Question

What is the distance between the planes 2x + 2y – z + 2 = 0 and 4x + 4y – 2z + 5 = 0?

Answer 1

Step-by-step explanation:

2x+2y-z+2=0 multiply it by 2 we get

4x+4y-2z+4=0 =>4x+4y-2z = -4.

Hence the distance between the planes 2x+2y-z+2=0 and 4x+4y-2z+5=0 =>

$\frac{(5 - 4)}{ \sqrt{ {4}^{2} + {4}^{2} + {2}^{2}}} = \frac{1}{ \sqrt{16 + 16 + 4} } = \frac{1}{ \sqrt{36}} = \frac{1}{6}$

Hope it helps you.