form a cone about $c$ with central angle $\pi/3$. Any third vector $b$ must lie in the intersection of both cones.

The intersection is $\{0\}$, since the closest the two cones come is in the plane of $a$ and $c$ and they each lie $\pi/6$ from their central vectors, which lie $5\pi/6$ apart.

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I think you should say "the intersection is empty". At first glance I interpreted your sentence to mean "the intersection has measure $0$", and thought, "Wait, there might still be one vector in that intersection..." – Rahul Jan 23 '12 at 15:58
The intersection has exactly one vector, $0$. Thanks - I'll fix the notation :) – Neal Jan 23 '12 at 18:05
Oh, I forgot to say that length of each vector is 1. – Meee Jan 23 '12 at 18:21
Since you're only concerned with angles, you don't need to worry about lengths of nonzero vectors at all. – Neal Jan 23 '12 at 19:23

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