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.
Best Answer
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.