# Find three vectors in ${R^3}$ such that the angle between all of them is pi/3?

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.

• Get the vectors from the vertices of an equilateral triangle, say $(0,0,0), (1,0,0), (1/2,\sqrt{3}/2,0)$. – Integrand Jun 4 at 18:44
• The dot product of any of the other vectors with the $(0,0,0)$ vector would be zero and would not satisfy the conditions. – InvestingScientist Jun 4 at 18:45
• @InvestingScientist The above comment is not listing vectors but the vertices of the equilateral triangle. – Peter Foreman Jun 4 at 18:47
• Do you know that diagonals of cube faces forms a tetrahedron? – Alexey Burdin Jun 4 at 18:52
• 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? – InvestingScientist Jun 4 at 19:22

Diagonals of cube faces forms a tetrahedron

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.

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 $$(-1,-1,2)\cdot (-1,2,-1)=-3$$.

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