Question

Verify that the two planes are parallel, and find the distance between the planes. $$ \begin{array}{l} 4 x-4 y+9 z=7 \\ 4 x-4 y+9 z=18 \end{array} $$

Short answer

The planes are indeed parallel and the distance between them is \(11 / \sqrt{113}\).

Step-by-step solution

Confirm if planes are parallel

Formularize the given plane equations as \(4x - 4y + 9z = c1\) and \(4x - 4y + 9z = c2\). Observe the coefficients of x, y, and z in both equations - they are identical (4, -4, 9), meaning that the planes' normal vectors are equal. This fact confirms that the planes are indeed parallel.

Calculate the magnitude of the normal vector

Calculate the magnitude of a normal vector using the formula \(\|n \| = \sqrt{a^2 + b^2 + c^2}\) where a, b, and c are the coefficients in the plane equation. So \(\|n \| = \sqrt{4^2 + (-4)^2 + 9^2} = \sqrt{16 + 16 + 81} = \sqrt{113}\).

Calculate distance between planes

Use the formula \(d = |c2 - c1| / \|n \|\) to determine the distance between the two planes. Substitute the appropriate values to yield \(d = |18 - 7| / \sqrt{113} = |11| / \sqrt{113} = 11 / \sqrt{113}\).