Problem 91 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), then the locus of the centroid of \(\Delta \mathrm{ABC}\) is: \([\) Online April 9, 2017] (a) \(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=1\) (b) \(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=3\) (c) \(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=\frac{1}{9}\) (d) \(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=9\)

Step-by-Step Solution

Answer

The locus is \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{9} \) (option c).

Step 1: Determine the Equation of the Plane

Since the plane is at a distance of 3 units from the origin, and it intersects the coordinate axes, we represent the plane equation as \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \). We know its distance from the origin is 3, so the equation becomes: \( \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 3 \).

Step 2: Simplify Distance Condition

From Step 1, we rearrange to get the condition on the intercepts: \( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{9} \). This describes the relationship between the intercepts as required by the problem.

Step 3: Find Coordinates of Points A, B, C

To find the intercepts where the plane cuts the axes, we set two coordinates to zero at a time: \(A(a,0,0)\), \(B(0,b,0)\), and \(C(0,0,c)\). These are the points where the plane intersects the x, y, and z axes.

Step 4: Determine the Centroid of ΔABC

The centroid (G) of a triangle formed by points \(A(a,0,0)\), \(B(0,b,0)\), and \(C(0,0,c)\) is calculated using the average of their coordinates: \(G\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)\).

Step 5: Express the Locus of the Centroid

The locus of the centroid \(G\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)\) will satisfy the equation obtained from the distance condition, but scaled down by a factor of 3 because the centroid is 1/3 of the intercepts: \( \frac{9}{x^2} + \frac{9}{y^2} + \frac{9}{z^2} = 1 \). Simplifying this gives the locus equation \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{9} \).

Key Concepts

Plane Geometry

In plane geometry, we often work with flat surfaces that extend infinitely in two dimensions. These planes can intersect with other objects, such as lines or other planes. In this exercise, we are dealing with a plane that intersects the coordinate axes. We visualize this by imagining a flat sheet cutting through the 3D space represented by the x, y, and z axes.
It's important to understand how such a plane can be defined by an equation, in this case, the intercept form: \[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \] This form is useful because it directly relates to where the plane cuts through the axes at the points A, B, and C. Here, 'a', 'b', and 'c' represent the respective points where the plane meets the x, y, and z axes.

Coordinate Geometry

Coordinate geometry, or analytic geometry, involves using algebraic equations to describe geometric figures. This approach combines algebra and geometry by using coordinates to represent positions in space. In a 3D space anchored by the x, y, and z axes, points and shapes can be precisely defined through equations. The exercise involves finding the centroid of a triangle that is formed by the intersection of a plane with these axes.
Understanding coordinate geometry allows us to find important geometric properties such as distances, midpoints, and other points of interest, such as the centroid, which is the average of the points' coordinates in a triangle. For the triangle in question, with vertices at \((a,0,0)\), \((0,b,0)\), and \((0,0,c)\), the centroid is calculated using: \[ G\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) \]

Locus Equation

A locus is a set of points that satisfy a certain condition. The locus equation in this exercise finds the path traced by the centroid's movement as the plane, with constant distance from the origin, changes its position. In simpler terms, as the plane slides around, the centroid moves in space.
The locus of a geometric object like a centroid is derived from its geometric properties and constraints. Here, the plane has a fixed distance from the origin, defined by: \[ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{9} \] This relationship is simplified for the centroid to show how its path relates to the variables in question, leading to the final simplified locus equation of: \[ \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{9} \]

Distance from Origin

Distance from the origin in a three-dimensional space is a measure from the point \((0,0,0)\) to any other point in space. In this problem, a plane is said to maintain a fixed distance of 3 units from the origin. This constant distance helps lock the position of the plane relative to the origin regardless of its orientation.
To express that a 3D plane is at a fixed origin distance, the problem utilizes the formula we've seen in the context of coordinates: \[ \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 3 \] This formula stems from the general principle that the orthogonal (perpendicular) distance from a point to a plane can be measured through the coefficients of the plane's equation, hence it is particularly useful in problems dealing with fixed distance constraints like this one.