Find the normal vector of the plane that is pointing into the part of the plane which has the origin.

Question

A plane called p has the cartesian equation

 

4x + 2y - 2z = 7

 

Given that the plane separates 3D space into two parts. Find the normal vector of the plane, pointing into the part which contains the origin. 

Answer 1

Dear Brian,

This question has two parts:

1) find the normal vector: that's pretty easy, see http://math.stackexchange.com/questions/352134/finding-the-vector-perpendicular-to-the-plane

it's (4,2,-2) or (2,1,-1) if you prefer, OR its opposite and

2) make sure it points towards the origin side of the plane. That is perhaps hard to imagine computing if you view it as starting at some unknown (x,y,z) IN THE PLANE and proceeding along the perpendicular vector until you hit the origin! So, instead, imagine starting AT THE ORIGIN and proceeding along one of the perpendicular vectors UNTIL YOU HIT THE PLANE. Now, what is true here? Sure, you will need to satisfy 4x + 2y - 2z = 7. But, you will be doing it with (x,y,z) composed of some multiple of the perpendicular vector,  i.e. let's say a(2,1,-1). So just chuck that into the equation for the plane: 4(2a) + 2(a) -2(-a) = 7 : 12a = 7, a=7/12 . That means, you went IN the (2,1,-1) direction to get to the plane; but, the vector you want goes THE OTHER WAY from the plane to the origin, hence (-2,-1,1).

By the way, that trick for just using the equation coordinates for the perpendicular vector also works for 2D geometry, of course -- we're just so used to dealing with slope-intercept form for lines, we usually just go right for that, because it's reflex to separate two variables to the two sides of the equation. Can't do that with three variables, there aren't three sides to the equation!

Hope this helped you along,

-- S.