Given planes $Ax + By + Cz – D = 0$ and $Ax + By + Cz – F = 0$ show that they are parallel and the distance between them is given by
$$\frac{D-F}{\sqrt{A^2 + B^2 + C^2}}.$$
I am afraid I don't know how to start to prove this, previously when finding the distance between two planes I had used the orthogonal projection formula, which is similar to the one given, but not the same. I also don't know how to prove they are parallel without real numbers, because those variables could represent any plane (right?).
Thanks in advance.
Best Answer
Let $\hat{n}$ represent the unit normal vector of a plane at any point. Let $d$ be the distance of the plane from origin. So the distance vector will be $d\hat{n}$ since $\vec{d}$ is along the normal.
Now, let $\vec{r}$ be the position vector of any point on the given plane. From the figure it's easy to observe that NP is perpendicular to ON and therefore: $\vec{NP}\cdot\vec{ON}=0$$$(r-d\hat{n})\cdot d\hat{n}=0$$Simplifying the above equation, you get:$$\vec{r}\cdot\hat{n}=d$$Which is known as the normal form equation of plane.(Note that unit vector of normal is required).
Hence, if $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$, normal vector is $\vec{n}=a\hat{i}+b\hat{j}+c\hat{k}$, and the distance of plane from origin is $d$, then first find unit vector along normal which comes out to be $\frac{\vec{n}}{|n|}$
Now equation of plane is $$r\cdot\frac{\vec{n}}{|n|}=d$$which can be written as$$ax+by+cz=d\cdot|n|$$This is the Cartesian form of the plane.
In your question, the normals for both planes are $$\vec{n}=A\hat{i}+B\hat{j}+C\hat{k}$$ and the distance of the first plane from origin is $$\frac{D}{|n|}=\frac{D}{\sqrt{A^2+B^2+C^2}}$$. (It's $\frac{D}{|n|}$ and not $D$.)
But the distance of both planes from origin is different. This is like parallel lines. Try find the distance between two parallel lines first and then use the same logic here.