
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0
(A). $\dfrac{5}{2}$
(B). $\dfrac{7}{2}$
(C). $\dfrac{9}{2}$
(D). $\dfrac{3}{2}$
Answer
510.9k+ views
Hint: Before attempting this question, one should have prior knowledge about the concept of planes and also remember to consider the equal of plane i.e. 4x + 2y + 4z + 5 = 0 as equation 2 and divide it by 2, using this information can help you to approach the solution of the question.
Complete step-by-step answer:
According to the given information we have 2 planes parallel to each other represented by the equations 2x + y + 2z = 8 and 4x + 2y + 4z = - 5
Taking 2x + y + 2z = 8 as equation 1 and 4x + 2y + 4z = -5 as equation 2
Now dividing equation 2 by 2 we get
$\dfrac{4}{2}x + \dfrac{2}{2}y + \dfrac{4}{2}z = \dfrac{{ - 5}}{2}$
$ \Rightarrow $$2x + y + 2z = \dfrac{{ - 5}}{2}$
As we know that distance between two parallel planes is given by \[d = \dfrac{{\left| {{D_2} - {D_1}} \right|}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}\]
Also, we know that A = 2, B = 1 and C = 2 also ${D_1} = 8$ and ${D_2} = \dfrac{{ - 5}}{2}$
Now substituting the values in the distance formula, we get
Distance between two parallel planes = \[\dfrac{{\left| {8 - \left( { - \dfrac{5}{2}} \right)} \right|}}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}} }}\]
$ \Rightarrow $ Distance between two parallel planes = \[\dfrac{{\left| {8 + \dfrac{5}{2}} \right|}}{{\sqrt {4 + 1 + 4} }}\]
$ \Rightarrow $ Distance between two parallel planes = \[\dfrac{{\left| {\dfrac{{16 + 5}}{2}} \right|}}{{\sqrt 9 }}\]
$ \Rightarrow $ Distance between two parallel planes = \[\dfrac{{21}}{{2 \times 3}}\]
$ \Rightarrow $ Distance between two parallel planes = \[\dfrac{{21}}{6} = \dfrac{7}{2}\]
Therefore, the distance between the two parallel planes \[\dfrac{7}{2}\]
Hence, option B is the correct option.
Note: In the above solution we came across the two terms ``plane” which can be explained as a flat surface which is a two-dimensional surface which is constructed by the combination of two axis such as the combination of x axis and y axis will be x-y plane. Equation of a plane is represented by ax + by + cz = d and this is named as the scalar equation of the plane.
Complete step-by-step answer:
According to the given information we have 2 planes parallel to each other represented by the equations 2x + y + 2z = 8 and 4x + 2y + 4z = - 5
Taking 2x + y + 2z = 8 as equation 1 and 4x + 2y + 4z = -5 as equation 2
Now dividing equation 2 by 2 we get
$\dfrac{4}{2}x + \dfrac{2}{2}y + \dfrac{4}{2}z = \dfrac{{ - 5}}{2}$
$ \Rightarrow $$2x + y + 2z = \dfrac{{ - 5}}{2}$
As we know that distance between two parallel planes is given by \[d = \dfrac{{\left| {{D_2} - {D_1}} \right|}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}\]
Also, we know that A = 2, B = 1 and C = 2 also ${D_1} = 8$ and ${D_2} = \dfrac{{ - 5}}{2}$
Now substituting the values in the distance formula, we get
Distance between two parallel planes = \[\dfrac{{\left| {8 - \left( { - \dfrac{5}{2}} \right)} \right|}}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}} }}\]
$ \Rightarrow $ Distance between two parallel planes = \[\dfrac{{\left| {8 + \dfrac{5}{2}} \right|}}{{\sqrt {4 + 1 + 4} }}\]
$ \Rightarrow $ Distance between two parallel planes = \[\dfrac{{\left| {\dfrac{{16 + 5}}{2}} \right|}}{{\sqrt 9 }}\]
$ \Rightarrow $ Distance between two parallel planes = \[\dfrac{{21}}{{2 \times 3}}\]
$ \Rightarrow $ Distance between two parallel planes = \[\dfrac{{21}}{6} = \dfrac{7}{2}\]
Therefore, the distance between the two parallel planes \[\dfrac{7}{2}\]
Hence, option B is the correct option.
Note: In the above solution we came across the two terms ``plane” which can be explained as a flat surface which is a two-dimensional surface which is constructed by the combination of two axis such as the combination of x axis and y axis will be x-y plane. Equation of a plane is represented by ax + by + cz = d and this is named as the scalar equation of the plane.
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