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Using the formula from this page: http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html how can you find the individual x,y, and z distances? I'm trying to figure it out but can't wrap my head around it.

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You can find the individual distances using the given formula. The individual distances between any point $(x,y,z)$ on the line and the given point is $ |x_0-x|,|y_0-y|$ and $|z_0-z|$. But we know that any point in the line is given by $(x_1+(x_2-x_1)t , y_1+(y_2-y_1)t, z_1+(z_2-z_1)t)$. From the given result we see that the distance is minimum at the point corresponding to ( taking $r=(x,y,z)$ ) $t^*=(r_1-r_0).(r_2-r_1)/|r_2-r_1|$. So, the individual distances are $ |x_0-x^*|,|y_0-y^*|$ and $|z_0-z^*|$ where $ x^*=x_1+(x_2-x_1)t^*, y^*=y_1+(y_2-y_1)t^*$ and $z^*=z_1+(z_2-z_1)t^*$.

Also there is another way of doing it. You can find the straight line passing through the given point $(x_0,y_0,z_0)$ and perpendicular to the given straight line. After that solve those two lines to find the point of intersection $(x_3,y_3,z_3)$. Then the individual distances are $ |x_0-x_3|,|y_0-y_3|$ and $|z_0-z_3|$

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How would I find that line you speak of? –  bakis2011 Aug 1 '12 at 20:45
    
Have a look at the edit. You can do it using the given formula! –  Kasun Fernando Aug 1 '12 at 20:56
    
Thanks a lot! I don't know how i missed that the first read through hahahaha –  bakis2011 Aug 1 '12 at 23:50
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