Calc 3 planes- not really understanding the problem/solution?

[SMA_01]

Homework Statement

The problem states to find an equation of the plane consisting of all points that are equidistant from (-1,3,-1) and (-2,-3,03) and having -1 as the coefficient of x.

The Attempt at a Solution

I actually got the solution by finding a vector using the two points and finding a new point on the plane by computing the midpoint of the 2 given points, and then plugged them into the plane equation. The thing is I don't understand how that worked out. I know that for the plane equation, you need a vector perpendicular to the plane, the normal vector right? So how come for this problem, it was sufficient to just compute a vector using the two given points which lie on the plane? Isn't this vector on the plane and not perpendicular to it?
Maybe I'm missing the obvious, an explanation would be appreciated.

Thanks

 

[lineintegral1]

No, the vector you computed will not be on the plane. Think about the geometric interpretation of this. How would such a plane be oriented such that the given conditions are true? It seems that you went through the correct procedure; can you picture it now?

 

[SMA_01]

@lineintegral1: Can you elaborate a bit more please? I'm sorry, but don't fully understand...

 

[lineintegral1]

@lineintegral1: Can you elaborate a bit more please? I'm sorry, but don't fully understand...

Sure. You are looking for a plane in which every point on the plane is the same distance from (-1, 3, -1) as they are from (-2, -3, 3). This will only happen if the plane is positioned such that it is between the two points and oriented perpendicular to the vector created between the two points. The final step asks you to calculate the plane with -1 as the x-coefficient. This simply tells you which point should be subtracted from the other to create the normal vector.

After reading this, see if you can draw it yourself.

 

[SMA_01]

I see where you're coming from, and will do thanks!