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Question
Find the distance between the planes 2x - y + 2z = 5 and 5x - 2.5y + 5z = 20
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Solution 1
`2/5 = - 1/(-2.5) = 2/5`
`2/5 = 2/5 = 2/5`
⇒ parallel planes
`5x - 5/2 y + 5z= 20 xx 2/5`
`=> 2x - y + 2z = 8` and `2x - y + 2z = 5`
`=> d = |(8-5)/sqrt(4+1+4)| = 3/sqrt9 = "1 unit"`
Solution 2
Consider the equations of planes, 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20.
Here, we can see the above two planes are parallel planes.
As, 5x – 2.5y + 5z = 20 can also be written as 2x – y + 2z = 8
If the equation of two parallel planes are
ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0
Then, distance between the two parallel planes is given by:
`d = |(d_2 - d_1)/sqrt(a^2 +b^2 +c^2)|`
let us take the two parallel planes be 2x – y + 2z = 5 and 2x – y + 2z = 8
Therefore the distance is given by:
`d = |(5-8)/(sqrt(2^2 + (-1)^2 +2^2))|`
= `|(-3)/(sqrt(2^2 + (-1)^2) + 2^2)|`
= `|(-3)/sqrt(4+1+4)|`
= `|(-3)/3|`
= 1
Hence the distance between the given two planes is 1 units.
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