orientation of vectors in 3d space

-1/3

Start with this link: http://mathcentral.uregina.ca/QQ/database/QQ.09.00/nishi1.html

Now visualize the triangle in the bottom step but solve for the central angle and use the law of cosines. The side lengths are sqrt(3), sqrt(3) and 2sqrt(2).

a^2 + b^2 - 2abcos y = c^2
3 + 3 - 6 cos y = 8
- 6 cos y = 2
cos y = -1/3

first of all why sides with sq.root3? i said "unit vectors" i.e. vectors with modulus
1. secondly if ur answer is correct, i am not getting how any two of the four vectors are inclined at the same value of angle

The sqrt of three comes from the example I linked to where they used a vector length of the sqrt of three for convience. The cosine of the angle is length independent so you can change the length if you desire.

If I read your problem correctly, all the vectors must be at the same angle to each other so the shape formed by connecting the ends of the vector will be a regular polyhedron. If you find one angle using a regular polyhedron as a base you should get all of them. The link has a nice geometrical explanation.

So divide by sqrt(3) and be done with it. The length of the vectors does not affect the angle between them, as long as it's the same for all.


Imagine a tetrahedron. Imagine your unit vectors originating from its center of mass and pointing to its corners. The angle of the vectors is the same because of the tetrahedron's symmetry. The value of the angle derives from multiple applications of the Pythagorean theorem. Good luck!

well thanks for the help!

thanks to you too