Find the equation of the plane through the intersection of the planes of $x-2y+z=1$ and $2x+y+z=8$ and parallel to the line:

$\frac{x-3}{1} = \frac{y-1}{2} = \frac{z-2}{1} $

I'm facing difficulties solving this problem- how do I combine the parallel attribute and the intersection of the two planes? As far as I know, there would be an infinite number of planes through the line that is the intersection of the two given planes.

(Not homework: preparing for a test)

How I can get the equation of the plane in the $a(x-x')+b(y-y')+c(z-z')=0$ form?


  • 1
    $\begingroup$ Use the two planes to find the line of intersection. The two lines intersect at some point, then use the two lines to get two other points in the desired plane. Then use the cross product (or however you want) to get the normal. $\endgroup$ – copper.hat Jan 16 '14 at 19:45
  • $\begingroup$ @Cameron Buie. I am sorry. I am wrong. I delete my answer. I had understood. Thanks $\endgroup$ – nadia-liza Jan 16 '14 at 21:04

You are absolutely correct that there will be infinitely-many planes through the line of intersection $U$ of the two given planes. However, given any line $V$ that isn't parallel to $U$, there is only one plane through $U$ parallel to $V$. In particular, the normal vector of that plane is necessarily orthogonal to the direction vectors of both $U$ and $V$. (I discuss this more in this related post.)

For brevity's sake, let's describe the two lines with vector equations. In particular, the line of intersection $U$ is comprised of all points $\langle x,y,z\rangle$ such that $x=7-3y$ and $z=5y-6.$ That is, $U$ can be characterized as the set of all points of the form $$\langle x,y,z\rangle=\langle 7-3s,s,5s-6\rangle=\langle7,0,-6\rangle+s\langle-3,1,5\rangle$$ for some real $s$.

Let $V$ be the other line, so by our above work, if $\langle x,y,z\rangle$ lies on $V,$ then $\frac{x-3}1=\frac{z-2}1$ (so $x=z+1$) and $\frac{y-1}2=\frac{z-2}1$ (which implies that $y=2z-3$). Hence, the points of $V$ are those of the form $$\langle x,y,z\rangle=\langle t+1,2t-3,t\rangle=\langle1,-3,0\rangle+t\langle1,2,1\rangle$$ for some real $t$.

Now, the normal vector to our plane should be orthogonal to both $\langle-3,1,5\rangle$ and $\langle1,2,1\rangle,$ so a convenient choice is the cross-product $$\langle a,b,c\rangle=\langle-3,1,5\rangle\times\langle1,2,1\rangle=\langle-9,8,-7\rangle.$$ Now we can choose any point $\langle x',y',z'\rangle$ on $U$--for simplicity, say $\langle x',y',z'\rangle=\langle7,0,-6\rangle$--and we have our plane equation $$-9(x-7)+8(y-0)-7(z+6)=0.$$


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