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.
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: $$ i_3 = i_1 + r\vec A = i_1 + \frac{|i_2 - i_1|}{\vec A \cdot \vec C}\vec A $$