Finding a point along a line in three dimensions, given two points I need to find a point along a line segment in three-dimensional space, given two points.  
For example: Find a point along a line segment between point $a(-2, -2, -2)$ and $b(3, 3, 3)$ which is at distance $3$ from point $a$, in the direction of point $b$.
I've been trying to figure out a way to do it with three parametric equations, or with similar triangles, but I can't seem to make sense of it.
Any help is greatly appreciated.
 A: The parametric equations joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are
$$\begin{align*}x&=(1-u)x_1+ux_2\\y&=(1-u)y_1+uy_2\\z&=(1-u)z_1+uz_2\end{align*}$$
where taking $u=1$ gives the point $(x_2,y_2,z_2)$. The distance from $(x_1,y_1,z_1)$ to a point given by the parametric equation above is $d|u|$, where $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. One now needs to solve $3=du$ and substitute the value of $u$ obtained in the parametric equations to obtain the desired point.
A: An alternative method even though the question has been answered:
Begin by creating a vector $\mathbf{\overrightarrow{\text{BA}}}$ by subtracting $\mathbf{A}$ from $\mathbf{B}$: $\langle 3, 3, 3 \rangle - \langle -2, -2, -2 \rangle = \langle 5, 5, 5 \rangle$. Then, normalize this vector by dividing $\mathbf{\overrightarrow{\text{BA}}}$ by its length, $ \frac{\mathbf{\overrightarrow{\text{BA}}}}{\lVert \mathbf{\overrightarrow{\text{BA}}} \rVert} = \frac{\langle 5, 5, 5 \rangle}{\sqrt{5^2+5^2+5^2}} = \langle 0.577, 0.577, 0.577 \rangle$. This new unit vector can then be scaled and added to $\mathbf{A}$ to find the point in space at the desired distance. In this case: $\mathbf{A} + 3\langle 0.577, 0.577, 0.577 \rangle = \langle -0.268, -0.268, -0.268 \rangle$.
Hope this is useful/correct!
