# Finding an equation of a plane

Find an equation of the plane that passes through the line of intersection of the planes $x-y=1$ and $y+2z=3$ and is perpendicular to the plane $x+y-2z=1$.

I'm having a hard time with this. I'm very confused by this section in the book, as the definitions and equations are all over the place. Can anyone give me any hints or tips on how to start this one and things I should be looking for?

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You can take a look here: physicsforums.com/archive/index.php/t-5658.html – anonymous Feb 3 '11 at 2:31

If the equation of a plane is given by $ax+by+cz+d=0$, then the normal vector for that plane is $(a,b,c)$, if $a$, $b$, $c$ are not all $0$. Since every line contained in a plane is perpendicular to the normal vector, you can cross the two normal vectors of $x-y=1$ and $y+2z=3$ to find the vector in the direction of the line of intersection, which is contained in your desired plane. The reason for crossing the normals is that the line of intersection is contained in both planes, and thus perpendicular to both normal vectors.

Now two planes are perpendicular if their normal vectors are perpendicular, so to find the normal vector of your desired plane, you can cross the vector in the direction of the line of intersection, and the normal vector of $x+y-2z=1$.

Lastly, a plane is determined by a point in the plane and a normal vector, so all that remains is to find a point in the plane. You should be able to find one in the line of intersection.

Edit: Your situation looks something like this at first. Maybe it'll help in visualizing the space in which you're working.

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So you say the line of intersection is contained in both both planes, which I agree with, and that it is perpendicular to the normal vectors. So if this line is perpendicular to the normal vectors, does this mean it is parallel to the planes? Also, how do we know at what angle the line intersects the two planes-does that matter? – maq Feb 3 '11 at 2:27
@mohabitar, If you're asking at what angle the line intersects the two planes, it is $0$, since it lies in both. I've added a picture to make this a little clearer, hopefully. Ultimately though, it does not matter, as you only need to find one point in a plane and a normal vector to describe a plane, no direct computation of angles is necessary. – yunone Feb 3 '11 at 2:37
I've been reading your instructions over and over to try to understand them, but I get stuck here "you can cross the two normal vectors of x−y=1 and y+2z=3 to find the vector in the direction of the line of intersection, which is contained in your desired plane". How does this work? Whats the reasoning? – maq Feb 3 '11 at 3:46
So I did the steps you said, without fully understanding all of them, and I came up with a final answer of 3x+3y+3z=9 or x+y+z=3. Does this look right? – maq Feb 3 '11 at 3:58
@mohabitar, the cross product two vectors returns a vector, and have you learned that the resulting vector is perpendicular to the two vectors you crossed? It should be somewhere in your text, and it is easily verified by dotting the resulting vector with each of the vectors you crossed and finding their dot product to be $0$. The line of intersection $L$ is contained in both planes, and thus must be perpendicular to both normals, correct, as all vectors in a plane are perpendicular to the normal of that plane. Crossing the normals gives a vector perpendicular to them both, precisely... – yunone Feb 3 '11 at 8:10