The question:

Given the points A(2, -3, 1) and B(8, 9, -5) in $\displaystyle R^3$, find:

The coordinatesdof point D on the line AB such that D is between A and B and $\displaystyle \vec{AD} = 2\vec{DB}$

My attempt:

First I found the vector $\displaystyle \vec{AB}$ by subtracting A from B to get (6, 12, -6). By drawing a small diagram, I realised I had to find the co-ordinate vector 2/3 the way to B from A. So I wrote the parametric form of a line with origin A (using the vector $\displaystyle \vec{AB}$ that I calculated). I subbed in 2/3 as the scalar multiple, and produced (6, 5, -3) as the answer to this question.

I checked this by finding $\displaystyle \vec{AD}$ and equating it to $\displaystyle 2\vec{DB}$ and both were equal to (4, 8, -4). I'm fairly sure my answer is correct, but my text has no solutions for practise exams, so I'd like to be sure.

Thanks!