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

  1. \(\frac{5}{{\sqrt 19}}\)
  2. \(\frac{7}{{\sqrt 19}}\)
  3. \(\frac{3}{{\sqrt 19}}\)
  4. \(\frac{9}{{\sqrt 19}}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{3}{{\sqrt 19}}\)
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Detailed Solution

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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.

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