Finding Points at a distance given a Plane equation and point of origin on a plane

[markraz]

Hi, if I have a plane equation

0.17677670x +0y -0.17677670z + 0.35355339 = 0

I know a world coordinate/point on the plane
P = (1.0 , 2.0 , 3.0)

What is the method(s) to find another point at a distance from the point p??
For instance I want to find another point (Q) from (P), that lies at a distance (with respect to the plane) of .25 in the X and .25 in the Y

FYI the answer of Q is:
X = 1.17677670 Y = 1.75000000 Z = 3.17677670

how do you go about finding Q only given the distances from P? and the plane equation?
any suggestions would be appreciated. Do you just avoid plugging into the plane equation altogether and just use position vectors?

I'm curious of finding the simplest method. Not because I'm lazy but I'm trying to write a program to do this with many points and I need to be fast
and to avoid division or minimize it.

Thanks

 

[Ishuda]

Hi, if I have a plane equation

0.17677670x +0y -0.17677670z + 0.35355339 = 0

I know a world coordinate/point on the plane
P = (1.0 , 2.0 , 3.0)

What is the method(s) to find another point at a distance from the point p??
For instance I want to find another point (Q) from (P), that lies at a distance (with respect to the plane) of .25 in the X and .25 in the Y

FYI the answer of Q is:
X = 1.17677670 Y = 1.75000000 Z = 3.17677670

how do you go about finding Q only given the distances from P? and the plane equation?
any suggestions would be appreciated. Do you just avoid plugging into the plane equation altogether and just use position vectors?

I'm curious of finding the simplest method. Not because I'm lazy but I'm trying to write a program to do this with many points and I need to be fast
and to avoid division or minimize it.

Thanks

For one interpretation of your question, you do no have enough information. If you are given a point P on the plane and a distance r from that point, there is a locus of points which satisfy those conditions. For example consider the plane z=0 and the point P=(0,0,0). The locus of points a distance of 1 from point P is the circle x²+y²=1.

From another interpretation, you trade memory for speed. That is given the plane
a x + b y + c z = d,
initialize the three quantities
a' = 1/a
b' = 1/b
c' = 1/c
accounting, of course, for any of a, b, or c equal zero. Then given the point P=(x₀, y₀, z₀) and the distance in two directions, for example a dx and dy, you would have Q=(x, y, z) where
x = x+dx
y = y+dy
z = c' * (d - (a * x + b * y) )

If neither of those are what you meant, please explain further.

 

[markraz]

Thanks appreciate it

Not sure if I put this together correctly:

plane 0.17677670x + 0y - 0.17677670z + 0.35355339 = 0

origin point on the plane p = (1.0 , 2.0 , 3.0)

dx = .25
dy = .25

a' = 1.0/ 0.17677670
b' = 0.0
c' = 1.0/-0.17677670

Q=(x, y, z) where
x = 1.0 + .25
y = 2.0 + .25
z = -0.17677670 * (0.35355339 - (0.17677670x * 1.0 + 0.0 * 2.0 ) )

this gives me
x=1.25
y=2.25
z=30

but the answer should be
X = 1.17677670 Y = 1.75000000 Z = 3.17677670


Thanks, appreciate your help

 

[Ishuda]

Thanks appreciate it

Not sure if I put this together correctly:

plane 0.17677670x + 0y - 0.17677670z + 0.35355339 = 0

origin point on the plane p = (1.0 , 2.0 , 3.0)

dx = .25
dy = .25

a' = 1.0/ 0.17677670
b' = 0.0
c' = 1.0/-0.17677670

Q=(x, y, z) where
x = 1.0 + .25 = 1.25
y = 2.0 + .25 = 2.25
z = -0.17677670 * (0.35355339 - (0.17677670x * 1.0 + 0.0 * 2.0 ) ) <<x=1.25; y=2.25

this gives me
x=1.25
y=2.25
z=30

but the answer should be
X = 1.17677670 Y = 1.75000000 Z = 3.17677670


Thanks, appreciate your help

Then dx is not 0.25. The x of Q (x_Q=1.17677670) minus the x of P (x_P=1.0) is equal to 0.17677670, not 0.25.

Note that when you say a y distance of 0.25 that the y of Q (y_Q=1.75) minus the y of P (y_P=2.0) is equal to -0.25 [0.25 in magnitude].

Also, notice that you will need to have a signed distance, not just a magnitude. For example both 1.75 and 2.25 are a distance of 0.25 from 2 however 2-1.75=0.25 (a positive signed distance from 1.75 to 2) whereas 2-2.25=-0.25 (a negative signed distance from 2.25 to 2). You could assign the signed distance in the opposite direction if you like, but you must be consistent.


EDIT: Note also that you computed z incorrectly for the new point [see the red above]. Also, speaking of consistency, you are saying a distance of 0.25 for both the x and y direction yet the signs are different so one must be a negative distance.

 

[markraz]

Then dx is not 0.25. The x of Q (x_Q=1.17677670) minus the x of P (x_P=1.0) is equal to 0.17677670, not 0.25.

Note that when you say a y distance of 0.25 that the y of Q (y_Q=1.75) minus the y of P (y_P=2.0) is equal to -0.25 [0.25 in magnitude].

Also, notice that you will need to have a signed distance, not just a magnitude. For example both 1.75 and 2.25 are a distance of 0.25 from 2 however 2-1.75=0.25 (a positive signed distance from 1.75 to 2) whereas 2-2.25=-0.25 (a negative signed distance from 2.25 to 2). You could assign the signed distance in the opposite direction if you like, but you must be consistent.

thanks
let me check my calculations over, I messed up

thanks

 

[markraz]

I just want to understand one thing... going a distance of .25 from world space point (1,2,3) in the x axis along the plane should only be an actual distance of .177, in world space with respect to the x axis, is this correct? so how does the equation x = x+dx work here?? It's almost as if this .177 is the 'adjacent' side of a right triangle?? I'm confused :sad:

thanks

 

[Ishuda]

I just want to understand one thing... going a distance of .25 from world space point (1,2,3) in the x axis along the plane should only be an actual distance of .177, in world space with respect to the x axis, is this correct? so how does the equation x = x+dx work here?? It's almost as if this .177 is the 'adjacent' side of a right triangle?? I'm confused :sad:

thanks

What you are talking about is the projection of one vector onto another, see
http://tutorial.math.lamar.edu/Classes/CalcII/DotProduct.aspx
for example.

It appears you are getting several things confused and not applying all of the (assumed by AutoCad?) information behind your problem. First, lets rewrite your plane equation in an easier form:
x - z + 2 = 0
To get this just divide your equation for the plane by 0.176777 [~$\displaystyle \frac{1}{4\sqrt{2}}$]

Your 0.25 in the x direction from point P [=(1, 2, 3)] seems to be: Holding y constant go in the positive x direction (dx is positive) staying in the plane so that you go a distance of 0.25. Well, if you hold y constant and stay in the plane, x and z must go in the same direction by the same amount, i.e. dz = dx, i.e. if the point is in the plane
x+dx - (z+dz) + 2 = 0 if dz=dx
The point P' [=(1+dx, 2 + dy, 3 + dz)] is a distance
||P' - P|| = $\displaystyle \sqrt{dx^2+dy^2+dz^2}=\sqrt{2 dx^2}=\sqrt{2} dx = 0.25$
since dx is positive dz=dx and dy is zero [remember that holding y constant]. Thus
dx ~ 0.176777
dz ~ 0.176777

Your 0.25 in the y direction in the plane seems to be: Holding x and z constant go in the negative y direction (dy is negative) staying in the plane. Well that's easy, since x and z are constant, and dy is negative
dy = -0.25

Therefore your point Q is (1+dx, 2+dy, 3+dx)=(1.176777, 1.75, 3.176777)

 

[markraz]

What you are talking about is the projection of one vector onto another, see
http://tutorial.math.lamar.edu/Classes/CalcII/DotProduct.aspx
for example.

It appears you are getting several things confused and not applying all of the (assumed by AutoCad?) information behind your problem. First, lets rewrite your plane equation in an easier form:
x - z + 2 = 0
To get this just divide your equation for the plane by 0.176777 [~$\displaystyle \frac{1}{4\sqrt{2}}$]

Your 0.25 in the x direction from point P [=(1, 2, 3)] seems to be: Holding y constant go in the positive x direction (dx is positive) staying in the plane so that you go a distance of 0.25. Well, if you hold y constant and stay in the plane, x and z must go in the same direction by the same amount, i.e. dz = dx, i.e. if the point is in the plane
x+dx - (z+dz) + 2 = 0 if dz=dx
The point P' [=(1+dx, 2 + dy, 3 + dz)] is a distance
||P' - P|| = $\displaystyle \sqrt{dx^2+dy^2+dz^2}=\sqrt{2 dx^2}=\sqrt{2} dx = 0.25$
since dx is positive dz=dx and dy is zero [remember that holding y constant]. Thus
dx ~ 0.176777
dz ~ 0.176777

Your 0.25 in the y direction in the plane seems to be: Holding x and z constant go in the negative y direction (dy is negative) staying in the plane. Well that's easy, since x and z are constant, and dy is negative
dy = -0.25

Therefore your point Q is (1+dx, 2+dy, 3+dx)=(1.176777, 1.75, 3.176777)

thank you... I must have searched net 100 times with "plot point from point on plane", "get point from other point on plane" etc....
Now I know that is called "projection of one vector onto another" this makes more sense... I still wouldn't have figured it out but you gave me great insight .. I thank you for your time

 

[markraz]

What you are talking about is the projection of one vector onto another, see
http://tutorial.math.lamar.edu/Classes/CalcII/DotProduct.aspx
for example.

It appears you are getting several things confused and not applying all of the (assumed by AutoCad?) information behind your problem. First, lets rewrite your plane equation in an easier form:
x - z + 2 = 0
To get this just divide your equation for the plane by 0.176777 [~$\displaystyle \frac{1}{4\sqrt{2}}$]

Your 0.25 in the x direction from point P [=(1, 2, 3)] seems to be: Holding y constant go in the positive x direction (dx is positive) staying in the plane so that you go a distance of 0.25. Well, if you hold y constant and stay in the plane, x and z must go in the same direction by the same amount, i.e. dz = dx, i.e. if the point is in the plane
x+dx - (z+dz) + 2 = 0 if dz=dx
The point P' [=(1+dx, 2 + dy, 3 + dz)] is a distance
||P' - P|| = $\displaystyle \sqrt{dx^2+dy^2+dz^2}=\sqrt{2 dx^2}=\sqrt{2} dx = 0.25$
since dx is positive dz=dx and dy is zero [remember that holding y constant]. Thus
dx ~ 0.176777
dz ~ 0.176777

Your 0.25 in the y direction in the plane seems to be: Holding x and z constant go in the negative y direction (dy is negative) staying in the plane. Well that's easy, since x and z are constant, and dy is negative
dy = -0.25

Therefore your point Q is (1+dx, 2+dy, 3+dx)=(1.176777, 1.75, 3.176777)

Hi I ran through this trying to find another point and I can't seem to get it to work as a generic plug and play formula.

How did exactly you derive at these values?
dx ~ 0.176777
dz ~ 0.176777

thanks

 

[Ishuda]

Hi I ran through this trying to find another point and I can't seem to get it to work as a generic plug and play formula.

How did exactly you derive at these values?
dx ~ 0.176777
dz ~ 0.176777

thanks

Suppose you have a point P [=(1, 2, 3)] in a plane given by
x - z + 2 = 0,
and you want to add a bit to x and z to get a new point P' =(1+dx, 2, 3+dz) and still stay in the plane so we also have
(1 + dx) - (3 + dz) + 2 = 0
Rearranging
(1 + dx) - (3 + dz) + 2 = (1 - 3 + 2) + dx - dz = dx - dz = 0
or
dz = dz.
You want the new point P' to be a distance 0.25 from P so
||P' - P|| = $\displaystyle \sqrt{dx^2+dy^2+dz^2}=\sqrt{2 dx^2}=\sqrt{2} |dx| = 0.25$
so
$\displaystyle \sqrt{2} |dx| = 0.25$
or
$\displaystyle |dx| = \frac{0.25}{\sqrt{2}}$ ~ 0.176777
Since dx is positive
dx = |dx|

 

[markraz]

Suppose you have a point P [=(1, 2, 3)] in a plane given by
x - z + 2 = 0,
and you want to add a bit to x and z to get a new point P' =(1+dx, 2, 3+dz) and still stay in the plane so we also have
(1 + dx) - (3 + dz) + 2 = 0
Rearranging
(1 + dx) - (3 + dz) + 2 = (1 - 3 + 2) + dx - dz = dx - dz = 0
or
dz = dz.
You want the new point P' to be a distance 0.25 from P so
||P' - P|| = $\displaystyle \sqrt{dx^2+dy^2+dz^2}=\sqrt{2 dx^2}=\sqrt{2} |dx| = 0.25$
so
$\displaystyle \sqrt{2} |dx| = 0.25$
or
$\displaystyle |dx| = \frac{0.25}{\sqrt{2}}$ ~ 0.176777
Since dx is positive
dx = |dx|

thanks sorry to bother you again
1. I'm still not sure how you derived
$\displaystyle \sqrt{2} |dx| = 0.25$
where did you get the 2?

2. and is the square root of 2 essentially finding the hypotenuse?

3.
||P' - P|| = $\displaystyle \sqrt{dx^2+dy^2+dz^2}=\sqrt{2 dx^2}=\sqrt{2} |dx| = 0.25$
I see this is the standard 3d "distance formula" but what is ||P' - P|| ??

thanks

 

[Ishuda]

thanks sorry to bother you again
1. I'm still not sure how you derived
$\displaystyle \sqrt{2} |dx| = 0.25$
where did you get the 2?

since dz=dx
dx² + dy² + dz² = dx² + 0 + dx² = 2 dx²

2. and is the square root of 2 essentially finding the hypotenuse?

Yes, in the plane with constant y.

3.
||P' - P|| = $\displaystyle \sqrt{dx^2+dy^2+dz^2}=\sqrt{2 dx^2}=\sqrt{2} |dx| = 0.25$
I see this is the standard 3d "distance formula" but what is ||P' - P|| ??

thanks

The notation ||P-P'|| is sometimes used as the norm (or magnitude) of P-P' or, equivalently, the distance between points P and P'. It is really, IMO, just an extension of the absolute value sign used to indicate that you may be dealing with vectors, functions, points, etc. instead of just plain numbers. if P is a real number ||P|| is just |P|

 

[markraz]

ok thanks..
I hate to keep bothering you thanks for your patience. I'm wondering if we can work out one more example ?

if I have a plane

1x + 0y + .064705z + .425936 = 0

my point of origin P will be (X=2.60236954 Y = 2.00000000 Z = 3.12940952)

same offsets as last time
offset x = .25
offset y = .25

the answer for this should be X=2.66707430 Y = 1.75000000 Z = 3.37089098
but for some reason when I use the formulas on previous pages I get X=2.771946 Y = 1.75000000 Z = 3.37089098

so I'm missing something obviously.

I will pay you or donate to your favorite charity for your time. I appreciate this

 

[Ishuda]

ok thanks..
I hate to keep bothering you thanks for your patience. I'm wondering if we can work out one more example ?

if I have a plane

1x + 0y + .064705z + .425936 = 0

my point of origin P will be (X=2.60236954 Y = 2.00000000 Z = 3.12940952)

same offsets as last time
offset x = .25
offset y = .25

the answer for this should be X=2.66707430 Y = 1.75000000 Z = 3.37089098
but for some reason when I use the formulas on previous pages I get X=2.771946 Y = 1.75000000 Z = 3.37089098

so I'm missing something obviously.

I will pay you or donate to your favorite charity for your time. I appreciate this

I'm not sure what you mean. This is not the same kind of problem. None of the three points indicated lie in the plane indicated. In fact no pair lie in the same plane at all.

As far as pay goes, pay it forward. I've had a lot of help in my life and am trying to do just that.

While we are speaking of another example, you need to lay out some specific rules for just exactly what you mean when talking about the dx, dy, and dz and the possible vector lengths [lengths between points]. Which points/lines/vectors have to lie in the plane, what is to be done if one (or two) of the components of the normal vector [the a, b, and c of the plane equation] happen to be zero, etc. For example, the locus of points equidistant from a world point is a sphere. You can narrow that down slightly by restricting the points to a plane but that (in most cases) still leaves a locus of point forming a circle. To narrow that down to a single point will require further restrictions. In your first example, the additional restrictions were
First: Hold y constant, stay in the plane, the circle has a radius of 0.25, go in the positive x direction.
Second: Hold the new x, and z constant and go in the negative y direction 0.25 to the final point.

I need to go right now but I'll be back later with other comments.

 

[markraz]

I'm not sure what you mean. This is not the same kind of problem. None of the three points indicated lie in the plane indicated. In fact no pair lie in the same plane at all.

As far as pay goes, pay it forward. I've had a lot of help in my life and am trying to do just that.

While we are speaking of another example, you need to lay out some specific rules for just exactly what you mean when talking about the dx, dy, and dz and the possible vector lengths [lengths between points]. Which points/lines/vectors have to lie in the plane, what is to be done if one (or two) of the components of the normal vector [the a, b, and c of the plane equation] happen to be zero, etc. For example, the locus of points equidistant from a world point is a sphere. You can narrow that down slightly by restricting the points to a plane but that (in most cases) still leaves a locus of point forming a circle. To narrow that down to a single point will require further restrictions. In your first example, the additional restrictions were
First: Hold y constant, stay in the plane, the circle has a radius of 0.25, go in the positive x direction.
Second: Hold the new x, and z constant and go in the negative y direction 0.25 to the final point.

I need to go right now but I'll be back later with other comments.

thanks.. I think I'm going to put together a more complete question with better pictures/examples to make it more clear, since I'm not even sure what I want

I really appreciate your help you are a savior

I will post back Thursday

thanks

 

[markraz]

Since I don't know mathematical terminology I'll give non mathematical overview of what I'm looking to accomplish.

I have a bunch of lines/vectors in 3D space. They make up a collection of different planes. In this specific example roof, door, wall, window overhang etc...

Say I want to locate points or lines with respect to these different planes. What is the best way to approach this if I know the plane equation and a point of origin on the plane?

So for example say I know the plane of the roof

R = 0x -.5y + .5z -.5 = 0

and I know a point on the roof

(lower left) point of roof P = 0 , 0 , 1

Now, say I want to plot points from this point P along the slope of the plane R

Essentially I want to treat each plane "object" as it's own 2D coordinate system?

What is the preferred mathematical approach to go about this?

Again I appreciate your help I can't thank you enough

 

[Ishuda]

Since I don't know mathematical terminology I'll give non mathematical overview of what I'm looking to accomplish.

I have a bunch of lines/vectors in 3D space. They make up a collection of different planes. In this specific example roof, door, wall, window overhang etc...

Say I want to locate points or lines with respect to these different planes. What is the best way to approach this if I know the plane equation and a point of origin on the plane?

So for example say I know the plane of the roof

R = 0x + -.5 + .5z + -.5 = 0

and I know a point on the roof

(lower left) point of roof P = 0 , 0 , 1

Now, say I want to plot points from this point P along the slope of the plane R

Essentially I want to treat each object as it's own 2D coordinate system?

What is the preferred mathematical approach to go about this?

Again I appreciate your help I can't thank you enough

Another formulation for a plane is
a (x-x₀) + b (y-y₀) + c (z-z₀) = 0
where (x₀, y₀, z₀) is a point in the plane and not all of a, b, and c are zero. Notice that if you were to write the 'usual equation' for the plane as
a x + b y + c z = d
then d = a x₀ + b y₀ + c z₀. Now you say you want to stay in the plane so you have to decide how to stay in the plane. That is how do you want to go "along the slope of the plane".

Do you want to choose a x and y and go along the line
z = z₀ + [a (x₀ - x) + b (y₀ - y)] / c?
[assuming, of course, that c is not zero]. This is essentially the first method I mentioned. Note the dx=x-x₀ and dy=y-y₀ are actual changes is x and y and not changes 'along the plane'. Possibly because the b in the original problem was zero is why AutoCad computed it the way it did.

Do you want to go a particular distance D in world space in the positive z direction holding x constant and staying in the plane. Then, since x is a constant x₀ the change in the points y and z have to satisfy
b dy + c dz = 0,
That is x=x₀, y=y₀+dy, and z=z₀+dz. That leads to
dy = - c dz / b.
The distance we want to go is
$\displaystyle \sqrt{dy^2\, +\, dz^2}\, =\, \sqrt{(\frac{c}{d})^2\, dz^2\, +\, dz^2}$
=$\displaystyle \sqrt{[(\frac{c}{d})^2\, +\, 1] dz^2}$
=$\displaystyle \sqrt{[(\frac{c}{d})^2\, +\, 1]}\, \, dz\, =\, D$
since dz is positive. Thus
x = x₀
y = y₀ - $\displaystyle \frac{c}{d}\, \frac{D}{\sqrt{[(\frac{c}{d})^2\, +\, 1]}}$
z = z₀ + dz = z₀ + $\displaystyle \frac{D}{\sqrt{[(\frac{c}{d})^2\, +\, 1]}}$
BTW: Note that there would be no solution for this if b were zero, so how you are allowed to go along the slope of the plane is determined in part by the plane itself.

One final comment for now. Note that the step size you choose for the dx and/or dy and/or dz should be predicated on the a, b, and c values if you don't want the series of points along the line to be too widely spaced.

 

[markraz]

Another formulation for a plane is
a (x-x₀) + b (y-y₀) + c (z-z₀) = 0
where (x₀, y₀, z₀) is a point in the plane and not all of a, b, and c are zero. Notice that if you were to write the 'usual equation' for the plane as
a x + b y + c z = d
then d = a x₀ + b y₀ + c z₀. Now you say you want to stay in the plane so you have to decide how to stay in the plane. That is how do you want to go "along the slope of the plane".

Do you want to choose a x and y and go along the line
z = z₀ + [a (x₀ - x) + b (y₀ - y)] / c?
[assuming, of course, that c is not zero]. This is essentially the first method I mentioned. Note the dx=x-x₀ and dy=y-y₀ are actual changes is x and y and not changes 'along the plane'. Possibly because the b in the original problem was zero is why AutoCad computed it the way it did.

Do you want to go a particular distance D in world space in the positive z direction holding x constant and staying in the plane. Then, since x is a constant x₀ the change in the points y and z have to satisfy
b dy + c dz = 0,
That is x=x₀, y=y₀+dy, and z=z₀+dz. That leads to
dy = - c dz / b.
The distance we want to go is
$\displaystyle \sqrt{dy^2\, +\, dz^2}\, =\, \sqrt{(\frac{c}{d})^2\, dz^2\, +\, dz^2}$
=$\displaystyle \sqrt{[(\frac{c}{d})^2\, +\, 1] dz^2}$
=$\displaystyle \sqrt{[(\frac{c}{d})^2\, +\, 1]}\, \, dz\, =\, D$
since dz is positive. Thus
x = x₀
y = y₀ - $\displaystyle \frac{c}{d}\, \frac{D}{\sqrt{[(\frac{c}{d})^2\, +\, 1]}}$
z = z₀ + dz = z₀ + $\displaystyle \frac{D}{\sqrt{[(\frac{c}{d})^2\, +\, 1]}}$
BTW: Note that there would be no solution for this if b were zero, so how you are allowed to go along the slope of the plane is determined in part by the plane itself.

One final comment for now. Note that the step size you choose for the dx and/or dy and/or dz should be predicated on the a, b, and c values if you don't want the series of points along the line to be too widely spaced.

Thank you Ishuda, this is exactly what I need. I can't thank you enough. All these scenarios you gave me really make it easy to understand now.
I'm curious, I scoured my calc III, Linear algebra and trig books.. and I couldn't find anything remotely like this. Is this something that is not usually dealt with for some reason? or do I just have the wrong books?

thank you again

 

[Ishuda]

Thank you Ishuda, this is exactly what I need. I can't thank you enough. All these scenarios you gave me really make it easy to understand now.
I'm curious, I scoured my calc III, Linear algebra and trig books.. and I couldn't find anything remotely like this. Is this something that is not usually dealt with for some reason? or do I just have the wrong books?

thank you again

I don't know of any books right off hand that have something like that in it but there are books and web pages which have bits and pieces of the methods used and, of course the general formulas for planes and lines are floating around 'everywhere'. For example the equation of a line is
$\displaystyle \overset{\rightarrow}{p}\, =\, \overset{\rightarrow}{p_0}\, +\, t\, \overset{\rightarrow}{v}$
where $\displaystyle \overset{\rightarrow}{p}$ is a point on the line, $\displaystyle \overset{\rightarrow}{p_0}$ is an initial point on the line, $\displaystyle \overset{\rightarrow}{v}$ is a direction vector, and t is a (scaled) distance. If the norm (magnitude) of $\displaystyle \overset{\rightarrow}{v}$ is 1, t is the actual world distance. So, given two points on your line, p₀=(x₀,y₀,z₀) and p₁=(x₁, y₁, z₁) and the distance D between them we can take $\displaystyle \overset{\rightarrow}{v}$ as
$\displaystyle v_1\, =\, \frac{x_1\, -\, x_0}{D}$,
$\displaystyle v_2\, =\, \frac{y_1\, -\, y_0}{D}$,
and
$\displaystyle v_3\, =\, \frac{z_1\, -\, z_0}{D}$

Suppose p₀ is in your initial plane and you find a p₁ in that plane by some method, say as discussed before here in this thread. Then you could construct a vector $\displaystyle \overset{\rightarrow}{v}$ as above. By construction, the norm of $\displaystyle \overset{\rightarrow}{v}$ is one so that in the equation above for a line, t would be the world space distance and you could compute a set of equidistant points on the line in the direction you want along the plane.

EDIT: Oh, and as far as books go, I would think looking for 'numerical methods' would be more productive than looking through Calc books. Also, you might look more under computer graphics