Thread: Equation of plane intersecting 2 planes and perpindicular to another

Find the equation of a plane that passes through the interesection of planes

x - z = 1
y + 2z = 3

and is perpendicular to x + y - 2z = 1

I already looked and understood the solution provided, however, I do not understand why my approach did not work. Here is what I did:

First, I solved for a point at the intersection of planes x - z = 1 and y + 2z = 3. One of the points happens to be (0, 5, -1)

Next, I take the plane x + y - 2z = 1 and find a vector that is coplanar to that particular plane. I do this by finding two points on the plane (0, 0, -1/2) and (1, 0, 0). With these two points, I form the vector <1, 0, 1/2>.

Finally, I use the vector equation of a plane to find the plane initially requested:

<1, 0, 1/2> . <x, y-5, z+1>

and get the cartesian equation: x + z/2 = -1/2

The answer in the solutions however is entirely different. (x + y + z = 4)

So what was the bad assumption that I made? Where did I go wrong? My approach seemed to make complete logical sense, unless of course I overlooked something. Thanks and any help is much appreciated.

Lets label the planes:
A : x - z = 1
B : y + 2z = 3
C : x + y - 2z = 1

You found one point in the intersection of A and B. Call this point P.
Now how many planes go through point P and is perpendicular to C? Infinitely many.
You used the vector <1, 0, 1/2> and that gives one solution, but think about rotating <1,0,1/2> in the plane of C. Any of these rotated vectors gives you another possible solution.

The solution is incorrect because you need to use the fact that the intersection of A and B is a line. So the solution should pass through at least 2 points in the intersection of A and B. This will reduce the number of possible planes perpendicular to C to just one.

Originally Posted by snowtea

Lets label the planes:
A : x - z = 1
B : y + 2z = 3
C : x + y - 2z = 1

You found one point in the intersection of A and B. Call this point P.
Now how many planes go through point P and is perpendicular to C? Infinitely many.
You used the vector <1, 0, 1/2> and that gives one solution, but think about rotating <1,0,1/2> in the plane of C. Any of these rotated vectors gives you another possible solution.

The solution is incorrect because you need to use the fact that the intersection of A and B is a line. So the solution should pass through at least 2 points in the intersection of A and B. This will reduce the number of possible planes perpendicular to C to just one.

Yes, I started drawing it in paint and soon realized my mistake. As mentioned, I looked over the fact that the intersection is a line, and that the vector I considered should be perpendicular to that line. Thanks for the help!