Question

Show that the distance between the parallel planes \(ax + by + cz + {d_1} = 0\) and \(ax + by + cz + {d_2} = 0\) is

\(D = \frac{{\left| {{d_1} - {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\)

Short answer

The separation between the parallel planes \(ax + by + cz + {d_1} = 0\) and \(ax + by + cz + {d_2} = 0\) is equal to \(D = \frac{{\left| {{d_1} - {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\).

Step-by-step solution

Concept used

The expression to find the distance from the point \(P\left( {{x_1},{y_1},{z_1}} \right)\) to the plane.

\(D = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\) ……. (1)

Here,

\({x_1},{y_1}\), and \({z_1}\) are the coordinates of the point \(P\left( {{x_1},{y_1},{z_1}} \right)\), which is \((1, - 2,4)\),

a, b, and \(c\) are the normal vector numbers to the plane and

\(d\)is the constant parameter.

The expression to find constant parameter \((d)\).

\(a{x_0} + b{y_0} + c{z_0} + d = 0\) ……. (2)

Here,

\({x_0},{y_0}\), and \({z_0}\) are the coordinates of any point \(P\left( {{x_0},{y_0},{z_0}} \right)\) in the plane.

Calculate the normal vector and constant parameter from the equation

As the planes are parallel, distance between planes is the distance from any point on one plane to the other plane. Therefore, choose any point on one plane and calculate its distance to the other plane.

Write the equation of first plane as follows.

\(ax + by + cz + {d_1} = 0\)

Rewrite the expression as follows.

\(ax + by + cz = - {d_1}\) ……. (3)

Consider the values of \(y\) - and \(z\)-coordinates are 0.

Substitute 0 for \(y\) and \(z\) in equation (3) and obtain the \(x\)-coordinate value.

\(\begin{array}{l}ax + b(0) + c(0) = - {d_1}\\ax = - {d_1}\\x = \frac{{ - {d_1}}}{a}\end{array}\)

Therefore, the point \(\left( {\frac{{ - {d_1}}}{a},0,0} \right)\) is on the first plane.

Write the equation of second plane as follows.

\(ax + by + cz + {d_2} = 0\)

\(ax + by + cz = - {d_2}\) ……… (4)

Write the normal vector from the equation of second plane.

\({\rm{n}} = \langle a,b,c\rangle \)

Compare equations \((2)\) and \((4)\),we get

The value of constant parameter \((d)\) is \({d_2}\).

Calculate the distance

Substituting the values in equation (1)

\(\begin{array}{l}D = \frac{{\left| {(a)\left( {\frac{{ - {d_1}}}{a}} \right) + (b)(0) + (c)(0) + {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\\ = \frac{{\left| { - {d_1} + 0 + 0 + {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\\ = \frac{{\left| { - \left( {{d_1} - {d_2}} \right)} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\\ = \frac{{\left| {{d_1} - {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\end{array}\)

Hence, the separation between the parallel planes \(ax + by + cz + {d_1} = 0\) and \(ax + by + cz + {d_2} = 0\) is equal to \(D = \frac{{\left| {{d_1} - {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\).