I need to find a line perpendicular to a known line.
I have the parametric equation for the known line and the point on the known line through which the perpendicular line should pass, how do I calculate what the vector of the perpendicular line should be?
I know its probably quite simple but I have learn t all I know about about this on YouTube.
Thanks for the speed response
Do I solve D x N = 0 for each vector ie X,Y,Z
my line is in the form [X,Y,Z] x t[Vx,Vy,Vz]
so for the perpendicular line
Vxd x Vxn = 0 and Vyd x Vyn = 0 and Vzd x Vzn = 0.
Did I understand this correctly?
I understand, you mean the dot product o f the vectors must equal 0. I did a bit more research.
Here is what I am trying to achieve.
Attached is a diagram of a possible route of a Cartesian robot.
I need to link path AB and path BC with an arc, to ensure smooth transition into and out of the arc paths AB & BC need to be tangents to the arc. The arc should be between points D & G. Length DB = BG
Using the cad software I have think I have found the solution but I do not know how to calculate it manually.
This is what I have done.
1. Draw 2 circles e & f with centers D & G and axis is path AB & BC respectively, the radius of the circles will be twice the distance of BD
2. Find the intersecting points of circles e & f. There will be 2. Lets call these points H & I
3. Find the midpoint of line HI. Lets call it J
4. J will then be the center point of the arc starting at point D and ending at point G.
I have tested this in 2d using the cad and replaced the circles with lines perpendicular to the path and it works perfectly so I am hoping this will work in 3d.
Like I said I have the theory of what I need to do but am struggling with how to actually do the calculation, the cad program does all the calculations for me.
I realize now that calculating the perpendicular line will not work as there are an infinite no. of these lines and I need the two to intersect.
Can you help me? I have started this as a new thread aswell so that I get a different heading.