A math question

[tipforeveryone]

I am not really good at math,

Can you show me how to solve this

Known:

• Position of Red and Blue (in Vec3)
• Length of edge between Blue and Black,
• Alpha Angle (in Degree)

I want to get the position of Black (in Vec3)

Thanks

[SpiderPig]

You would use Sin () and Cos () to find the x and y position.

Is alpha a constant?  Like 90°?

Do the red and blue positions vary in the third dimension as well? Or are you looking for the 2D position only of the black?

[tipforeveryone]

Thanks for you reply

All of Red, Blue, Black are in 3D and alpha is not a constant

Can you help me with a formula, it is bad that I can't remember these basic formulas for triangle calculation

• red:GetPosition(true)
• blue:GetPosition(true)
• u:Length()
• alpha
• black:GetPosition(true) = ???

[macklebee]

Spiderpig is asking about the clarification of angle alpha in relation to what plane it is on. Is that angle value based on an orthographic view (looking down from top) or is the angle along the plane that would exist if the 3 points were on different vertical points in 3D space. Assuming that we are looking at the triangle above in the XZ plane and the Y axis is going into the screen. If its the latter, i don't think we have enough information about what determines the orientation of the plane between the 3 points.

[tipforeveryone]

Here you are

[macklebee]

I don't think you have given us enough information to determine how the plane is oriented in 3d space to figure it out. We would need alpha broken down into orthographic components at the very least. 

 

[macklebee]

Take a look at the picture to see the problem.

Think of the line between Red and Blue as a rotating shaft. Think of the green line of known length at known angle as a fixed limb coming off Blue. As the Red/Blue shaft rotates, the Black dot could be anywhere along the red circular path and still hold true to the information that has been provided. 

[tipforeveryone]

Oh I see, Thanks for pointing it out.

If there is one more vector (a) which have known Orange point position, and it is on the same plane with vector u and w (as figure image),
and a new beta angle which = 1/2 * alpha angle

will it be posible to calculate the Black position ?

[macklebee]

Should be possible as it gives the orientation of the plane. 

[tipforeveryone]

Just now, macklebee said:

Should be possible as it gives the orientation of the plane. 

It sound sweet, then please help me with a simple fomula or steps to archive my calculation

[macklebee]

Quote

It sound sweet, then please help me with a simple fomula or steps to archive my calculation

I don't know how simple it will be but I will try tomorrow after work if someone does not help you before then. I needed to go to bed two hours ago. 

[Ma-Shell]

In the following I will use the following terminology:

A -> blue point, B -> red point, C -> yellow point, D -> black point (point to find); I will further use o to specify the dot-product of two vectors and use v1, v2, v3 to specify the individual components of v.

Overview:

You need to mirror B on the line AC. This will yield a point B' which is on the line AD. From there you can calculate D by going |u| from A in the direction of AB'.

Step by step:

The line going through A and C can be written as X=A+p*a with p being a scalar variable.

We need to find a vector Z such that B+Z is on that line (i.e. there exists a value for p such that B+Z = A+p*a -> Z = A+p*a-B -> Z = p*a-w) and that is perpendicular to a (i.e. Z o a = 0). Together that is:

0 = (p*a-w) o v

0 = (p*a1-w1)*a1 + (p*a2-w2)*a2 + (p*a3-w3)*a3

-> p = (a1*w1+a2*w2+a3*w3)/(a1*a1+a2*a2+a3*a3)

-> p = (a o w) / |a|

With this we can calculate Z as Z = p*a-w

Now we can mirror B as B' = B+2*Z

-> B' = B+2*(p*a-w)

Now we can calculate D as D = A + |u| * (B' - A) / |(B' - A)|

Implementation:

Simply hack in the formulas I marked as fat and you should be good (p is a float and B' and D are both vectors). Let me know if you need any further help with that!

[macklebee]

Ma-Shell obviously knows his vector math but like the OP, vector math has always been a struggle for me though I can handle 3D geometry fair enough . I have to break vector math down into steps that I can understand and follow, So even though you have an answer already (but past my ability to follow), I decided to take a crack at this. I used multiple references and I will post them with the steps as this was a fun learning experience for me.

So the problem was to find pointD given this information:

 

Given: 

• 3 points given are all on the same plane, points A, B, & C 
• The angle between VectorA-B and VectorA-C is equal to angle between  VectorA-B and VectorA-D
• The distance from point A to point D is known

Find: point D 

Assumptions:

• The three given points are not collinear to allow for determining the plane equation
• Due to equal angles between vectors, VectorA-MirroredPoint is the reflected vector of VectorA-C across the mirror plane that is along VectorA-B

 

Solution:

To reflect a point across the mirror plane, we must first find the plane that the three given points (A,B, and C) are located. Once we have planeABC, then the mirror plane will be orthogonal to that plane and pass through the line or vector AB. Once the mirror plane is determined, then we can mirror the point. And finally determine the point along the line where the fixed length would occur.

1) Find the plane that contains points A, B, and C. Using the 3 points, you can calculate two vectors:

    1a) Find the two vectors given the 3 points - A, B, & C. --Finding a 3D line through 2 points --https://www.youtube.com/watch?v=JlRagTNGBF0
          Vab = FindVector( pointA, pointB ) --vector AB
          Vac = FindVector( pointA, pointC ) --vector AC

    1b) Get the normal vector to two vectors. --Equation of a Plane Passing Through 3 Points --https://www.youtube.com/watch?v=0qYJfKG-3l8                           

          normalvector = FindNormalOfTwoVectors( Vab, Vac )

    1c) Get the Plane Equation using the normal vector and a point --Note that you can use any of the 3 points above and get same answer.

          plane = GetPlaneEquation( normalvector, pointA )

2) To find orthogonal plane for mirror/reflection, we need two vectors and point on the plane -Just like before!
     --Find the Equation of Plane Containing a Line and Orhtogonal to a Given Plane --https://www.youtube.com/watch?v=aupy5wwAyb8
    2a) The normal of planeABC is a vector on the Mirror plane

          Vn = FindNormalVectorOfPlane(plane) 

    2b) We want the mirror/reflection plane to also contain vector AB - which we have already found = Vab. Find the normal vector to the two vectors.

          Mirror_normalvector = FindNormalOfTwoVectors(Vn, Vab)

    2c) Now get the Mirrored Plane's Equation using pointA or pointB and its normal vector

          Mirror_plane = GetPlaneEquation(Mirror_normalvector, pointA)

3) Now that we have the reflection plane that that runs through line AB, lets mirror pointC about it

   --Find Mirror Image of a Point Reflected on Vector Plane
   --https://www.youtube.com/watch?v=fvXG-DVx6z0

     pointMP = FindMirroredPoint(Mirror_plane, pointC)

4) Find pointD that is Dist_AD from pointA

   --Finding a point along a line in 3D given 2 points
   --Used alternative method shown at bottom of page
   --https://math.stackexchange.com/questions/83404/finding-a-point-along-a-line-in-three-dimensions-given-two-points

     pointD = FindPointAlongLine(pointA, pointMP, Dist_AD)

 

Scripts:

• Vector_Plane_PointReflection_Equations.lua contains the functions shown above       Vector_Plane_PointReflection_Equations.lua
• Main.lua contains the example script that shows how to use the functions                    Main.lua

 

[tipforeveryone]

Thanks macklebee, your explaination is clear as daylight ! so detail. Finally I did it,

It took time to understand math in English ^^