Question

find the distance between the given parallel planes. 6z = 4y - 6x, 9z = 3 - 9x + 6y

Explanation:

Step1: Rewrite plane equations in general form

The first plane $6z = 4y-6x$ can be rewritten as $6x - 4y+6z = 0$. Divide by 2 to get $3x - 2y + 3z=0$.
The second plane $9z = 3-9x + 6y$ can be rewritten as $9x-6y + 9z-3 = 0$. Divide by 3 to get $3x - 2y+3z - 1=0$.

Step2: Use the distance formula for parallel planes

The distance $d$ between two parallel planes $Ax+By + Cz+D_1 = 0$ and $Ax+By + Cz+D_2 = 0$ is given by $d=\frac{\vert D_1 - D_2\vert}{\sqrt{A^{2}+B^{2}+C^{2}}}$. Here, $A = 3$, $B=-2$, $C = 3$, $D_1 = 0$, $D_2=-1$.
So $d=\frac{\vert0-(-1)\vert}{\sqrt{3^{2}+(-2)^{2}+3^{2}}}=\frac{1}{\sqrt{9 + 4+9}}=\frac{1}{\sqrt{22}}$.

Answer:

$\frac{1}{\sqrt{22}}$