TITLE: Find three-dimensional vectors $\vec{v_1},\vec{v_2},\vec{v_3}$ satisfying
QUESTION [5 upvotes]: Find three-dimensional vectors $\vec{v_1},\vec{v_2},\vec{v_3}$ satisfying 
$\vec{v_1}.\vec{v_1}=4$,
$\vec{v_1}.\vec{v_2}=-2$,
$\vec{v_1}.\vec{v_3}=6$,
$\vec{v_2}.\vec{v_2}=2$,
$\vec{v_2}.\vec{v_3}=-5$,
$\vec{v_3}.\vec{v_3}=29$.

Since $\vec{v_1}.\vec{v_1}=4,\vec{v_2}.\vec{v_2}=2,\vec{v_3}.\vec{v_3}=29$.
So $|\vec{v_1}|=2,|\vec{v_2}|=\sqrt2,|\vec{v_3}|=\sqrt{29}$.
Let the angle between $\vec{v_1},\vec{v_2}$ is $\alpha$ and the angle between $\vec{v_2},\vec{v_3}$ is $\beta$ and the angle between $\vec{v_3},\vec{v_1}$ is $\gamma$.
So i found $\cos\alpha=\frac{-1}{\sqrt2},\cos\beta=\frac{-5}{\sqrt{58}},\cos\gamma=\frac{3}{\sqrt{29}}$
Then i got stuck and could not solve further.
Please help me.Thanks.

REPLY [4 votes]: Chose $\vec v_1=(2,0,0)^T$.
From:  $\vec v_1\cdot \vec v_2=-2$ we have:
$$
(2,0,0)(x,y,z)^T=2x=-2 \Rightarrow x=-1
$$
and from $\vec v_2 \vec v_2=2$ we have 
$$
(-1,y,z)(-1,y,z)^T=1+y^2+z^2=2 \Rightarrow z=\pm \sqrt{1-y^2}
$$
so $\vec v_2=(-1,t,\pm \sqrt{1-t^2})^T$
Using the analogous relations for $\vec v_1$ and $\vec v_3$ we find:
$$
\vec v_3=(3,u,\pm \sqrt{20-u^2})^T
$$
Now we can use the last relation $\vec v_2 \cdot \vec v_3=-5$
To find $t$ as a function of $u$ or vice versa.