Problem 65 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the plane \(2 x-y+2 z+3=0\) has the distances \(\frac{1}{2}\) and \(\frac{2}{3}\) units from the planes \(4 x-2 y+4 z+\lambda=0\) and \(2 x-y+2 z+\) \(\mu=0\), respectively, then the maximum value of \(\lambda+\mu\) is equal to : [April 10, 2019 (II)] (a) 9 (b) 15 (c) 5 (d) 13

Step-by-Step Solution

Answer

The solution does not match any provided option.

Step 1: Understand the formula for the distance between two planes

The distance between two parallel planes of the form \(Ax+By+Cz+D_1=0\) and \(Ax+By+Cz+D_2=0\) is given by the formula \(\frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}\). This formula will be used to find the distances given in the problem.

Step 2: Find \(\lambda\) using the known distance from the first plane

The given distance from \(2x-y+2z+3=0\) to \(4x-2y+4z+\lambda=0\) is \(\frac{1}{2}\). The planes are parallel, so using the formula: \(\frac{|3 - \lambda|}{\sqrt{4^2 + (-2)^2 + 4^2}} = \frac{1}{2}\). Calculate the left side: \(\sqrt{16 + 4 + 16} = \sqrt{36} = 6\). Thus, \(|3 - \lambda| = 3\). This gives us two equations: \(3 - \lambda = 3\) or \(3 - \lambda = -3\). Thus, \(\lambda = 0\) or \(\lambda = 6\).

Step 3: Find \(\mu\) using the known distance from the second plane

The given distance from \(2x-y+2z+3=0\) to \(2x-y+2z+\mu=0\) is \(\frac{2}{3}\). The planes are parallel, so using the formula: \(\frac{|3 - \mu|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{2}{3}\). Calculate the left side: \(\sqrt{4 + 1 + 4} = \sqrt{9} = 3\). Thus, \(|3 - \mu| = 2\). This gives us two equations: \(3 - \mu = 2\) or \(3 - \mu = -2\). Thus, \(\mu = 1\) or \(\mu = 5\).

Step 4: Maximize \(\lambda + \mu\)

The possible values for \(\lambda\) are 0 and 6, while for \(\mu\) they are 1 and 5. To find the maximum value of \(\lambda + \mu\), test the sums of possible pairs. The sums \(\lambda + \mu\) are: 0 + 1 = 1, 0 + 5 = 5, 6 + 1 = 7, 6 + 5 = 11. The maximum value is 11.

Step 5: Check provided options

None of the provided options (9, 15, 5, 13) match the result of 11. Double-check calculations and the problem statement to ensure the solution finds the correct or intended alternative answer.

Key Concepts

Plane Geometry

Plane Geometry is a branch of mathematics that deals with shapes and figures on a flat surface. In this concept, a plane is a flat, two-dimensional surface that extends infinitely in all directions. Planes are fundamental in geometry, as they represent the spaces in which geometric shapes live.

When we talk about parallel planes, it means that these planes never intersect. They have the same orientation and are equidistant from each other at any point. Understanding this helps us analyze problems involving calculations of distances or interactions between planes, making it easier to solve problems involving real-world scenarios, such as architectural design or physics simulations.

Key features of plane geometry include:

• Planes being defined by three non-collinear points or by equations in three-dimensional space, like the forms seen in the exercise.
• Parallel planes having equal coefficients in their linear equations, which prevent them from intersecting.
• The linear equation of a plane generally being written as $Ax + By + Cz + D=0$, where $A$, $B$, and $C$ are constants defining the plane’s orientation.

Recognizing the properties of parallel planes and understanding the mathematical definitions are essential for solving analytical geometry problems related to distance.

Distance Formula

The distance formula is an essential tool used in various branches of geometry, including plane geometry and analytical geometry. It helps find the shortest distance between two points, lines, or surfaces in space.

For planes, especially parallel planes as seen in the problem, the distance formula simplifies significantly. The formula for the distance between two parallel planes \(Ax + By + Cz + D_1 = 0\) and \(Ax + By + Cz + D_2 = 0\) is \[\text{Distance} = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}.\]This formula is derived from the general distance formula involving vectors and the Pythagorean theorem.

To use it efficiently:

• Identify \(D_1\) and \(D_2\), which represent the constant terms of the plane equations. These determine the offset of each plane from the origin.
• Use coefficients \(A\), \(B\), and \(C\) from the linear equations to calculate the denominator, representing the plane's normal vector magnitude.
• Plug the values into the formula to calculate the distance accurately.

Understanding how to apply this formula correctly allows students to solve complex geometry problems involving planes, enhancing their comprehension of spatial relationships.

Analytical Geometry

Analytical Geometry, also known as coordinate geometry, uses a coordinate system to describe geometric figures and analyze their properties mathematically. It bridges algebraic equations and geometric figures, enabling more complex geometric reasoning and solving.

In the context of the problem, analytical geometry plays a crucial role in deriving and applying equations to find relationships between geometric figures, such as the distance between planes. By interpreting and manipulating algebraic plane equations, the geometry becomes more accessible and formulaic.

Important points about Analytical Geometry include:

• It allows a combination of algebra and geometry to solve distance, area, and intersection problems.
• Translates geometric problems into algebraic terms, making solutions systematic and analytical.
• Helps visualize complex geometric problems with coordinate systems, transforming abstract ideas into more concrete terms.

Through this approach, students can solve problems like finding maximum values, verifying solutions, or discovering new relationships, without merely relying on geometric constructions. Analytical geometry thus turns complex spatial problems into solvable algebraic tasks, as in the original problem's calculations.