Question

Find the distance between the planes 2x - y +  2z = 5 and 5x - 2.5y + 5z = 20

Answer 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"`

Answer 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.