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The equation represents the distance between two parallel planes. These two planes are given as:

\[
2x + 4y - z = 16 \quad \text{(Plane 1)}
\]
and
\[
2x + 4y - z = 39 \quad \text{(Plane 2)}
\]

The general formula for the distance \( d \) between two parallel planes of the form \( Ax + By + Cz = D_1 \) and \( Ax + By + Cz = D_2 \) is:

\[
d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}
\]

For the given planes:
- \( A = 2 \), \( B = 4 \), and \( C = -1 \),
- \( D_1 = 16 \) and \( D_2 = 39 \).

Let's compute the distance.The distance between the two planes is approximately \( d = 5.02 \).

Would you like more details or explanations?

Here are some related questions to explore:
1. How do you derive the formula for the distance between parallel planes?
2. What is the geometric significance of the normal vector in this problem?
3. How can we find the shortest distance from a point to a plane?
4. What happens if the planes are not parallel? How would we calculate the distance then?
5. How would the formula change if the coefficients of the planes were different?

**Tip:** The distance between two parallel planes depends only on the difference in their constant terms and the direction of the normal vector.

Find the distance (d) between the planes 2x + 4y - z = 16 and 2x + 4y - z = 39.

Learn how to calculate the distance between two parallel planes, 2x + 4y - z = 16 and 2x + 4y - z = 39, using the formula for the distance between parallel planes. This problem involves vector and plane geometry concepts, ideal for high school students.

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