# calculate vector between 2 vectors in 3d sphere

I have a sphere with known radius $R$ and 2 vectors $\vec a$ and $\vec b$ within this sphere both starting from $(0,0,0)$ and ending on the edges of this sphere so $\left\lVert\vec a \right\rVert = R$ and $\left\lVert\vec b \right\rVert = R$.

The angle between these 2 vectors is $\alpha = 90^o$ but their coordinates are unspecified( known but they can be anywhere in the sphere)

How can I calculate vector $\vec c$ coordinates if this vector also starts at $(0,0,0)$, its length is also equal to spheres radius : $\left\lVert\vec c \right\rVert = R$ , and is somewhere between $\vec a$ and $\vec b$ specified by the known angle of $\beta$ that is an angle between $\vec a$ and $\vec c$ and is $0^o \lt\beta \lt 90^o$

If $\vec{c}$ lies between $\vec{a}$ and $\vec{b}$, it lies in the plane spanned by them. So there are $s,t$ so that $\vec{c}=s\vec{a}+t\vec{b}$. If we know the angle between $\vec{c}$ and $\vec{a}$ is $\beta$, we know that $$tR^2 = \vec{c} \cdot \vec{a} = R^2\cos{\beta},$$ so $t=\cos{\beta}$. But then the angle between $\vec{b}$ and $\vec{c}$ is $90^{\circ}-\beta$, so $$sR^2 = \vec{c} \cdot \vec{b} = \cos{(90^{\circ}-\beta)} = \sin{\beta},$$ so $$\vec{c} = \vec{a}\cos{\beta}+\vec{b}\sin{\beta}.$$ We can verify that $\vec{c} \cdot \vec{c} = \vec{a} \cdot \vec{a} \cos^2{\beta} + \vec{a} \cdot \vec{b} \sin{\beta}\cos{\beta} + \vec{b} \cdot \vec{b} \sin^2{\beta} = R^2$ since $\vec{a} \cdot \vec{b} = 0$.