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I'm trying to find this green vector and I've got three points A, B, C. How can I find that vector in Unity code-wise?

enter image description here

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2 Answers 2

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Given the input vectors

   Vector3 AB =  Vector3(Bx, By, Bz) - Vector3(Ax, Ay, Az);
   Vector3 Caxis = Vector3(Cx, Cy, Cz) - Vector3(Ax, Ay, Az); // This is the vector from A to C

The direction of the result vector C we seek is the same direction as (AB cross Caxis) cross AB

   Vector3 Cdir = Vector3.Cross(Vector3.Cross(AB, Caxis), AB).Normalize();

Finding the magnitude is more involved. We know that starting at the point Cz and going in the direction Cdir must intersect along AB

   Caxis - alpha * Cdir == beta * AB

and we need to find unknowns alpha and beta. We can find a vector perpendicular to Caxis by projecting AB onto the plane to which Caxis is normal

   Vector3 Cperp = Vector3.ProjectOnPlane(Caxis, AB).Normalize();

so our equation with alpha and beta can be reduced to one unknown

  - alpha * (Cdir . Cperp)/(AB . Cperp) == beta

So solving for alpha (which is in fact the magnitude of our final vector C)

   Caxis == alpha * (Cdir - AB* (Cdir . Cperp)/(AB . Cperp) )

This is a vector equation, but we can solve for alpha by taking the dot product with Caxis on both sides. Writing code again this is

   float num = Caxis.magnitude*Caxis.magnitude;
   float denom1 = Vector3.Dot(Cdir, Caxis);
   float denom2 = Vector3.Dot(AB, Caxis) * Vector3.Dot(Cdir, Cperp) / Vector3.Dot(AB, Cperp);

   float Cmag = num/(denom1 - denom2); 

C is then, using the sign convention from your diagram,

   Vector3 C = Cmag*Cdir;

I'm leaving it without checks for cases where you get a divide by zero, such as when points B and C are identical.

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  • \$\begingroup\$ perfect! Thank you! \$\endgroup\$ Jul 23 at 11:38
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You can do this more simply. This vector in green is just AC with the portion parallel to AB removed.

So:

Vector3 AC = C - A;
Vector3 AB = B - A;

return AC - Vector3.Project(AC, AB);
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  • \$\begingroup\$ Also, thank you for naming your points in such a way that the answer includes AC, AB 😆 \$\endgroup\$
    – DMGregory
    Jul 23 at 13:27

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