# Finding the point of intersection of two lines

Gold Member
[SOLVED]Finding the point of intersection of two lines

Hi, I would really, really appreciate it if someone could help me with this.

## Homework Statement

Find the point of intersection between the lines:
$$R_1(\lambda)=[1,\hspace {4} 0 \hspace {4} ,-1] + \lambda[1, \hspace {4} 1,\hspace {4} 1]$$

$$R_2(\lambda)=[1, \hspace {4}2,\hspace {4} 1] +\mu[4, \hspace {4}2, \hspace {4}2,]$$

## The Attempt at a Solution

I'm not really sure how to do this, my line of thought was that I need to somehow get rid of the constants variable $$\lambda and \mu$$ by equating

$$R_1(\lambda)=R_2(\mu)$$
I got:

$$\mu[4,2,2]-\lambda[1,1,1]=[0,-2,-2]$$

This is Where I think I'm stuck. I think I need a second equation that relates lambda and mu but I can't seem to find one, I think that it may be something to do with the angles of two intersecting lines before and after the point of intersection being the same, but really I'm just grasping at straws.

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HallsofIvy
Homework Helper
Just set up the equations for the individual x, y, z, coordinates:

In $R_1(\lambda)= [1, 0 ,-1] + \lambda[1, 1, 1]$
$x= 1+ \lambda$, $y= \lambda$ and $z= -1+ \lambda$

In $R_2(\mu)= [1, 2, 1]+ \mu [4, 2, 2]$
$x= 1+ 4\mu$, $y= 2+ 2\mu$,and $z= 1+ 2\mu$
Now set those equal:
$1+ \lambda= 1+ 4\mu$, $\lambda= 2+ 2\mu$, and $-1+\lambda= 1+ 2\mu$.

That gives you three equations for the two unknown numbers $\lambda$ and $\mu$. "In general", you can't solve three equations for two unknowns because, "in general" two lines in three dimensions are "skew"- they don't intersect. Go ahead and solve 2 of the equations for $\lambda$ and $\mu$, then put those values into the third equation to see if they satisfy that equation. If they do, those values of $\lambda$ and $\mu$ give the point of intersection. If they don't then the lines do not intersect.

Gold Member
Just set up the equations for the individual x, y, z, coordinates:

In $R_1(\lambda)= [1, 0 ,-1] + \lambda[1, 1, 1]$
$x= 1+ \lambda$, $y= \lambda$ and $z= -1+ \lambda$

In $R_2(\mu)= [1, 2, 1]+ \mu [4, 2, 2]$
$x= 1+ 4\mu$, $y= 2+ 2\mu$,and $z= 1+ 2\mu$
Now set those equal:
$1+ \lambda= 1+ 4\mu$, $\lambda= 2+ 2\mu$, and $-1+\lambda= 1+ 2\mu$.

That gives you three equations for the two unknown numbers $\lambda$ and $\mu$. "In general", you can't solve three equations for two unknowns because, "in general" two lines in three dimensions are "skew"- they don't intersect. Go ahead and solve 2 of the equations for $\lambda$ and $\mu$, then put those values into the third equation to see if they satisfy that equation. If they do, those values of $\lambda$ and $\mu$ give the point of intersection. If they don't then the lines do not intersect.

Thanks a bundle!

I got them to $$\lambda = 4$$ and $$\mu = 1$$ and they satisfied the third equation. But I don't really understand why $$\lambda$$ and $$\mu$$ give the point of intersection . . .

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cristo
Staff Emeritus
Thanks a bundle!

I got them to $$\lambda = 4$$ and $$\mu = 1$$ and they satisfied the third equation. But I don't really understand why $$\lambda$$ and $$\mu$$ give the point of intersection . . .

You've found the values of $\lambda$ and $\mu$ such that $R_1(\lambda)=R_2(\mu)$, so plugging your value for $\lambda$ into R_1 will give you a point. Plugging your value for $\mu$ into R_2 will give you another point. But necessarily, these points will be the same. Hence this is the point of intersection.

Gold Member
You've found the values of $\lambda$ and $\mu$ such that $R_1(\lambda)=R_2(\mu)$, so plugging your value for $\lambda$ into R_1 will give you a point. Plugging your value for $\mu$ into R_2 will give you another point. But necessarily, these points will be the same. Hence this is the point of intersection.

Ah, I understand it now, thanks a million guys :D