Plane intersect
Author
If I have a point in space and a normal with that, how can I calculate where on a plane (what point in space) it would hit if that vector was of infinite length?
plane equation :
ax+by+cz+d = 0
line equation:
x = x0 + t*dx;
y = y0 + t*dy;
z = z0 + t*dz;
substitute, solve for t, you''re all set.
ax+by+cz+d = 0
line equation:
x = x0 + t*dx;
y = y0 + t*dy;
z = z0 + t*dz;
substitute, solve for t, you''re all set.
Author
Thanks, just to make sure
dx, dy, dz = the normal
x0, y0, z0 = the starting point
That resulted in the following equation:
[a*(x0+t*dx)] + [b*(y0+t*dy)] + [c*(z0+t*dz)] + d = 0
To solve t I got two passed two values of t into the equation, using these to calculate the k value of a linear equation.
(y = kx + m)
f(t) = [a*(x0+t*dx)] + [b*(y0+t*dy)] + [c*(z0+t*dz)]
y1 = f(0)
y2 = f(1)
k = y2 - y1
m = f(0)
t = (0 - m) / k
Then passed t back into the given line equation...
This seems like the long way around? Is there an easier way?
[edited by - JohanK on March 26, 2002 3:56:36 AM]
[edited by - JohanK on March 26, 2002 6:06:05 AM]
dx, dy, dz = the normal
x0, y0, z0 = the starting point
That resulted in the following equation:
[a*(x0+t*dx)] + [b*(y0+t*dy)] + [c*(z0+t*dz)] + d = 0
To solve t I got two passed two values of t into the equation, using these to calculate the k value of a linear equation.
(y = kx + m)
f(t) = [a*(x0+t*dx)] + [b*(y0+t*dy)] + [c*(z0+t*dz)]
y1 = f(0)
y2 = f(1)
k = y2 - y1
m = f(0)
t = (0 - m) / k
Then passed t back into the given line equation...
This seems like the long way around? Is there an easier way?
[edited by - JohanK on March 26, 2002 3:56:36 AM]
[edited by - JohanK on March 26, 2002 6:06:05 AM]
[a*(x0+t*dx)] + [b*(y0+t*dy)] + [c*(z0+t*dz)] + d = 0
Factor t out of the expression :
t*(a*dx+b*dy+c*dz)+(a*x0+b*y0+c*z0+d)=0
Factor t out of the expression :
t*(a*dx+b*dy+c*dz)+(a*x0+b*y0+c*z0+d)=0
This topic is closed to new replies.
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