Problem 69 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Distance between two parallel planes \(2 x+y+2 z=8\) and \(4 x+2 y+4 z+5=0\) is [2013] (A) \(\frac{5}{2}\) (B) \(\frac{7}{2}\) (C) \(\frac{9}{2}\) (D) \(\frac{3}{2}\)

Step-by-Step Solution

Answer

None of the provided options match the calculated distance. Check question or options.

Step 1: Recognize the equations of the planes

The given planes are \( 2x + y + 2z = 8 \) and \( 4x + 2y + 4z + 5 = 0 \). We observe that the second plane equation can be simplified to \( 2x + y + 2z = -\frac{5}{2} \), confirming that both planes share the same normal vector \((2, 1, 2)\), meaning they are parallel.

Step 2: Calculate distance between parallel planes

The distance \( D \) between two parallel planes of the form \( ax + by + cz + d_1 = 0 \) and \( ax + by + cz + d_2 = 0 \) is given by \[ D = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}} \]. Here, \( d_1 = -8 \) and \( d_2 = -\frac{5}{2} \).

Step 3: Substitute values into distance formula

Substituting the known values into the distance formula, we get \[ D = \frac{|-\frac{5}{2} + 8|}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{|\frac{16}{2} - \frac{5}{2}|}{\sqrt{9}} = \frac{|\frac{11}{2}|}{3} = \frac{11}{6}. \]

Key Concepts

Equations of Planes

Understanding how to write equations of planes is essential in solving geometry problems involving 3D space. A plane is defined by an equation in the form \(ax + by + cz + d = 0\), where \(a\), \(b\), and \(c\) are the coefficients of \(x\), \(y\), and \(z\) respectively, and \(d\) is a constant term.
This equation describes all the points \((x, y, z)\) that form the plane. To work with these, it is important to identify the normal vector of the plane, which is denoted by \((a, b, c)\). This vector is perpendicular to the plane and crucial for determining properties such as parallelism.

In the given exercise, the equations were \(2x + y + 2z = 8\) and \(4x + 2y + 4z + 5 = 0\). It was found that, once simplified, both equations share the same normal vector \((2, 1, 2)\), indicating parallel planes.

Distance Formula

The distance formula for parallel planes is a tool used to calculate how far apart two planes are when they are parallel. Parallel planes have identical normal vectors. The generic form of their equations are: \(ax + by + cz + d_1 = 0\) and \(ax + by + cz + d_2 = 0\).

The formula to find the distance \(D\) between these planes is:

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

This formula effectively uses the difference in the constant terms (\(d_2 - d_1\)) and divides it by the magnitude of the normal vector, \(\sqrt{a^2 + b^2 + c^2}\).

Applying this to the problem, with \(d_1 = -8\) and \(d_2 = -\frac{5}{2}\), we substitute these in to find the distance \(D = \frac{| -\frac{5}{2} + 8 |}{\sqrt{9}} = \frac{\frac{11}{2}}{3}\).
This result gives us a simplified distance.

Parallel Planes

Parallel planes in geometry are planes that never intersect, similar to parallel lines in two dimensions. They share the same normal vector. This means that if you take two plane equations and their normal vectors (\((a, b, c)\)), and they are identical or scalar multiples, the planes are parallel.

In 3D space, visualizing this concept might be a bit challenging, but imagine two sheets of paper laid flat on a table, one above the other, never touching. They represent parallel planes.

In the exercise, we confirmed the planes were parallel because both equations had the same normal vectors, \((2, 1, 2)\), by simplifying \(4x + 2y + 4z + 5 = 0\) to a form matching \(2x + y + 2z = 8\) with a proportional constant term. This essential characteristic allows us to apply the distance formula specifically designed for parallel planes.