Distance between the parallel planes 2x − y + 3z + 4 = 0 and 6x − 3y + 9z − 3 = 0 is:

Distance between the parallel planes 2x − y + 3z + 4 = 0 and 6x − 3y + 9z − 3 = 0 is:

• A. 5/√(14)
• B. 15/√(14)
• C. 5/(3√(14))
• D. 1/(3√(14))

Answer: A) 5/√(14)

Explanation: Divide the second equation by 3 to normalize: 2x − y + 3z − 1 = 0. Distance between ax+by+cz+d₁=0 and ax+by+cz+d₂=0 is |d₁ − d₂|/√(a²+b²+c²) = |4 − (−1)| / √(4+1+9) = 5/√(14).