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Given: $$Line\ l \leftrightarrow \frac{2x-3}{2} = \frac {y}{2}=\frac{2z+1}{6}$$ $$Line\ m \leftrightarrow 3x-5=0 \vert 3y+3z+2=0$$

Question find the point L on l and the point M on m such that LM is parallel to p with following equation: $$p \leftrightarrow x= \frac {y}{6}=\frac {z}{2}$$

I solved this by chosing two points $L (x1,y1,z1)$ and $M (x2,y2,z2)$. I know that $M-L // p$. So i have the following system of equations: $$\frac{5}{3}-(\lambda+\frac{3}{2})=k$$ $$\mu-2\lambda=6k$$ $$-\mu-\frac{2}{3}-(3\lambda-\frac{1}{2})=2k$$ And then I filled in the parametric equation of m for M and that of l for L.

As a solution I got $L (9/5;3/5;-1/5)$ and $M (5/3;-1/5;-7/15)$. Is this correct or did I do something wrong?

Michthan
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    https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – John Wayland Bales Feb 04 '18 at 17:01
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    Have you tried checking your own work? It’s a simple matter to verify whether or not the conditions of the problem are met by your proposed solution. – amd Feb 04 '18 at 19:37
  • @amd I already found another solution, but none of both solutions fit the problem – Michthan Feb 05 '18 at 18:11
  • @John Wayland Bales, is the question better now? – Michthan Feb 05 '18 at 19:06
  • Your original solution had the problem that it wasn’t parallel to the given line. Your current solution does produce a line with the correct direction, but point $L$ doesn’t lie on line $l$. Without seeing the steps of your solution it’s impossible to say for sure where you’re going wrong. – amd Feb 05 '18 at 19:37
  • Yes, thanks for using MathJax. – John Wayland Bales Feb 06 '18 at 00:25
  • Looks like you might be getting $y$ and $z$ mixed up when you set up the final system of equations. That said, the solution you have at the end of the question doesn’t satisfy the system of equations just above it. – amd Feb 09 '18 at 18:23
  • @amd, I find $\mu$ and $\lambda$ out of the equation and then I use these to find L and M – Michthan Feb 10 '18 at 18:24

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One way to start is by finding parametric equations for $l$ and $m$. For the first line, this can be done by setting the given equation equal to $\lambda$ and then solving for $x$, $y$ and $z$. Doing this should give you something like $$L=\lambda\left(1,2,3\right)+\left(\frac32,0,-\frac12\right).$$

For line $m$, one way to find a parameterization is to switch to homogeneous coordinates. The line is essentially given as the intersection of two planes. If we homogenize those equations, then the line can be described as the null space of the matrix $$\begin{bmatrix}3&0&0&-5\\0&3&3&2\end{bmatrix}.$$ This matrix is already in echelon form, so one can read from it that its null space is spanned by $[5,-2,0,3]^T$ and $[0,-1,1,0]^T$. Converting that back into inhomogeneous Cartesian coordinates yields $$M=\mu\left(0,-1,1\right)+\left(\frac53,-\frac23,0\right).$$ If there’s another way of finding this parametric equation that makes more sense to you, use it, but you should end up with something equivalent.

Now, make use of the condition that $(L-M)$ is parallel to $p$. Taking advantage of the fact that you’re working in $\mathbb R^3$, this condition can be expressed as $(L-M)\times \mathbf p=0$, where $\mathbf p$ is the line’s direction vector. If you expand this out using the above parameterizations of $L$ and $M$, you’ll end up with a system of linear equations in $\lambda$ and $\mu$, which I expect you know how to solve.

amd
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  • With x being a vector product? Because I solved the system of equations you can see on top. (I'll edit it to show the full system in a bit) – Michthan Feb 09 '18 at 12:18
  • @Michthan That’s right: $\times$ stands for cross product. – amd Feb 09 '18 at 17:47