I've been having a little trouble visualising this so I post here to try to get some help. I need to find 4 vectors in 3-space so the dot product of any two is negative. I thought there could only be a maximum of three vectors as if we evenly place four vectors around in 3-space the maximum separation angle is pi/2, which gives a non-negative dot product. I guess my assumption that the maximum separation is pi/2 is probably wrong...
2026-03-28 07:36:30.1774683390
arranging vectors in 3-space
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If the separation is $\frac \pi 2$ the dot product is zero. You need the separations to be strictly greater than $\frac \pi 2$. The maximum for four vectors is $\frac \pi 2$ in two dimensions, but a third dimension gives more flexibility. Think of the four vectors from the center of a regular tetrahedron to the corners, for example.