Distance between planes Find the distance between the planes $$x + 2y +2z = 4$$ $$z= -\frac12 (x-1)-(y-2)+3$$
First of all how do you check if they are parallel?
The integers in plane two are leading me astray? How do I handle those integers. I know that using the dot product would tell me if those planes are perpendicular or not.
Then I use the cos formula.
 A: Write both equations in the form $ax+by+cz+d=0$. Then a normal to the plane is given by $(a,b,c)$. The normal for the first is $(1,2,2)$, the normal for the second is $(−\frac{1}{2},−1,−1)$, which is a multiple of the first. Hence they are parallel. 
Note that the above equation can be written as $\langle (a,b,c), (x,y,z) \rangle = -d$, so the plane consists of all points whose inner product with the normal $(a,b,c)$ equals $-d$.
You can also see that if you multiply the defining equation by a non-zero constant, the set of satisfying points remains the same. Hence if the normals of two planes are collinear (ie, one is a non-zero multiple of the other), then their defining equation can be written in the same form, except the $d$ values may be different. The $d$ values determine whether or not the two planes are the same or parallel and different.
A: Two planes $ax+by+cz=d$ and $a'x+b'y+c'z=d'$ are parallel if and only if $(a',b',c')$ is a non-zero constant times $(a,b,c)$.
This is because the vector $(a,b,c)$ is perpendicular to the plane with the first equation, and $(a',b',c')$ is perpendicular to the second.
A: Equation of the first plane is $\vec r . ( 1, 2, 2) = 4 $ and the equation of the second is  $\vec r . ( 1, 2, 2) = 11 $.
From this you can see that the planes are parallel, and, since their distances from the origin are $\frac{4}{\sqrt 9}$ and $\frac{11}{\sqrt 9}$ respectively, their distance from each other is $\frac{11 - 4}{\sqrt 9} = \frac{7}{\sqrt 9}$
A: If $ax+by+cz=d$ is an equation of a plane then $(a,b,c)^T$ is a normal vector of this plane. We can verify easly that the two given plane have the same normal vector $u=(1,2,2)^T$
Now to determine the distance $\delta$ between the plane, we pick two points of the plane say $A=(0,1,1)$ for the first plane and $B=(1,2,3)$ for the second so we have
$$\delta=|\vec{AB}\frac{u}{||u||}|=\frac{7}{\sqrt{9}}.$$
