CODEWITHSUNDEEP
unity-game-engine
user2523112
user2523112

Reputation: 191

how do i know position of point, perpendicular to line

enter image description here

There are 2 Point A(a,b,c), B(d,e,f)

enter image description here

There is the line AB, connecting A and B

Line CC' is perpendicular to line AB

I want to Point C 's position on line CC'

If length of line CC' is 1, what is C 's position?

How do I calculate C's position with unity?

A : Vector3(a,b,c)

B : Vector3(d,e,f)

AC' length = BC' length

CC' length = 1

AB is perpendicular to CC'

-> C Vector3(?,?,?)

Upvotes: 0

Views: 1749

Answers (1)

Homar
Homar

Reputation: 1529

As you describe it, the problem is undetermined. There are infinitely many solutions for C because you are working in 3D space. To visualize, imagine the point C orbiting around the axis AB. Regardless of where C is in its orbit, it will always be orthogonal to AB and length(CC') = 1. You need to further constrain the problem.

The Mathematics

First, we calculate the point C'. To do this, we take the vector from A to B, which is AB = B - A. Then, to get to C' we just travel half the distance of AB from point A:

C' = A + AB/2

Now we need to find a vector orthogonal to AB. Here we encounter the problem that I described initially. There are infinitely many such vectors so we need to further constrain the problem. Suppose that we can choose a vector v that is not colinear to the lines AB or C'C. Now we can find C'C by finding a vector orthogonal to both AB and v.

We know that the cross-product of two linearly independent vectors produces a vector that is orthogonal to both vectors. So all that is left to do is normalize the result so that the length is 1:

C'C = normalize(AB x v)

Finally, we can find point C by traveling from C' along the vector C'C:

C = C' + C'C

Unity Code

Here, I provide some untested code that simply implements the mathematics described above. I'm not awfully familiar with Unity so it is quite possible that there exists some built-in functions that would relieve some of the work:

Vector3 v = new Vector3(0, 0, 1); // Choose this as you wish
Vector3 AB = B - A;
Vector3 C_prime = A + AB / 2;
Vector3 C = C_prime + Vector3.Normalize(Vector3.cross(AB, v));

Upvotes: 4

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