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I've found this challenge to solve that I have no idea where to start. We are supposed to find the symmetric line to this one

$r: (10,1,2) + \lambda (3,1,1)$

with regard to the plane

$\alpha \ \{ 3x+y-z-2 = 0$

Where should I begin? The info that I've extracted are:

Intersection point of $r$ and $\alpha$: $(1,-2,-1)$

Vector of $\alpha$: (3,1,-1)

Vector of line $r$: $(3,1,1)$

Thank you.

bru1987
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  • What does a "symmetric line ... with regard to the plane" mean? –  Sep 08 '15 at 21:20
  • @Bye_World sorry for the confusion. The lines are a reflection one of the other, as if the plane was a mirror – bru1987 Sep 08 '15 at 21:23
  • Oh OK, the line reflected through the plane. Have you taken linear algebra? –  Sep 08 '15 at 21:25
  • @Bye_World es that Is correct. Not really, we are supposed to use only vectors, projections and general concepts of lines and planes. – bru1987 Sep 08 '15 at 21:31
  • Can you figure out how to decompose your direction vector, $(3,1,1)$, into a component parallel to your plane and a component orthogonal to your plane? –  Sep 08 '15 at 21:34
  • @Bye_World Yes I can, that can be done. After that, how can I find the vector of the new line? – bru1987 Sep 08 '15 at 21:39

1 Answers1

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HINT: Your line will be of the form $$\mathbf r_{\text{reflect}}(t) = \mathbf vt+\mathbf b$$

Because you found the intersection of $\mathbf r$ and $\mathbf \alpha$, you can use that for your postion vector $\mathbf b$. Why? To find $\mathbf v$ just decompose the direction vector of $\mathbf r$, $(3,1,1)$, into a vector parallel to the plane and a vector perpendicular to the plane. $$(3,1,1) = \mathbf u = \mathbf u_\| + \mathbf u_\perp$$

$\mathbf v$ will simply be $\mathbf v = \mathbf u_\| - \mathbf u_\perp$. Why?