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Given planes:

$$\pi_1: 3x+12y-3z-5=0$$ $$\pi_2: 3x-4y+9z+7=0$$

and lines $$l_1: x=3-2t, y=-1+3t, z=2+3t$$ $$l_2: x=-5+2t, y=3-4t, z=-1+3t$$ find a line that is parallel to the planes and intersects with these two lines.

For the new line to be parallel to two planes its direction vector has to be cross product of $[-2,3,3]$ and $[2,-4,3]$. But then I can choose only one point for the parametric form of this line. But lines $l_1$ and $l_2$ do not intersect. What should I do? Thanks for your help.

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  • $\begingroup$ did you already draw everything? $\endgroup$ – janmarqz Apr 4 '18 at 17:20
  • $\begingroup$ There’s nothing in the problem that says that all three lines intersect at a single point. The line you‘re asked to find intersects $l_1$ and $l_2$ separately. $\endgroup$ – amd Apr 4 '18 at 20:01
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The required line is in the direction of $(3,12,-3)\times (3,-4,9)=12(8,-3,-4)$

Suppose that line meets $l_1$ at $(3-2s,-1+3s,2+3s)$ and $l_2$ at $(-5+2t,3-4t,-1+3t)$. Then

$$(3-2s+5-2t,-1+3s-3+4t,2+3s+1-3t)\parallel (8,-3,-4)$$

$$\frac{8-2s-2t}{8}=\frac{-4+3s+4t}{-3}=\frac{3+3s-3t}{-4}$$

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  • $\begingroup$ Is there a way to express it using only one parameter? $\endgroup$ – maq Apr 5 '18 at 8:26
  • $\begingroup$ Once you have solved the equations and found $s$ and $t$, you have two points on the required line. You can subtract them to obtain the direction vector of the line. The line can be expressed in one parameter. $\endgroup$ – CY Aries Apr 5 '18 at 8:31

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