For each of your line segments, you can convert to point-vector form by using either point, and the difference between the two points as the direction vector.
(x ,y ,z ) = (x1,y1,z1) = t*( x2-x1), (y2-y1), (z2-z1)) for the first line. t=0 gives (x1,y1,z1) and t=1 gives (x2,y2,z2)
0
Modern healthcare requires a modern cloud. View this brief video to understand how the Concerto Cloud for Healthcare can help your organization.
Sorry for getting back to this so late.
In the example provided, the final answer is unclear to me (i.e. the (X,Y,Z) location of the intersection point). Is the final intersection point (X,Y,Z) = (5,6,2)?
Also as an aside, I'm not getting the cross product he is getting in his example (see below)
from article "Using Bensegueni's method, we find
a(V1 x V2) = a(-10,-11,-13)"
I get a(V x V2) = a (-10, 11, -13) (i.e. middle term positive).
In 2D, if two lines aren't parallel, it exists, for sure, an interception point.
In 3D there is another requirement: besides to be not parallel, the lines must be coplanar, say, they must be in a same plane.
Assuming the lines are defined by their end points (x1, y1, z1) and (x2, y2, z2) lets use the general parametric equation:
x = x1 + (x2 - x1)*t
y = y1 + (y2 - y1)*t
z = z1 + (z2 - z1)*t
if lines are
Line A --> (xa1,ya1, za1, xa2,ya2, za2) and
Line B --> (xb1,yb1, zb1, xb2,yb2, zb2),
first define the parametric equations
Line A
x = xa1 + (xa2 - xa1) * ta
y = ya1 + (ya2 - ya1) * ta
z = za1 + (za2 - za1) * ta
Line B
x = xb1 + (xb2 - xb1) * tb
y = yb1 + (yb2 - yb1) * tb
z = zb1 + (zb2 - zb1) * tb
To check if they are parallel, get the vectors for each line.
Line A
(the components of the vector are the coefficients of ta):
M1 = (xa2 - xa1) * i + (ya2 - ya1) * j + (za2 - za1) * k
Line B
(do the same as for Line A, by using the coefficients of tb):
M2 = (xb2 - xb1) * i + ...
If M1 and M2 are equal or proportional, the lines are parallel, so there is no interception point.
If not, go ahead.
Now set the x and y values.
Of course, x and y must be the same in both lines equations, so,
x = xa1 + (xa2 - xa1) * ta = xb1 + (xb2 - xb1) * tb
and
y = ya1 + (ya2 - ya1) * ta = yb1 + (yb2 - yb1) * tb
By solving tha above equations, we have values for ta and tb.
Now, just apply these values in the z equations to chek is they are true.
If so, you found the interception point, if not, they don't intercept each another.
Jose
0
Featured Post
Is your cloud always on? With an Always On cloud you won't have to worry about downtime for maintenance or software application code updates, ensuring that your bottom line isn't affected.