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this is my first post.. I hope this good

I have 1 triangle in space (3D)... and I know all data except the coordinates of 3er point(vertex)... for example this:

triangle data example

then:

Right Triangle 

    x     y      z

v1 = (1.42, 0.239, 0.032)

v2 = (2.97, 1.500, 0.032)

v3 = (2.33, ???, ????)

A = 2

B = 0.99 = 1

C = 2.232

a = 63.664°

b = 26.336°

c= 90°

with all this data I thought that with a simple trigonometric identities I could achieve the goal...BUT NO!

then I check this:

How to find the third coordinate of a right triangle given 2 coordinates and lengths of each side

and this: Find the third vertex of a triangle in $3D$

3d geometry: triangle 2 points known, find 3rd point

and all information in

How to find coordinates of 3rd vertex of a right angled triangle when everything else is known?

with all that information I found this: http://awaismunir.net/universal/tangents/3rd-third-vertext-calculate-right-angled-triangle.gif

then I deduced the following:

v3cox = ((-v1.y * opposite) + (v2.y * opposite) + (adjacent * v1.x)) / adjacent

v3coy = ((v1.co * opposite) - (v2.x * opposite) + (adjacent * v1.y)) / adjacent

and this work in 2d!....

then I tried with:

v3coz = ((v1.z * opposite) - (v2.z * opposite) + (adjacent * v1.z)) / adjacent
or
v3coz = ((v1.z * opposite) + (v2.z * opposite) - (adjacent * v1.z)) / adjacent
or
v3coz = ((v1.z * opposite) + (v2.z * opposite) + (adjacent * v1.z)) / adjacent

but any of this "coz" work

I hope you can help me

----------------------------- UPDATE

AFTER @Martigan IDEAS I understand that I have infinite number of points that will satisfy the same conditions:

enter image description here

what's up if I know some data about v3.. for example v3x = 2.33 v3y = ?? and v3.z =??

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Diego Quevedo
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    You don't have all the data needed. You have an infinite number of points that will satisfy the same conditions. This points are on the circle of center $v_2$, of radius $B$ and obtained by rotation around the line $v_1v_2$. – Martigan Jan 21 '15 at 14:29
  • @Martigan thanks for answere... which other data I need??? – Diego Quevedo Jan 21 '15 at 14:31
  • @Martigan I thing that I have all data because I change any of that the v3 coordinates will change – Diego Quevedo Jan 21 '15 at 14:35
  • You need to know at least one of the three coordinates of $v_3$ (either $x,y,z$). – Martigan Jan 21 '15 at 14:55
  • You're not understanding what @Martigan wrote. If you know 2 points, and all the lengths, you can rotate around the side joining the two known points, and get different third points. – Gerry Myerson Jan 21 '15 at 14:56
  • @Martigan if I have v3x or v3y.... what I need to do in order to obtain the missing coordinates?? – Diego Quevedo Jan 21 '15 at 17:30
  • The scalar product between $v-3v-2$ and $v_2v_1$ is equal to zero, which gives one equation. An other steps from the lenght of the segment $v_3v_2$. – Martigan Jan 21 '15 at 18:18
  • @Martigan thanks for reply.. I'm a little confused...

    v−3v−2 in this case ... who is v??? v3???

    following your steps I do:

    (v−3v−2) * (v2 - v1)=0 and

    (v−3v−2) * (v3-v2)=0

    is it correct? but what?? I need to find new "v"...yes???..

    sorry I'm not a mathematician

    thanks

     – Diego Quevedo Jan 21 '15 at 20:51
  • @Martigan,@Gerry Myerson, The other idea is with rotation mentioned, but it's the same when looking for the intersection of the circles in 3d http://prntscr.com/5vd1on – Diego Quevedo Jan 21 '15 at 20:58
  • @Martigan I update the information... for example...
    what's up if I know some data about v3.. for example v3x = 2.33 v3y = ?? and v3.z =??     – Diego Quevedo Jan 21 '15 at 21:49
  • @Martigan may be can you help me please? I have the same problem but I don't have any v3 data, BUT I have the Normal of the plane where the triangle is.... Is posible obtain the v3 coordinates? I should open another thread with this question? .. thanks again – Diego Quevedo Jun 05 '17 at 23:12
  • @DiegoQuevedo It might be better to open an other thread. However, the basic answer is yes, you can find v3 (or at least two possible solutions. – Martigan Jun 06 '17 at 08:53
  • @Martigan I follow your advice and I add some new thread link thanks for all – Diego Quevedo Jun 07 '17 at 16:11

1 Answers1

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Let $\overrightarrow{B} . \overrightarrow{A}=0$

I call $x_3$ the $x$ coordinate of vertex $v_3$, for convenience purpose.

That is $(x_3-x_2)(x_2-x_1)+(y_3-y_2)(y_2-y_1)+(z_3-z_2)(z_2-z_1)=0$

Given that you know $x_1,x_2,x_3,y_1, y_2, z_1,z_2$ it gives you a relationship between $y_3$ and $z_3$ (or, easier to use - see afterwards, a relationship between $(y_3-y_2)$ and $(z_3-z_2)$.

Then you know that the lenght of $\overrightarrow{B}$ is $0.99$

Then you have $(x_3-x_2)^2+(y_3-y_2)^2+(z_3-z_2)^2=0.99^2$

Granted, this gives you a potential of two values for $y_3$, but you should be able to rule out one of the two if needed.

You could have used the scalar product between $\overrightarrow{B}$ and $\overrightarrow{C}$, that would have given you a non-quadratic equation, but that I feel more complex to compute.

Martigan
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  • Amazing Answer thanks for all the help.... I don't know how to close the topic. – Diego Quevedo Jan 22 '15 at 22:40
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    @DiegoQuevedo You don't have to "close" the question. Other answers might be added, it is also a reference for other questions. However, if you liked my answer, you should accept it (on the left hand side of the text, the arrow to upvote and downvote, and, for you, an acceptation button). – Martigan Jan 23 '15 at 08:20