CODEWITHSUNDEEP
c++mathvector3dgeometry
Mohammad Sohaib
Mohammad Sohaib

Reputation: 575

Finding point perpendicular to two intersecting lines in 3D

I have two lines (L1, L2) in 3D.

L has the following starting and ending coordinates : P1(x1, y1, z1) and P2(x2, y2, z2).

L2 has P3(x3, y3, z3) and P2(x2, y2, z2). Notice how both L1's and L2's P2 are same: meaning they intersect at this specific point.

Now I want to find a point P(x,y,z) at any distance from P2 that line (P,P2) is perpendicular to plane on which points (P1,P2,P3) are placed.

Upvotes: 1

Views: 2040

Answers (3)

Chexxor
Chexxor

Reputation: 416

The cross product is the way of calculating perpendicularity relative to your two lines. You need to make vectors of your line parameters, simples way would be this:

vecL1 = (x1-x2, y1-y2, z1-z2) and
vecL2 = (x3-x2, y3-y2, z3-z2)

Cross product you can google how to calculate, but in this scenario:

//Replacing the new x,y,z's with i, j, k to avoid naming confusion.
vecL3 = vecL1 x vecL2 = (j1*k2 - j2*k1, k1*i2 - k2*i1, i1*j2 - j1*i2)

Now the cross product per definition is a new vector (line) that is strictly perpendicular to the two lines/vectors you used to calculate this with. But vectors lack position, so you need to add the intersection point to this vector in order to find some point.

//i3, j3, k3 being the third vector's parameters
P3(i3+x2, j3+y2, k3+z2)

PS: The distance from your P2 to P3 is per definition (how cross products work) to the area of a parallelogram that the two lines are sides of, I found a link to illustrate:
Cross product
Normalizing the 3rd vector will make the distance equal to 1 from P2.

Upvotes: 1

Adl A
Adl A

Reputation: 183

If you have any line AB then an arbitrary point C will always be perpendicular to AB IFF triangle ABC has no angle larger than π/2

That means there will always be a point D on line AB such that CD is perpendicular to AB

Upvotes: 0

melak47
melak47

Reputation: 4850

The cross product will give you a perpendicular vector to the plane described by two other vectors, in pseudo-code:

normal = cross(normalize(P1-P2), normalize(P3-P2))

Since you've defined P2 as the point of intersection, you can simply add this normal vector to P2 to get your perpendicular point.

Upvotes: 1

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