I assume you know how to find vector which connect between center of two particles.
if one is (x1,y1,z1),another one is (x2,y2,z1). then the vector \vec{r} is (x2-x1,y2-y1,z2-z1).
Assume velocity of particle 1 isvec{v}= (vx1,vy1,vz1),
The velocity component in the vector vec{r} direction can be calculated as \vec{v_p}=(\vec{v}\cdot\vec{r})/r^2 \vec{r},where r is the length of velocity \vec{r}=\sqrt{(x2-x1)^2+(y2-y1)^2+(z2-z1)^2}

vx1'=((vx1*(x2-x1)+vy1*(y2-y1)+vz1*(z2-z1))/r^2) (x2-x1)
vy1'=((vx1*(x2-x1)+vy1*(y2-y1)+vz1*(z2-z1))/r^2) (y2-y1)
vz1'=((vx1*(x2-x1)+vy1*(y2-y1)+vz1*(z2-z1))/r^2) (z2-z1)
And the velocity compendicular to vec{r} is \vec{v_n}=\vec{v}-\vec{v_p}

P.S. The inner product between \vec{A} and unit vector \hat{n} will give you magnitude of projection of vector \vec{A} in the direction of \hat{n}=\vec{r}/r, so the component in the direction \hat{n} is (\vec{A}\cdot\hat{n}) \hat{n}