Thread: Find perpendicular line given parametric eqn and point

Hi all

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

Originally Posted by drcronik

Hi all
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.

If you have a given line then you know a point, $\displaystyle P$, on that line and the direction vector, $\displaystyle D$.
Now pick any vector, $\displaystyle N$, that is perpendicular to $\displaystyle D$. That is $\displaystyle N\cdot D=0$
The line you want is $\displaystyle P+tN.$

Hi Plato

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?

Thanks

Originally Posted by drcronik

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.

You are making far too much out of this.
Say the line is $\displaystyle <3,-2,5>+t<1,-3,4>$
The direction vector is $\displaystyle <1,-3,4>$.
Pick any vector perpendicular to that vector.
$\displaystyle <-1,1,1>$ will work. $\displaystyle <4,0,-1>$ will also work.
Infinitely many will work. All that is necessary is $\displaystyle <-1,1,1>\cdot<1,-3,4>=0$.

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.

Thanks