# Find three-dimensional vectors $\vec{v_1},\vec{v_2},\vec{v_3}$ satisfying

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.

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.

Since you only know their length and relative orientation, the system is underdetermined. So pick the first vector to be something simple, like 2i, make the second vector to also lie in the xy plane and calculate third relative to those.