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Math Help - Coordinates of a point in 3d space

  1. #1
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    Coordinates of a point in 3d space

    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.
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  2. #2
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    Re: Coordinates of a point in 3d space

    A vector going in the direction of \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 \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 \begin{align*} \frac{\sqrt{3}}{3} \end{align*} to each component of your starting point \displaystyle \begin{align*} (1, 1, 1) \end{align*}.

    So your point \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*}.
    Thanks from Antonio144
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  3. #3
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    Re: Coordinates of a point in 3d space

    Quote Originally Posted by Antonio144 View Post
    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).
    Each point on \overline{BA} has coordinates (4t+1,4t+1,4t+1) where 0\le t\le 1~.

    Use the distance formula to find the value of t so that the length of \overline{BI} is one.
    Thanks from Antonio144
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    Re: Coordinates of a point in 3d space

    Thank you!
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