How can we determine the intersection point of two lines with given equations?

[ineedhelpnow]

determine whether the lines $L_1$ and $L_2$ are parallel, skew, or intersecting. if they intersect, find the point of intersection.

$L_1: \frac{x-2}{1}= \frac{y-3}{-2}= \frac{z-1}{-3}$

$L_2: \frac{x-3}{1}= \frac{y+4}{3}= \frac{y+4}{3}= \frac{x-2}{-7}$

i understand if one vector is the scalar multiple of the other then the lines are parallel. if their dot product is equal to 0 then they intersect. according to my book they intersect but i can't seem to understand how. the two vectors we're dealing with here are $\left\langle 1,-2,-3 \right\rangle$ and $\left\langle 1,3,-7 \right\rangle$, right?

 

[Prove It]

determine whether the lines $L_1$ and $L_2$ are parallel, skew, or intersecting. if they intersect, find the point of intersection.

$L_1: \frac{x-2}{1}= \frac{y-3}{-2}= \frac{z-1}{-3}$

$L_2: \frac{x-3}{1}= \frac{y+4}{3}= \frac{y+4}{3}= \frac{x-2}{-7}$

i understand if one vector is the scalar multiple of the other then the lines are parallel. if their dot product is equal to 0 then they intersect. according to my book they intersect but i can't seem to understand how. the two vectors we're dealing with here are $\left\langle 1,-2,-3 \right\rangle$ and $\left\langle 1,3,-7 \right\rangle$, right?

Not quite.

The lines are parallel if they have the same direction vectors - or at least if they go in the same direction, so like you said, a scalar multiple of each other.

Two lines intersect if there is some point where they have the same x, y, z values (i.e. they pass through the same co-ordinate). So determine each line's parametric equations and set them equal.

Skew lines are two lines that do not intersect and are not parallel.

Just a side note: If the dot product of the two direction vectors is 0, then the lines are PERPENDICULAR.

 

[ineedhelpnow]

perpendicular. meaning they intersect, right?

 

[Dethrone]

Perpendicular does not imply that the vectors intersect, it just means that the vectors are 90 degrees to each other. You have correctly identified the direction vectors, and clearly, they're not scalar multiplies of each other as you have correctly surmised. We have two other options; either the lines are skewed, or intersecting. How can we determine that?

 

[ineedhelpnow]

im not sure.

 

[Dethrone]

If the two lines intersect, then they both must share the same point $(x, y, z)$. Which form of equation (parametric, vector, symmetric) allows us to easily compare the x's, y's, and z's?

 

[ineedhelpnow]

parametric

 

[Dethrone]

Yep :D Now if we have both equations in parametric form and we equate them, what do we find?

 

[ineedhelpnow]

by equating them do you mean solve for a variable in the first one and then plug it into the second equation?

 

[Dethrone]

If we have both equations in parametric form, we can equate the equations of $x$, $y$, and $z$ of each equation. For example, $L_1$ has $x=t+2$, and $L_2$ has $x=s+3$. If they indeed share the same x-coordinate, then we can equate the two x-equations: $t+2=s+3$ Then, we will have three equations and two unknowns, $t$ and $s$ which you will have to determine. If there exists such parameter $t$ and $s$ that satisfies the above equations, then they intersect; if there isn't, then they lines don't intersect.

The difference between points in $R^2$ and $R^3$ is that in $R^2$, if the lines aren't parallel, they must intersect at some point. This is not true for $R^3$. If the lines aren't parallel and do not intersect, we call them skewed.

 

[ineedhelpnow]

oh i see what you mean. sorry. not familiar with the term "equating"

 

[ineedhelpnow]

so i subtract the second equation $4-t=3-2s$ from the first one $3+2t=1+4s$ but that still doesn't help me solve for s. or t.

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um oops. ignore that lost post. that's a totally different problem. lol now that i posted it anyways, in that case would the lines be skew because you can't solve for neither s nor t?

 

[Dethrone]

I don't see where you're getting your equations. I've previously said that $t+2=s+3$. Looking at $y$, we have $y=-2t+3$ and $y=3s+4$, so equating gives $-2t+3=3s+4$. Do you have all three equations? Are you saying the equations in your first post is skewed, or the equations you just mentioned?

 

[ineedhelpnow]

I don't see where you're getting your equations. I've previously said that $t+2=s+3$. Looking at $y$, we have $y=-2t+3$ and $y=3s+4$, so equating gives $-2t+3=3s+4$. Do you have all three equations? Are you saying the question posted on this thread is skewed, or the equations you just mentioned?

so sorry. i was looking at the problem before this one in my book

 

[Dethrone]

You are right that the lines are skewed. :D

 

[ineedhelpnow]

is the reason right? because you can't find their points on intersection?

 

[Dethrone]

Yes, but if we can find the parameters $s$ and $t$ that satisfies both equations, then we can plug one of the parameters to its respective parametric equation. It doesn't matter which parameter you plug it, they will both retrieve the same coordinate of intersection. (Yes)

 

[ineedhelpnow]

but since we cannot find them that means they do not intersect? :)

 

[Dethrone]

Yes, it means there's no point of intersection. :D

 

[ineedhelpnow]

ok so back to the original problem. i believed you made a mistake for one of the equations. it should be $3-2t=-4+3s$ not $3-2t=4+3s$ which is what you said. even by doing this and subtracting the first two equations i don't understand how to find the point of intersection because I am still not being given a value for s or t

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$L_1: \frac{x-2}{1}= \frac{y-3}{-2}= \frac{z-1}{-3}$

$L_2: \frac{x-3}{1}= \frac{y+4}{3}= \frac{y+4}{3}= \frac{x-2}{-7}$

i just realized that for the second line i put an extra $\frac{y+4}{3}$. its supposed to be $L_2: \frac{x-3}{1}= \frac{y+4}{3}= \frac{x-2}{-7}$

 

[Dethrone]

Your second symmetric equation doesn't make sense. Where's the $z$? Sorry about the mistake, I was using a pen that was kind of drying out...and the negative sign didn't go through on the paper...:D I'm glad you caught it, it means you're putting your weight into the question as well, and not just taking my word for it.

 

[ineedhelpnow]

oh wow you actually do it on paper first? man that's real dedication :) my second symmetric equation is $\frac{x-3}{1}= \frac{y+4}{3}= \frac{x-2}{-7}$ in the original question i mistakenly posted the y part of it twice.

 

[Dethrone]

So we have the three equations:

$$t+2=s+3$$
$$-2t+3=3s-4$$
$$-3t+1=-7s+2$$

Agree or disagree?

I'm sure you mean $\frac{x-3}{1}= \frac{y+4}{3}= \frac{z-2}{-7}$?

my second symmetric equation is $\frac{x-3}{1}= \frac{y+4}{3}= \frac{x-2}{-7}$

 

[ineedhelpnow]

yeah that's correct. and for the symmetric equation (i apologize), i didnt realize that i was putting x instead of z. it is supposed to be z instead

 

[Dethrone]

We have to isolate for $t$ and $s$. Although the method of substitution works, I think elimination works better. What do you think?

 

[ineedhelpnow]

its been a long long long long time since I've done elimination. do you mind showing me solve for the first two equations so i can see how it's done?

 

[Dethrone]

The goal of elimination is to eliminate one variable. Examining the three equations, it seems like it would be easiest if we multiply equation $1$ by 2, and add it to equation $2$ so that the $t$'s cancel.

We have equation $1$:

$$t+2=s+3$$

Multiply both sides by $2$:

$$2t+4=2s+6$$

What do we get when we add that equation to equation $2$, which is $-2t+3=3s-4$?

 

[ineedhelpnow]

oh wait that's what I've been doing the whole time (turns out I've dome elimination more than i thought) only i didnt multiply by 2

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s=1

 

[Dethrone]

Use $s$ to determine $t$. Now check if $s$ and $t$ satisfies all three of the equations. We are given three equations but two unknowns, meaning that we are given more equations than we need. Using elimination means that equations $1$ and $2$ are satisfied, but not necessarily the 3rd one.

To check further, you can plug both $s$ and $t$ into their respective parametric equations. They must give the same coordinate.

 

[ineedhelpnow]

t=3

 

[Dethrone]

Yep, now apply the methods of checking I mentioned above. (Nod)

 

[ineedhelpnow]

im actually not getting the same answers. i think i made a mistake.

 

[ineedhelpnow]

t=2

 

[ineedhelpnow]

it does satisfy the third equation as well. so now i plug in s and t into the parametric equations and and get x,y,z which turns out to be (4,-1,-5) and i checked with the back of the book. it's right :)