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We have the point $A(1,2,1)$ and the lines:

d1: $$x+2y-z+1 = 0 , x-y+z-1=0$$ d2: $$2x-y+z=0, x-y+z=0$$

Write the equation of the plane that passes through $A$ and is parallel to the two lines.

I'm really lost on this one. Got no clue whatsoever on what I have to do here.

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  • $\begingroup$ First step, can you calculate the parametric equations of each line? $\endgroup$ – imranfat Jan 26 '16 at 14:50
  • $\begingroup$ No, that's what really bugs me. I'm not used to dealing with line equations in this form. $\endgroup$ – MikhaelM Jan 26 '16 at 14:55
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Here is an outline. I am renaming your known point to $P$ so I can use the standard notation in my step 4.

  1. Find parametric equations for each line. One way to do that is to take $z=t$ then treat $t$ as a constant. That gives you two equations in two unknowns $x$ and $y$ which you can solve by the usual techniques. You will get linear expressions for $x$ and $y$ in terms of $t$, which are your parametric equations.
  2. From those parametric equations, get a vector parallel to each line. Just use the coefficients in front of $t$ in the expressions for $x$, $y$, and $z$.
  3. Find the cross-product of those two vectors. That vector, perpendicular to the two lines, will also be perpendicular to your desired plane, so it is a normal vector for the plane.
  4. The coordinates of that cross-product are the coefficients of the general linear equation $Ax+By+Cz=D$ for your desired plane. Substitute the coordinates of your known point $P$ to find the constant $D$. Now you are done.

Let me know if you have trouble in any of those steps.

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  • $\begingroup$ Thank you!!! I got the same answer as my answer key, which is $x-y+z=0$ as the plane equation. I just wasn't used to working with those line equations, but once I converted them to parametric form, it all made sense. $\endgroup$ – MikhaelM Jan 26 '16 at 15:16

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