Question

What is the Distance between parallel planes \[2x - 2y + z + 3 = 0\] and \[4x - 4y + 2z + 5 = 0\]?
A. \[\dfrac{2}{3}\]
B. \[\dfrac{1}{3}\]
C. \[\dfrac{1}{6}\]
D. 2

Answer 1

Hint: Here, two equations of the parallel planes are given. First, simplify the equation of the plane \[2x - 2y + z + 3 = 0\] by multiplying it by 2. Then, apply the formula of the distance between two parallel planes. In the end, substitute the values in the formula and solve it to get the required answer.

Formula used: The distance between the two parallel planes \[ax + by + cz + {d_1} = 0\] and \[ax + by + cz + {d_2} = 0\] is: \[D = \left| {\dfrac{{{d_1} - {d_2}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\]

Complete step by step solution: The equations of the parallel planes are:
\[2x - 2y + z + 3 = 0\] \[.....\left( 1 \right)\]
\[4x - 4y + 2z + 5 = 0\] \[.....\left( 2 \right)\]

Multiply the equation \[\left( 1 \right)\] by 2.
\[4x - 4y + 2z + 6 = 0\] \[.....\left( 3 \right)\]
We have to calculate the distance between the given planes.
So, apply the formula of the distance between the two parallel planes \[ax + by + cz + {d_1} = 0\] and \[ax + by + cz + {d_2} = 0\].
Comparing the given information, we get
\[a = 4,b = - 4,c = 2,{d_1} = 6\] and \[{d_2} = 5\]
Substitute the above values in the formula \[D = \left| {\dfrac{{{d_1} - {d_2}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].
\[D = \left| {\dfrac{{6 - 5}}{{\sqrt {{4^2} + {{\left( { - 4} \right)}^2} + {2^2}} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{1}{{\sqrt {16 + 16 + 4} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{1}{{\sqrt {36} }}} \right|\]
\[ \Rightarrow D = \dfrac{1}{6}\] units
Thus, the distance between the parallel planes \[2x - 2y + z + 3 = 0\] and \[4x - 4y + 2z + 5 = 0\] is \[\dfrac{1}{6}\] units.

Thus, Option (C) is correct.

Note: We can solve this problem by using the following formula.
The distance between the two parallel planes \[ax + by + cz + {d_1} = 0\] and \[px + qy + rz + {d_2} = 0\] is: \[\left| {\dfrac{{{d_1}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }} - \dfrac{{{d_2}}}{{\sqrt {{p^2} + {q^2} + {r^2}} }}} \right|\] .