# Coordinates of a point in 3d space

• Jun 16th 2013, 06:54 AM
Antonio144
Coordinates of a point in 3d space
Hello.

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).

http://img46.imageshack.us/img46/394/sw9g.jpg

Your help is greatly appreciated. :)
• Jun 16th 2013, 07:26 AM
Prove It
Re: Coordinates of a point in 3d space
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*}.
• Jun 16th 2013, 07:39 AM
Plato
Re: Coordinates of a point in 3d space
Quote:

Originally Posted by Antonio144
Each point on $\displaystyle \overline{BA}$ has coordinates $\displaystyle (4t+1,4t+1,4t+1)$ where $\displaystyle 0\le t\le 1~.$
Use the distance formula to find the value of $\displaystyle t$ so that the length of $\displaystyle \overline{BI}$ is one.