The distance between the parallel planes 3x + y + 3z = 8 and 9x + 3y + 9z = 15 is:

Question

Question

Answer 1

Concept:

Distance between two parallel plane ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is \(\rm |\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}|\)

Calculation:

Here, 3x + y + 3z = 8 and 9x + 3y + 9z = 15

Divide 9x + 3y + 9z = 15 by 3 we get

3x + y + 3z = 5

Now, distance between 3x + y + 3z = 8 and 3x + y + 3z = 5

\(\rm= |\frac{8-5}{\sqrt{3^2+1^2+3^2}}|\\ =\frac{3}{\sqrt{19}}\)

Hence, option (3) is correct.