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Is there a simple way to do this? I have found that ${(a.b)/|a||b|}$ must be equal to ${1/2}$ but from there I am stuck how to proceed. Any help?

P.s. This is from the MIT 2016 Linear Algebra course and is not homework.

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    $\begingroup$ Get the vectors from the vertices of an equilateral triangle, say $(0,0,0), (1,0,0), (1/2,\sqrt{3}/2,0)$. $\endgroup$ – Integrand Jun 4 at 18:44
  • $\begingroup$ The dot product of any of the other vectors with the $(0,0,0)$ vector would be zero and would not satisfy the conditions. $\endgroup$ – InvestingScientist Jun 4 at 18:45
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    $\begingroup$ @InvestingScientist The above comment is not listing vectors but the vertices of the equilateral triangle. $\endgroup$ – Peter Foreman Jun 4 at 18:47
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    $\begingroup$ Do you know that diagonals of cube faces forms a tetrahedron? $\endgroup$ – Alexey Burdin Jun 4 at 18:52
  • $\begingroup$ Hello all, sorry misinterpreted. Thank you for that it was most helpful. I was stuck for ages. Other than using that method, is there any other way to find such vectors? $\endgroup$ – InvestingScientist Jun 4 at 19:22
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Diagonals of cube faces forms a tetrahedron

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Take an orthonormal basis $(i,j,k)$ of $\mathbb R^3$, $v_1 = \cos \frac{\pi}{6} i + \sin \frac{\pi}{6} j$ and $v_2 = \cos \frac{\pi}{6} i - \sin \frac{\pi}{6} j$. We have by construction $\angle(v_1, v_2) = \frac{\pi}{3}$.

Now let's find $\alpha$ such that $v_3 = \cos \alpha i + \sin \alpha k$ solve the problem.

That will be the case providing that $\angle(v_1,v_3) = \frac{\pi}{3}$, i.e. if $\cos \frac{\pi}{6} \cos \alpha = \cos \frac{\pi}{3}$, i.e. $\cos \alpha = \frac{1}{\sqrt 3}$ and $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \frac{\sqrt2}{\sqrt{3}}$.

Finally $$\begin{cases} v_1 &= \frac{\sqrt 3}{2} i + \frac{1}{2}j\\ v_2 &= \frac{\sqrt 3}{2} i - \frac{1}{2}j\\ v_3 &= \frac{1}{\sqrt 3}(i + \sqrt 2 k) \end{cases}$$

is a solution.

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The position vectors of the vertices of the equilateral triangle $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ with respect to the center of the triangle $(1/3,1/3, 1/3)$. We can also multiply the obtained vectors by $3$. Therefore we get $(3,0,0)-(1,1,1) = (2,-1,-1)$, $(-1,2,-1)$, $(-1,-1,2)$

Check: all of the norms are $\|(-1,-1,2)\|=\sqrt{6}$, and the dot products are $(-1,-1,2)\cdot (-1,2,-1)=-3$.

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  • $\begingroup$ Oh, the angles are $2\pi/3$... $\endgroup$ – orangeskid Jun 4 at 20:30

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