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I'm having way to much issue with this, I would think it's not super hard, but I'm getting no where with it, and I need to slove it to progress with the thing I'm making.

Anyways here is my problem, in the image below I want to figureout position $i_3$, all points are in 3d space but its a triangle so its coplanar, I have the $xyz$ of point $i_1$ and point $i_2$ I have the distance between $i_1$ and $i_2$ I have the unit vectors of all sides and I know at $i_2$ its a $90°$ angle, with this information how can I get pos $xyz$ of $i_3$, the solutuion I can think of is line intersection in 3d space, but I dont know how to pull that off. I really really appreciating any help, and hopefully a somewhat simple solution that don't go over my head. Thanks in advance.

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Maffred
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Patrik Fröhler
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  • Can you write down the equations of the lines in parametric form and solve them simultaneously? – David Quinn Jan 07 '16 at 19:18
  • @DavidQuinn This is just like the last step of solving a larger problem and how I get the position and vectors etc is partly by ray tracing and some other stuff witch mean for example position and direction of i1 I get from the raytracer and i know i2 is perpendicular to i1 and 50 distance from it unit vector A i know go thru i1 but all of this just makes things more confusing so I've tried simplified the problem as much as possible, so this smaller problem is part of the larger simplifed problem I posted here: http://math.stackexchange.com/questions/1602431/collision-point-of-circle-and-line – Patrik Fröhler Jan 07 '16 at 19:33

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Note that if $\theta$ is the angle at $i_1$, then: $$ \cos\theta = \frac{\vec A \cdot \vec C}{|\vec A||\vec C|} = \vec A \cdot \vec C $$ Now compute the distance $d$ between $i_1$ and $i_2$: $$ d = |i_2 - i_1| $$ Then we can use simple primary trig ratios to get the distance $r$ between $i_1$ and $i_3$: $$ d = r\cos\theta \implies r = \frac{|i_2 - i_1|}{\vec A \cdot \vec C} $$ Then since $\vec A$ is a unit vector, we can just plug $r$ into the usual line formula: $$ i_3 = i_1 + r\vec A = i_1 + \frac{|i_2 - i_1|}{\vec A \cdot \vec C}\vec A $$

Adriano
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