How to find the equation of a straight line in 3D?

If I have data points (x1,x2), (y1,y2) and (z1,z2) how can I find the equation between these points(it is a straight line)? Thank you!

Answer 1

[David Goodmanson]

Hi Federico,

Are you looking for a parametric description? It's constructive.

p1 = rand(1,3)
p2 = rand(1,3)
u = (p2-p1)/norm(p2-p1);   % unit vector, p1 to p2
d = (-1:.1:1)';            % displacement from p1, along u
xyz = p1 + d*u; 
plot3(xyz(:,1),xyz(:,2),xyz(:,3),'o-')

[AFHG]

Thanks, exactly what i was looking for. How would d change for p1 = [1 5 7] p2 = [10 7 4]? Whats the logic behind d, what is it doing?

[Matt J]

@Federico, you should already know what d is doing. Otherwise, how do you know that the code is "exactly what you were looking for"?

[AFHG]

Thats because i tried to run it and its working for the rand coordinates.

[David Goodmanson]

Hi Federico,

It's certainly best not to accept an answer unless you know what it is doing. Just because the code pops out an answer is not justification enough.

As far as d is concerned, draw two points p1 and p2, separated by a distance of, say, 3. Draw a vector of length 1 starting at p1 and pointing toward p2. That's u*. Then the vector p1 + d*u is a point d/3 of the way from p1 to p2. Varying d traces out the line. Values d>3 go past point p2 and negative values of d create points heading away from p2.

*u is the unit vector pointing in the direction (p2-p1) and is its own entity. To create u, the base of u is at p1, but later for other purposes u could be moved to any location.

[AFHG]

Ok thank you!

Answer 2

[Matt J]

If

 p1=[x1,y1,z1];
 p2=[x2,y2,z2];

then equations for the line are

(x-p1)*null(p2-p1)=0  %equations in x

[AFHG]

for the equation overall, what would it be?

[AFHG]

and by the way, your code gives the error: Incorrect use of '=' operator. To assign a value to a variable, use '='. To compare values for equality, use '=='.

When == is used it says matrix dimensions must agree.

[Matt J]

for the equation overall, what would it be?

Sorry, what would what be? I didn't give you a partial answer. A 3D line satisfies a system of 2 equations and that's what I gave you. Any point x on the line that you plug in will satisfy them and any point not on the line will not satisfy them.

and by the way, your code gives the error: Incorrect use of '=' operator.

It isn't a command. It's a system of equations. So you shouldn't be trying to execute it.

[AFHG]

So how do i run it if i can't execute it. I need to be able to use it within a loop;

[Matt J]

What does it mean to "run" an equation? An equation isn't something that has an output. It is a relationship that something (here, a point on the line) must satisfy.

[AFHG]

So how to i make it into something that has an output?

[Matt J]

You would have to tell us what you're inputting to the code what you expect the output to be.

[AFHG]

Suppose there are particles moving. There is a loop which changes the x,y,z position values each time it runs (as time passes). i need to form the equation of the line from the precedent and current coordinate which changes when the direction of the particle changes. I need this equation to perform further calculations within the loop.

[James Tursa]

Please show us your code, or a small sample of it, so we can see how you intend to use these linear equations downstream in your code. E.g., are you going to use them to determine if a particle has "turned"? To interpolate or predict particle positions or collisions? Or ...?