Shortest Distance between two vectors in 3D space

To find the shortest (perpendicular) distance between two vectors O and V in 3 dimensions.

[SD Angle]= shortestdistance(O,V,Q);

To find the shortest (perpendicular) distance between two vectors O and V in 3 dimensions. Q is a vector joining O and V. One point on each vector also needs to be known to comupte Q (Q=Point1-Point2)

SD is the shortest distance returned by the function.
Angle is the angle between the two vectors.

Example:
O = [-0.012918 0.060289 0.998097];
V = [47.9083 -3.8992 65.6425];
Point1 = [35.4 5.6 -49.4];
Point2 = [37.4 5.8 32.8];
Q = Point1 - Point 2;
[SD Angle]= shortestdistance(O,V,Q);

The Algorithm:

In 3D space, the shortest distance between two skew lines is in the direction of the common perpendicular.

To find a vector, P=(Px,Py,Pz), perpendicular to both vectors (O and P), we need to solve the two simultaneous equations, O.P=0 and V.P=0.

Although two equations in three unknowns cannot generally be solved analytically, these homogenous equations can be transformed into a series of two equations in two unknowns by using the ratios Px/Pz and Py/Pz, which can then be solved using normal methods.

Choosing an arbitrary value for Pz, we can determine a valid P, as well as the corresponding unit vector U = P /|P|.

Then to find the shortest distance, the scalar product can be used to find the projection of any vector Q (connecting the two skew lines and Q can be computed by knowing one point on each of the skew lines )onto the unit vector U. Thus the shortest distance SD= Q .U ,

The angle between the two vectors is computed by taking the arccosine of the scalar product of V and O divided by the product of the magnitudes of V and O , i.e. Angle = acos( (V.0)/|V||O| )*180/pi

Copyright 2013 Mathew Philip.
Date: 2013/09/04