Question

Verify that the two planes are parallel, and find the distance between the planes. $$ \begin{array}{l} 2 x-4 z=4 \\ 2 x-4 z=10 \end{array} $$

Short answer

The planes \(2x - 4z = 4\) and \(2x - 4z = 10\) are indeed parallel and the distance between them is \(\frac{3\sqrt{5}}{5}\).

Step-by-step solution

Verify that the planes are parallel

Two planes are parallel if their normal vectors are parallel. The normal vector of a plane in the standard form: \(Ax+By+Cz=D\) is determined by the coefficients A, B, C. The vectors are (2, 0, -4) for both planes. Since they are the same, the two planes are parallel.

Calculate the distance between the two planes

After confirming the planes are parallel, we can calculate their distance using the formula: \[d = \frac{|D2-D1|}{\sqrt{A^2 + B^2 + C^2}}\] Substituting for A, B, C, D1, and D2 we get: \[d = \frac{|10-4|}{\sqrt{2^2 + 0 + (-4)^2}} = \frac{6}{\sqrt{20}} = \frac{6}{2\sqrt{5}} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}\]

Write down and check the solution

The two planes \(2x - 4z = 4\) and \(2x - 4z = 10\) are parallel and the distance between them can be expressed as \(\frac{3\sqrt{5}}{5}\). You can check the solution by substituting back the values into the respective formulae to ensure everything balances out correctly