# Distance between two lines

#### Petrus

##### Well-known member
Hello MHB,
This is a problem from my book,
calculate distance between line $$\displaystyle l_1: (x,y,z) = (1,1,3)+t(1,-2,-3)$$ and (hope it work's read $$\displaystyle l_2$$ cause I don't know how to write that in latex)
I am stuck when I got $$\displaystyle l_2$$ in plane.
I start with $$\displaystyle l_1$$
$$\displaystyle P=(1+t,1-2t,3-3t)$$
what shall I do with $$\displaystyle l_2$$?

Regards,
$$\displaystyle |\pi\rangle$$

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello MHB,
This is a problem from my book,
calculate distance between line $$\displaystyle l_1: (x,y,z) = (1,1,3)+t(1,-2,-3)$$ and (hope it work's read $$\displaystyle l_2$$ cause I don't know how to write that in latex)
I am stuck when I got $$\displaystyle l_2$$ in plane.
I start with $$\displaystyle l_1$$
$$\displaystyle P=(1+t,1-2t,3-3t)$$
what shall I do with $$\displaystyle l_2$$?

Regards,
$$\displaystyle |\pi\rangle$$
Hey I pi rangle! Which methods do you know to determine the distance between 2 lines?

What you can do for instance with $$\displaystyle l_2$$ 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.

#### Petrus

##### Well-known member
Hey I pi rangle! Which methods do you know to determine the distance between 2 lines?

What you can do for instance with $$\displaystyle l_2$$ 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 dont prefer. The method I always use that we is calculate the distance between a vector $$\displaystyle |PQ|$$
this parameter I am unsure with what I shall use.

Regards,
$$\displaystyle |\pi\rangle$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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 dont prefer. The method I always use that we is calculate the distance between a vector $$\displaystyle |PQ|$$
this parameter I am unsure with what I shall use.

Regards,
$$\displaystyle |\pi\rangle$$
Can you solve for instance x and y from the set of equations (leaving z as an unknown)?

#### Petrus

##### Well-known member
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 $$\displaystyle x=4-3z$$

Regards,
$$\displaystyle |\pi\rangle$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello I like Serena
I get y=5 and $$\displaystyle x=4-3z$$

Regards,
$$\displaystyle |\pi\rangle$$
Let's see...

We have $x−y+3z−1=0$.

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

That can't be right!

#### Petrus

##### Well-known member
Let's see...

We have $x−y+3z−1=0$.

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

That can't be right!
Hello I like Serena
$$\displaystyle y=5$$ and $$\displaystyle x=6-3z$$

Regards,
$$\displaystyle |\pi\rangle$$

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello I like Serena
$$\displaystyle y=5$$ and $$\displaystyle x=6-3z$$

Regards,
$$\displaystyle |\pi\rangle$$
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?

#### Petrus

##### Well-known member
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
$$\displaystyle PQ=(5-3z-t,4+2t,z-3+3t)$$
We know that $$\displaystyle PQ*(1,-2,-3)=0$$ and $$\displaystyle PQ*(-3,0,1)=0$$
then I get this equation
$$\displaystyle 6t+10z=-18$$
$$\displaystyle 14t+6z=6$$
and we get $$\displaystyle t=-\frac{6}{13}$$, $$\displaystyle z=\frac{27}{13}$$
that means $$\displaystyle PQ=(-\frac{10}{13},\frac{40}{13},-\frac{30}{13})$$
but this is wrong, and I can't see where..

Regards,
$$\displaystyle |\pi\rangle$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
We got that
$$\displaystyle PQ=(5-3z-t,4+2t,z-3+3t)$$
We know that $$\displaystyle PQ*(1,-2,-3)=0$$ and $$\displaystyle PQ*(-3,0,1)=0$$
Good.

then I get this equation
$$\displaystyle 6t+10z=-18$$
$$\displaystyle 14t+6z=6$$
I don't. How did you get it?

#### Petrus

##### Well-known member
Good.

I don't. How did you get it?
Hello I like Serena
$$\displaystyle 5-3z-t-2(4+2t)-3(z-3+3t)=0$$
$$\displaystyle -3(5-3z-t)+z-3+3t=0$$

Regards,
$$\displaystyle |\pi\rangle$$

#### Petrus

##### Well-known member
Hello,
Now I get this equation if I have not made any misscalculation...
$$\displaystyle -6z-14t=-6$$
$$\displaystyle -8z+6t=18$$
$$\displaystyle t=\frac{39}{37}$$, $$\displaystyle z=-\frac{54}{37}$$

Regards,
$$\displaystyle |\pi\rangle$$

##### Active member
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?

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#### Petrus

##### Well-known member
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,
$$\displaystyle |\pi\rangle$$

##### Active member
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,
$$\displaystyle |\pi\rangle$$
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 in to finding the distance between lines?

#### Petrus

##### Well-known member
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 in to finding the distance between lines?
We wanna 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,
$$\displaystyle |\pi\rangle$$

##### Active member
We wanna 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,
$$\displaystyle |\pi\rangle$$
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

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

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.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello,
Now I get this equation if I have not made any misscalculation...
$$\displaystyle -6z-14t=-6$$
$$\displaystyle -8z+6t=18$$
$$\displaystyle t=\frac{39}{37}$$, $$\displaystyle z=-\frac{54}{37}$$

Regards,
$$\displaystyle |\pi\rangle$$
Your equation is still not quite right.

Let's clean it up a bit.

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

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?

#### Petrus

##### Well-known member
Your equation is still not quite right.

Let's clean it up a bit.

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

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,
$$\displaystyle 6z+14t=6$$
$$\displaystyle 10z+9t=18$$
that means $$\displaystyle t=-\frac{24}{43}$$ and $$\displaystyle z=\frac{99}{43}$$
that means $$\displaystyle PQ=(-\frac{58}{43},\frac{124}{43},-\frac{102}{43})$$
then we get the distance is and that is not the same as facit..

Regards,
$$\displaystyle |\pi\rangle$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello I like Serena,
$$\displaystyle 6z+14t=6$$
$$\displaystyle 10z+9t=18$$
that means $$\displaystyle t=-\frac{24}{43}$$ and $$\displaystyle z=\frac{99}{43}$$
that means $$\displaystyle PQ=(-\frac{58}{43},\frac{124}{43},-\frac{102}{43})$$
then we get the distance is and that is not the same as facit..

Regards,
$$\displaystyle |\pi\rangle$$
Can you try to simplify $(−15−3)+z(9+1)−t(−3−3)=0$ again?