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I'm having a lot of trouble trying to figure this out. Our prof hasn't covered this and we our assignment is due in two days. I'm trying to figure this out on my own:

Consider the two lines $L_1:x=-2t, y=1+2t, z=3t$ and $L_2:x=-1+4s,y=3+2s,z=5+s$. Find the point $P:(x,y,z)$ of intersection of the two lines.

Any ideas?

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  • $\begingroup$ Any ideas – yes, I know the answer. Do you have any ideas? Do you understand what the line definitions mean? Do you understand what a point of intersection is? $\endgroup$ – Ben Millwood Feb 25 '13 at 0:37
  • $\begingroup$ very likely he first equation in $L_1$ should read $x=-2t.$ Edited to that. $\endgroup$ – Will Jagy Feb 25 '13 at 0:51
  • $\begingroup$ Yea, it was a mis-type from his part...thanks for noticing that. $\endgroup$ – Dimitri Feb 25 '13 at 0:54
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At the point of intersection, the coordinates must satisfy the equations of both lines. Thus: $x_1=x_2, y_1=y_2,$ and $ z_1=z_2$. You have the following system of $t,s$:

$$-2t=-8+4s $$ $$1+2t=3+2s$$ $$3t=5+s$$

This system has a unique solution: $t=2,s=1$. Therefore, the point of intersection is: $P(-4,5,6)$

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A point on both lines will satisfy both equations. You can then write $$-2t=-8+4s \\ 1+2t = 3 + 2s \\ 3t=5+s$$

You can pick any two, solve them for $s,t$ and plug into the lines. You should get the same point no matter which line you plug into-a good check.

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  • $\begingroup$ I did exactly that which gave me s = -3 and t = 2/3 However, plugging those values back, I get the wrong answer for some reason. $\endgroup$ – Dimitri Feb 25 '13 at 0:44
  • $\begingroup$ Adding the first 2 equations together gives, 1=-5+6s which would imply s=1. $\endgroup$ – JB King Feb 25 '13 at 0:49
  • $\begingroup$ Got it! Thanks a lot! $\endgroup$ – Dimitri Feb 25 '13 at 0:54
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To complement the other solutions, you can also note that the system of equations you obtained is equivalent to the matrix equation $$ \begin{bmatrix} 4 & 2 \\ 2 & -2 \\ 1 & -3 \end{bmatrix} \begin{bmatrix} s \\ t \end{bmatrix} = \begin{bmatrix} 8 \\ -2 \\ -5 \end{bmatrix}. $$

This matrix equation is also equivalent to saying that the right hand side (which represents a vector from a point on one line to a point on the other line) is a linear combination of the direction vectors of the lines. It is a worthwhile exercise to see that this condition is equivalent to the two lines intersecting.

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Set $x=x$, $y=y$, etc. Then solve for $s$ and $t$. Then plug back into one of the line equations to get $x$, $y$, and $z$.

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