How Do You Calculate the Distance Between Two Skew Lines?

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Discussion Overview

The discussion revolves around calculating the distance between two skew lines represented in parametric and vector forms. Participants explore various methods to approach the problem, including vector calculations and the use of equations derived from the lines' definitions.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant presents the equations for line $$l_1$$ and seeks assistance with line $$l_2$$, which is given in a set of equations.
  • Another participant suggests finding a support vector and a direction vector for line $$l_2$$ to define a point similar to point $$P$$ on line $$l_1$$.
  • There is a discussion about different methods to calculate the distance between the lines, with one participant expressing a preference for calculating the distance between vectors rather than using point-to-plane methods.
  • Participants attempt to solve for variables $$x$$ and $$y$$ from the equations of line $$l_2$$, leading to some confusion and corrections regarding the values obtained.
  • One participant describes a method involving the cross product of direction vectors to find a unit vector perpendicular to both lines, which is then used to calculate the distance.
  • Another participant expresses uncertainty about their method and seeks clarification on the calculations involving parameters $$z$$ and $$t$$, which are intended to define the vector $$PQ$$ between points on the two lines.
  • There is a back-and-forth regarding the validity of different approaches, with participants sharing their calculations and questioning each other's methods.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to calculate the distance between the skew lines. Multiple competing views and methods are presented, and there is ongoing uncertainty about the correctness of specific calculations.

Contextual Notes

Some participants express confusion over the steps taken in their calculations, indicating potential miscalculations or misunderstandings of the methods discussed. The discussion includes various assumptions about the relationships between points on the lines and the vectors connecting them.

Petrus
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Hello MHB,
This is a problem from my book,
calculate distance between line $$l_1: (x,y,z) = (1,1,3)+t(1,-2,-3)$$ and
b49o3l.png


(hope it work's read $$l_2$$ cause I don't know how to write that in latex)
I am stuck when I got $$l_2$$ in plane.
I start with $$l_1$$
$$P=(1+t,1-2t,3-3t)$$
what shall I do with $$l_2$$?

Regards,
$$|\pi\rangle$$
 
Last edited:
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Petrus said:
Hello MHB,
This is a problem from my book,
calculate distance between line $$l_1: (x,y,z) = (1,1,3)+t(1,-2,-3)$$ and
b49o3l.png


(hope it work's read $$l_2$$ cause I don't know how to write that in latex)
I am stuck when I got $$l_2$$ in plane.
I start with $$l_1$$
$$P=(1+t,1-2t,3-3t)$$
what shall I do with $$l_2$$?

Regards,
$$|\pi\rangle$$

Hey I pi rangle! ;)

Which methods do you know to determine the distance between 2 lines?

What you can do for instance with $$l_2$$ is find a support vector and a direction vector, so you can define a point Q similar to your point P with 1 parameter.

If you click Reply With Quote you can see how I put your set of equations in $\LaTeX$.
$$l_2:
\left\{
\begin{aligned}
x-\phantom{1}y+3z-1 &= 0 \\
-2x + 3y -6z - 3 &= 0 \\
\end{aligned}
\right.$$
 
I like Serena said:
Hey I pi rangle! ;)

Which methods do you know to determine the distance between 2 lines?

What you can do for instance with $$l_2$$ is find a support vector and a direction vector, so you can define a point Q similar to your point P with 1 parameter.

If you click Reply With Quote you can see how I put your set of equations in $\LaTeX$.
$$l_2:
\left\{
\begin{aligned}
x-\phantom{1}y+3z-1 &= 0 \\
-2x + 3y -6z - 3 &= 0 \\
\end{aligned}
\right.$$
Hello I like Serena!
Thanks for showing me the latex code!:)
I know two method one when you do from point to plane which I don't prefer. The method I always use that we is calculate the distance between a vector $$|PQ|$$
this parameter I am unsure with what I shall use.

Regards,
$$|\pi\rangle$$
 
Petrus said:
Hello I like Serena!
Thanks for showing me the latex code!:)
I know two method one when you do from point to plane which I don't prefer. The method I always use that we is calculate the distance between a vector $$|PQ|$$
this parameter I am unsure with what I shall use.

Regards,
$$|\pi\rangle$$

Can you solve for instance x and y from the set of equations (leaving z as an unknown)?
 
I like Serena said:
Can you solve for instance x and y from the set of equations (leaving z as an unknown)?
Hello I like Serena
I get y=5 and $$x=4-3z$$

Regards,
$$|\pi\rangle$$
 
Petrus said:
Hello I like Serena
I get y=5 and $$x=4-3z$$

Regards,
$$|\pi\rangle$$

Let's see...

We have $x−y+3z−1=0$.
Substituting your solution, I get:

$$(4-3z)-(5)+3z-1 =0 \\
-2=0 $$

That can't be right!
 
I like Serena said:
Let's see...

We have $x−y+3z−1=0$.
Substituting your solution, I get:

$$(4-3z)-(5)+3z-1 =0 \\
-2=0 $$

That can't be right!
Hello I like Serena
I made some typo sorry!
$$y=5 $$ and $$x=6-3z$$

Regards,
$$|\pi\rangle$$
 
Last edited:
Petrus said:
Hello I like Serena
I made some typo sorry!
$$y=5 $$ and $$x=6-3z$$

Regards,
$$|\pi\rangle$$

Good!

Now point $Q$ can be expressed for instance as $(6-3z, 5, z)$.
Or I might also say that $l_2$ is $(6,5,0)+s(-3,0,1)$.

There are a couple of ways to proceed.
Perhaps you already know a way?
 
I like Serena said:
Good!

Now point $Q$ can be expressed for instance as $(6-3z, 5, z)$.
Or I might also say that $l_2$ is $(6,5,0)+s(-3,0,1)$.

There are a couple of ways to proceed.
Perhaps you already know a way?
Hello I like Serena,
Is my method not valid or I am doing something wrong cause I don't get same answer as facit...
We got that
$$PQ=(5-3z-t,4+2t,z-3+3t)$$
We know that $$PQ*(1,-2,-3)=0$$ and $$PQ*(-3,0,1)=0$$
then I get this equation
$$6t+10z=-18$$
$$14t+6z=6$$
and we get $$t=-\frac{6}{13}$$, $$z=\frac{27}{13}$$
that means $$PQ=(-\frac{10}{13},\frac{40}{13},-\frac{30}{13})$$
but this is wrong, and I can't see where..

Regards,
$$|\pi\rangle$$
 
  • #10
Petrus said:
We got that
$$PQ=(5-3z-t,4+2t,z-3+3t)$$
We know that $$PQ*(1,-2,-3)=0$$ and $$PQ*(-3,0,1)=0$$

Good.

then I get this equation
$$6t+10z=-18$$
$$14t+6z=6$$

I don't. How did you get it?
 
  • #11
I like Serena said:
Good.
I don't. How did you get it?
Hello I like Serena
$$5-3z-t-2(4+2t)-3(z-3+3t)=0$$
$$-3(5-3z-t)+z-3+3t=0$$

Regards,
$$|\pi\rangle$$
 
  • #12
Hello,
Now I get this equation if I have not made any misscalculation...
$$-6z-14t=-6$$
$$-8z+6t=18$$
$$t=\frac{39}{37}$$, $$z=-\frac{54}{37}$$

Regards,
$$|\pi\rangle$$
 
  • #13
This would be my way of continuing:

We begin with the two lines
$$l_1: (x,y,z) = (1,1,3)+t(1,-2,-3)$$
$$l_2: (x,y,z) = (6,5,0)+s(-3,0,1)$$

Now, take any two points on the line, and find the vector going from one to the other. Let's define $P_1=(1,1,3)$ and $P_2=(6,5,0)$. The vector connecting them is then
$$v = P_2 - P_1 = \langle 5,4,-3\rangle$$

Now, you need the component of this vector that is perpendicular to either line. In order to do so, you need some unit vector perpendicular to both $l_1$ and $l_2$. An easy way to find one is to take and normalize the cross product of the line vectors
$$n = \langle 1,-2,-3\rangle \times \langle -3,0,1\rangle
= \langle -2,8,-6 \rangle$$
$$u = n/ \| n \| = \frac{1}{\sqrt{26}} \langle -1,4,-3 \rangle $$

To finish, we have

$$dist = \left| v \cdot u \right| = \left| \frac{1}{\sqrt{26}} ((-1)(5)+(4)(4)+(-3)(-3)) \right|
\\= \frac{20}{\sqrt{26}} \approx 3.92 $$

Is that anything like how you're supposed to do it?
 
Last edited:
  • #14
TheBigBadBen said:
This would be my way of continuing:

We begin with the two lines
$$l_1: (x,y,z) = (1,1,3)+t(1,-2,-3)$$
$$l_2: (x,y,z) = (6,5,0)+s(-3,0,1)$$

Now, take any two points on the line, and find the vector going from one to the other. Let's define $P_1=(1,1,3)$ and $P_2=(6,5,0)$. The vector connecting them is then
$$v = P_2 - P_1 = \langle 5,4,-3\rangle$$

Now, you need the component of this vector that is perpendicular to either line. In order to do so, you need some unit vector perpendicular to both $l_1$ and $l_2$. An easy way to find one is to take and normalize the cross product of the line vectors
$$n = \langle 1,-2,-3\rangle \times \langle -3,0,1\rangle
= \langle -2,8,-6 \rangle$$
$$u = n/ \| n \| = \frac{1}{\sqrt{26}} \langle -1,4,-3 \rangle $$

To finish, we have

$$dist = \left| v \cdot u \right| = \left| \frac{1}{\sqrt{26}} ((-1)(5)+(4)(4)+(-3)(-3)) \right|
\\= \frac{20}{\sqrt{26}} \approx 3.92 $$

Is that anything like how you're supposed to do it?
Hello,
Your method works fine so should my, I have done some with my method and it works, I don't understand what I am doing wrong

Regards,
$$|\pi\rangle$$
 
  • #15
Petrus said:
Hello,
Your method works fine so should my, I have done some with my method and it works, I don't understand what I am doing wrong

Regards,
$$|\pi\rangle$$

Well, I can't tell you what you're doing wrong because I don't really understand your method. Why are you solving for z and t? What information are you looking for, and how does it tie into finding the distance between lines?
 
  • #16
TheBigBadBen said:
Well, I can't tell you what you're doing wrong because I don't really understand your method. Why are you solving for z and t? What information are you looking for, and how does it tie into finding the distance between lines?
Hello TheBigBadBen,
We want to find z and t so we know PQ, notice that P is on line 1 and Q is on line 2 so with other words Distance between two line is same as |PQ| the length of PQ

Regards,
$$|\pi\rangle$$
 
  • #17
Petrus said:
Hello TheBigBadBen,
We want to find z and t so we know PQ, notice that P is on line 1 and Q is on line 2 so with other words Distance between two line is same as |PQ| the length of PQ

Regards,
$$|\pi\rangle$$

All right, now I see. You're saying that for any point P on line 1 and point Q on line 2, we can write the vector between those two points as

$$PQ=(5-3z-t,4+2t,z-3+3t)$$

Which is correct. Presumably, for the right choice of P and Q, we'd be able to say that the length of that vector is the distance between the two lines. Now, what you need to do (and I now see that this is what you were trying to do) is find such a PQ that is perpendicular to both of the lines. Thus, we need to satisfy
$$ PQ \cdot (1,-2,-3) = 0\\
PQ \cdot (-3,0,1) = 0$$
Which will yield a unique PQ connecting the lines. Finding the length of that should yield the correct answer.

Perfect. Now that I understand, I'll try to see where that went wrong.
 
  • #18
Petrus said:
Hello,
Now I get this equation if I have not made any misscalculation...
$$-6z-14t=-6$$
$$-8z+6t=18$$
$$t=\frac{39}{37}$$, $$z=-\frac{54}{37}$$

Regards,
$$|\pi\rangle$$

Your equation is still not quite right.

Let's clean it up a bit.

$$\vec{PQ} = \begin{pmatrix}5\\4\\-3\end{pmatrix} + z \begin{pmatrix}-3\\0\\1\end{pmatrix} - t \begin{pmatrix}1\\-2\\-3\end{pmatrix}$$

Since $\vec{PQ} \perp \begin{pmatrix}1\\-2\\-3\end{pmatrix}$, we get:
$$(5-8+9)+z(-3-3)-t(1+4+9)=0$$
And since $\vec{PQ} \perp \begin{pmatrix}-3\\0\\1\end{pmatrix}$, we get:
$$(-15-3)+z(9+1)-t(-3-3)=0$$

Which equations do you get from that?
 
  • #19
I like Serena said:
Your equation is still not quite right.

Let's clean it up a bit.

$$\vec{PQ} = \begin{pmatrix}5\\4\\-3\end{pmatrix} + z \begin{pmatrix}-3\\0\\1\end{pmatrix} - t \begin{pmatrix}1\\-2\\-3\end{pmatrix}$$

Since $\vec{PQ} \perp \begin{pmatrix}1\\-2\\-3\end{pmatrix}$, we get:
$$(5-8+9)+z(-3-3)-t(1+4+9)=0$$
And since $\vec{PQ} \perp \begin{pmatrix}-3\\0\\1\end{pmatrix}$, we get:
$$(-15-3)+z(9+1)-t(-3-3)=0$$

Which equations do you get from that?
Hello I like Serena,
$$6z+14t=6$$
$$10z+9t=18$$
that means $$t=-\frac{24}{43}$$ and $$z=\frac{99}{43}$$
that means $$PQ=(-\frac{58}{43},\frac{124}{43},-\frac{102}{43})$$
then we get the distance is and that is not the same as facit..

Regards,
$$|\pi\rangle$$
 
  • #20
Petrus said:
Hello I like Serena,
$$6z+14t=6$$
$$10z+9t=18$$
that means $$t=-\frac{24}{43}$$ and $$z=\frac{99}{43}$$
that means $$PQ=(-\frac{58}{43},\frac{124}{43},-\frac{102}{43})$$
then we get the distance is and that is not the same as facit..

Regards,
$$|\pi\rangle$$

Can you try to simplify $(−15−3)+z(9+1)−t(−3−3)=0$ again?
 

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