Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
    
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.