Problem 21 - A Complete Resource Book in Mathematics for JEE Main

Question

A variable plane moves so that the sum of reciprocals of its intercepts on the three coordinate axes is constant \(\lambda\). It passes through a fixed point, which has coordinates \(\begin{array}{ll}\text { (A) }(\lambda, \lambda, \lambda) & \text { (B) }\left(\frac{1}{\lambda}, \frac{1}{\lambda}, \frac{1}{\lambda}\right)\end{array}\) (C) \((-\lambda,-\lambda,-\lambda)\) (D) \(\left(-\frac{1}{\lambda},-\frac{1}{\lambda},-\frac{1}{\lambda}\right)\)

Step-by-Step Solution

Answer

Option (C) \((-\lambda, -\lambda, -\lambda)\) passes through the plane.

Step 1: Understand the Equation of the Plane

The general equation of a plane is \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \), where \(a, b, c\) are the intercepts on the x, y, and z axes respectively.

Step 2: Set Up the Condition Given in the Problem

The problem states that the sum of the reciprocals of the intercepts is constant \( \lambda \). This means \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \lambda \).

Step 3: Analyze the Plane Equation Condition Further

Given that \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \lambda \), substitute this into the plane equation: \( \lambda abc = bc + ac + ab \).

Step 4: Determine the Point Satisfying the Condition

Substitute the possible fixed point coordinates into the equation from Step 3 to see which one satisfies it. Start by testing the points.

Step 5: Test Option (A) \((\lambda, \lambda, \lambda)\)

Substitute \(a = b = c = \lambda\) into \( \lambda abc = bc + ac + ab \). You get \( \lambda (\lambda \lambda \lambda) = \lambda^2 + \lambda^2 + \lambda^2 \), which simplifies to \( \lambda^3 = 3\lambda^2 \). This is only true if \( \lambda = 3 \). Hence, (A) is incorrect.

Step 6: Test Option (B) \(\left(\frac{1}{\lambda}, \frac{1}{\lambda}, \frac{1}{\lambda}\right)\)

Substitute \(a = b = c = \frac{1}{\lambda}\) into \( \lambda abc = bc + ac + ab \). The equation becomes \( \lambda \left(\frac{1}{\lambda}\right)^3 = \left(\frac{1}{\lambda}\right)^2 + \left(\frac{1}{\lambda}\right)^2 + \left(\frac{1}{\lambda}\right)^2 \). This simplifies to \( \frac{1}{\lambda^2} = \frac{3}{\lambda^2} \), not true for any \(\lambda\). Hence, (B) is incorrect.

Step 7: Test Option (C) \((-\lambda, -\lambda, -\lambda)\)

Substitute \(a = b = c = -\lambda\) into \( \lambda abc = bc + ac + ab \). The equation becomes \( \lambda (-\lambda)(-\lambda)(-\lambda) = \lambda^2(-\lambda) + \lambda^2(-\lambda) + \lambda^2(-\lambda) \). Hence it simplifies to \(-\lambda^3 = -3\lambda^2 \), which is true for any \( \lambda eq 0\). Thus, (C) is a solution.

Step 8: Test Option (D) \(\left(-\frac{1}{\lambda}, -\frac{1}{\lambda}, -\frac{1}{\lambda}\right)\)

Substitute \(a = b = c = -\frac{1}{\lambda}\) into \( \lambda abc = bc + ac + ab \). The equation becomes \( \lambda \left(-\frac{1}{\lambda}\right)^3 = 3\left(\frac{1}{\lambda^2}\right) \). This simplifies to \(-\frac{1}{\lambda^2} = \frac{3}{\lambda^2} \), not true for any \( \lambda\). Hence, (D) is incorrect.

Key Concepts

Equation of a Plane

In three-dimensional geometry, the equation of a plane can be expressed in a simple form once we know its intercepts on the coordinate axes. Imagine a plane that slices through the 3D space, intersecting the x, y, and z axes at points \((a, 0, 0)\), \((0, b, 0)\), and \((0, 0, c)\) respectively. The general form of a plane's equation is given by \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \). This equation tells us how far the plane is positioned from each of the three axes.
When a condition like the sum of the reciprocals of the intercepts is constant, such as in \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \lambda \), the behavior and orientation of the plane change. In this case, \( \lambda \) acts as a specific parameter, symbolizing a constant sum that the plane's intercepts maintain even as it moves. Understanding these sorts of equations helps us grasp the mathematical relationship the plane has with the axes and guides us in exploring the solution space of points satisfying the plane's setup.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, utilizes a coordinate system, typically the Cartesian coordinate system, to study geometric figures and their properties. In 3D coordinate geometry, points are denoted as \((x, y, z)\), extending the concept of 2D geometry where points are represented by just two coordinates, \((x, y)\). By using coordinates, we can describe shapes mathematically.
The beauty of coordinate geometry is its ability to combine algebra and geometry. This allows us to not only visualize geometric figures but also perform calculations to determine a variety of properties such as distances, angles, and intersections. In the context of planes, specifying intercepts and equations involves transforming geometric conditions into algebraic expressions easily handled by algebraic methods.
With the ability to analyze from an algebraic point of view, we understand the geometric implications of algebraic expressions, enabling problem-solving techniques that bridge both subjects. This is crucial as it allows us to validate whether a point, such as those given as possible options in the exercise, satisfies certain conditions described by the plane's equation.

Intercepts on Axes

Intercepts are critical in understanding the positions and orientations of geometric figures like lines and planes in space. An intercept is where a geometric figure crosses or touches one of the coordinate axes. In 3D geometry, we can talk about intercepts along the x-axis, y-axis, and z-axis. The intercepts \((a, b, c)\) of a plane imply that the plane meets the x-axis at point \((a, 0, 0)\), the y-axis at \((0, b, 0)\), and the z-axis at \((0, 0, c)\).
Intercepts simplify the equation of the plane to \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \), clearly showing how the plane aligns with each axis. They provide intuitive insight into how the plane relates to the three-dimensional space around it.
A particularly interesting scenario is when the sum of the reciprocals of these intercepts remains constant, forming a specific family of planes that share a unique configuration in space. This condition adds constraints to the problem, focusing our analysis to derive specific properties or satisfy given requirements, such as passing through a fixed point.