Step 1: Understanding the Concept:
We need to find the shortest distance between two planes. First, we must check if the planes are parallel. If they are, we can use a standard formula for the distance.
Step 2: Key Formula or Approach:
Two planes $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$ are parallel if their normal vectors $\langle A_1, B_1, C_1 \rangle$ and $\langle A_2, B_2, C_2 \rangle$ are proportional.
The distance between two parallel planes $Ax+By+Cz+D_1=0$ and $Ax+By+Cz+D_2=0$ is given by the formula:
\[ \text{Distance} = \frac{|D_1 - D_2|}{\sqrt{A^2+B^2+C^2}} \]
Step 3: Detailed Explanation:
The equations of the planes are:
Plane 1: $2x+y+2z-8=0$
Plane 2: $4x+2y+4z+5=0$
The normal vector for Plane 1 is $\vec{n_1} = \langle 2, 1, 2 \rangle$.
The normal vector for Plane 2 is $\vec{n_2} = \langle 4, 2, 4 \rangle$.
Since $\vec{n_2} = 2\vec{n_1}$, the planes are parallel.
To use the distance formula, we need to make the coefficients of $x, y, z$ the same in both equations. Let's divide the equation of Plane 2 by 2:
Plane 2 (modified): $\frac{4x+2y+4z+5}{2} = 0 \implies 2x+y+2z+\frac{5}{2}=0$.
Now we can compare the two equations in the form $Ax+By+Cz+D=0$:
Plane 1: $2x+y+2z-8=0 \implies A=2, B=1, C=2, D_1=-8$.
Plane 2 (modified): $2x+y+2z+\frac{5}{2}=0 \implies A=2, B=1, C=2, D_2=\frac{5}{2}$.
Now, apply the distance formula:
\[ \text{Distance} = \frac{|D_1 - D_2|}{\sqrt{A^2+B^2+C^2}} = \frac{|-8 - \frac{5}{2}|}{\sqrt{2^2+1^2+2^2}} \]
\[ = \frac{|-\frac{16}{2} - \frac{5}{2}|}{\sqrt{4+1+4}} = \frac{|-\frac{21}{2}|}{\sqrt{9}} \]
\[ = \frac{21/2}{3} = \frac{21}{2 \times 3} = \frac{21}{6} = \frac{7}{2} \]
Step 4: Final Answer:
The distance between the two planes is $7/2$.