Distance between the two planes: 2x+3y+4z = 4 and 4x+6y+8z = 12 is
Distance between the two planes: 2x+3y+4z = 4 and 4x+6y+8z = 12 is
2 units
- 4 units
- 8 units
\(\frac {2}{\sqrt 29}\) units
The Correct Option is D
Solution and Explanation
The equation of the planes are
2x+3y+4z = 4 ...(1)
4x+6y+8z = 12
⇒2x+3y+4z = 6 ...(2)
It can be seen that the given planes are parallel. It is known that the distance between two parallel planes, ax+by+cz=d1 and ax+by+cz=d2, is given by,
D = \(|\frac {d_2-d_1}{√a^2+b^2+c^2}|\)
⇒ D = \(|\frac {6-4}{√2^2+3^2+4^2}|\)
D = \(\frac {2}{\sqrt 29}\)
Thus, the distance between the lines is \(\frac {2}{\sqrt 29}\) units.
Hence, the correct answer is D.
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