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I will fully disclose that this is a homework question. I would prefer not to be given an answer directly, and am looking for more of an indication as to whether I am on the right track. The problem with the courses I am working with is that they only show examples, and do not explain exactly how it "works".

Given $l_1 = (6,-1,0)+t(3,1,-4)$ and $l_2 = (4,0,5)+s(-1,1,5)$ find the intersection of $l_1$ and $l_2$.

First I took $d_1 = (3,1,-4)$ and $d_2 = (-1,1,5)$,

Then I made sure that they did not have the same ratio. (if they have the same ratio this would indicate that they are either coincident or parallel) they do not have the same ratio, so they either intersect at a point or are skew.

Then I made parametric equations:

$l_1:$

$$\begin{align} x & = 6 + 3t\\ y & = -1 + t\\ z & = -4t\\ \end{align}$$

$l_2:$

$$\begin{align} x & = 4 - s\\ y & = s\\ z & = 5 + 5s\\ \end{align}$$

Then I equated them to eachother:

$$\begin{align} 6 + 3t & = 4 - s\\ -1 + t & = s\\ -4t &= 5 + 5s\\ \end{align}$$

I moved the unknowns to one side:

$$\begin{align} 3t + s & = -2 \qquad & \text{(we'll call this equation $1$)}\\ t - s & = 1 \qquad & \text{(we'll call this equation $2$)}\\ -4t - 5s & = 5 \qquad & \text{(we'll call this equation $3$)}\\ \end{align}$$

This is where it gets tricky. If I take equation $(1)$ and $(2)$, I can cancel out the $s$ value, but the values both become strange, where $t$ is $\frac 34$ and $s$ is $-2 (\frac 34)$, obviously the left and right hand sides don't match.

But I I take equation $(2)$ and $(3)$, the left and right hand sides do match, and then if I go to find the point of intersection I get decimal values for coordinates (why would that be the case?)

Any help would be great. I just want to know what I'm doing wrong. Please don't just give me the answer.

Edit:

I am not sure why people are digging this up to down-vote it, and would appreciate a comment explaining your down-vote.

  • Note that these are lines in space so even though they are not parallel, they may not intersect. – hmakholm left over Monica Apr 03 '17 at 17:34
  • @HenningMakholm They certainly could be skew, you're correct, I'll include that in my OP. –  Apr 03 '17 at 17:36
  • Take any two equations and solve for $s$ and $t$. Put the value in third eqn. If it satisfies, hurrah! – jonsno Apr 03 '17 at 17:37
  • @samjoe I'm doing that, but usually the LS and RS are whole numbers, I'm getting crazy numbers like 8.75 and 17.75. I was hoping someone could explain to me if that's okay, if the numbers can be crazy numbers. –  Apr 03 '17 at 17:38
  • As long as they are equally crazy... – jonsno Apr 03 '17 at 17:40
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    @Dodsy: It's not wrong to get non-whole numbers, just more calculation work. – hmakholm left over Monica Apr 03 '17 at 17:44
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    +1 for your honesty and your work! By the way, welcome to the site! If you want some basic information about writing math on MSE see here, here, here and here. –  Apr 03 '17 at 17:46
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    @InfiniteMonkey Thank you for this! I will remember this for any questions I ask in the future. –  Apr 03 '17 at 20:03
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    No problem! There is a very handy little tool called Detexify which lets you draw any symbol you are looking for and it will tell you what is the TeX command for it. Also, if you use Mathematica, a useful command is TeXForm[expr] which converts the Wolfram input expr into TeX. –  Apr 03 '17 at 20:12

2 Answers2

2

Your approach is correct so far. Since equation 2 has a nice coefficient of $1$ in front of the $t$, we can use that to easily eliminate the $t$ from the two others:

$$ Eq_1 - 3Eq_2: \qquad\qquad 4s = -5 \\ Eq_3 + 4Eq_2: \qquad \qquad -9s = 9 $$ These require different values of $s$, so they can never be satisfied at the same time. So the system of equations has no solutions, and the two lines do not intersect.

hmakholm left over Monica
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  • Thanks, I see now that the reason I was getting crazy decimal numbers was because of my choice of using equation 2 and 3 and then subbing the s value into 3. If I sub the s value into 2 instead, I get 0, then when I find the LS and RS, it comes out to 6 and 5- obviously a skew. I can't thank everyone enough for the confirmation that I was on the right track, as this was the biggest obstacle in my way with this question.

    Thanks for your help.

     –  Apr 03 '17 at 19:58
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Here is another way to find the distance between two lines in any number of dimensions.

If the lines intersect, the distance will be zero.

Find shortest distance between lines in 3D

marty cohen
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