Hello.
Need your help.
I need to find coordinates of a point I that lies on AB
What I know is:
-point A(5,5,5),
-point B(1,1,1),
-distance between B and I (in this case 1.0).
Your help is greatly appreciated.
A vector going in the direction of $\displaystyle \displaystyle \begin{align*} \mathbf{AB} = (5, 5, 5) - (1, 1, 1) = (4, 4, 4) \end{align*}$.
You want to move along by 1 unit, so a unit vector going in that direction is
$\displaystyle \displaystyle \begin{align*} \hat{\mathbf{AB}} &= \frac{1}{\sqrt{ 4^2 + 4^2 + 4^2 }} \, (4, 4, 4) \\ &= \frac{1}{4\,\sqrt{3}} \, (4, 4, 4) \\ &= \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \\ &= \left( \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3} \right) \end{align*}$
Therefore, to get to your point I, you need to add $\displaystyle \displaystyle \begin{align*} \frac{\sqrt{3}}{3} \end{align*}$ to each component of your starting point $\displaystyle \displaystyle \begin{align*} (1, 1, 1) \end{align*}$.
So your point $\displaystyle \displaystyle \begin{align*} I = \left( 1 + \frac{\sqrt{3}}{3}, 1 + \frac{\sqrt{3}}{3}, 1 + \frac{\sqrt{3}}{3} \right) = \left( \frac{3 + \sqrt{3}}{3} , \frac{3 + \sqrt{3}}{3} , \frac{ 3 + \sqrt{3}}{3} \right) \end{align*}$.