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Find the equation of a plane which is perpendicular to the plane $\pi\equiv x+2y-2z+3=0$ and it intersects it through the line that lies in the XOZ plane.

Normal vector of the given plane is $\overrightarrow{n}=(1,2,-2)$ and since the line lies in the XOZ plane its direction vector is $\overrightarrow{c}=(a,0,b)$. Normal vector of the plane which I'm looking for should be perpendicular to both of these vectors and when I apply cross product I get $\overrightarrow{m}=(2b,-(b+2a),-2a)$. And now I'm stuck.

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    $\begingroup$ There are infinitely many planes perpendicular to a given one. The only condition to be perpendicular is that the normal vector of one of them is a direction of the other. $\endgroup$
    – mfl
    May 13, 2014 at 19:16
  • $\begingroup$ @ManuelFdzLpz If the normal vector of one of them is a direction of the other that is if they both have the same normal vector, then they should be parallel. $\endgroup$
    – Nick
    May 13, 2014 at 19:27

1 Answer 1

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The points on the line of intersection $\cal L$ of $\pi$ and the $xz$-plane satisfy $x + 2(0) - 2z + 3 = 0$, so $x - 2z = -3$. In particular, points $(0,0,\frac{3}{2})$, and $(1,0,2)$ are on the line, so $\cal L$ has direction vector $<1,0,\frac{1}{2}>$, or $<2,0,1>$. So the plane perpendicular to $\pi$ through $\cal L$ should have normal vector that is perpendicular to both $<1,2,-2>$ and $<2,0,1>$, and the plane passes through $(1,0,2)$, so this should be enough to finish up the problem.

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