6.7. PLANES 97

6.7 PlanesYou have an idea of what a plane is already. It is the span of some vectors. However, it canalso be considered geometrically in terms of a dot product. To find the equation of a plane,you need two things, a point contained in the plane and a vector normal to the plane. Letp0 = (x0,y0,z0) denote the position vector of a point in the plane, let p = (x,y,z) be theposition vector of an arbitrary point in the plane, and let n denote a vector normal to theplane. This means that

n·(p−p0) = 0

whenever p is the position vector of a point in the plane. The following picture illustratesthe geometry of this idea.

p0p

n

Expressed equivalently, the plane is just the set of all points p such that the vectorp−p0 is perpendicular to the given normal vector n.

Example 6.7.1 Find the equation of the plane with normal vector n= (1,2,3) containingthe point (2,−1,5) .

From the above, the equation of this plane is just

(1,2,3) · (x−2,y+1,z−3) = x−9+2y+3z = 0

Example 6.7.2 2x+4y−5z = 11 is the equation of a plane. Find the normal vector and apoint on this plane.

You can write this in the form 2(x− 11

2

)+ 4(y−0)+ (−5)(z−0) = 0. Therefore, a

normal vector to the plane is 2i+4j−5k and a point in this plane is( 11

2 ,0,0). Of course

there are many other points in the plane. The thing which makes perfect sense is the anglebetween two vectors. The angle between two planes requires some definition. If you thinkabout it geometrically, you could imagine infinitely many angles between two lines both ofwhich lie in one of the planes and which intersect at a point on a line of intersection of twoplanes.

Definition 6.7.3 Suppose two planes intersect in a line. The angle between the planes isdefined to be the angle which is no more than π/2 between normal vectors to the respectiveplanes.

Example 6.7.4 Find the angle between the two planes x+2y− z = 6 and 3x+2y− z = 7.