Problem 92 - 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

Option (d): \( \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=9 \).

Step 1: Understand the Problem

We need to find the locus of the centroid of a triangle formed by the intersection points of a plane with the coordinate axes. The plane is a distance of 3 units from the origin.

Step 2: Set the Plane Equation

The general equation of a plane in intercept form is \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \). Here, \( OA = a, OB = b, OC = c \). The distance of the plane from the origin is given by \( \frac{|d|}{\sqrt{1^2 + 1^2 + 1^2}} = 3 \), which simplifies to \( d = 3\sqrt{3} \).

Step 3: Determine Intersections Points with Axes

The plane intersects the x-axis at \( x = a, y = 0, z = 0 \), the y-axis at \( x = 0, y = b, z = 0 \), and the z-axis at \( x = 0, y = 0, z = c \). Thus, the points are \( A(a,0,0) \), \( B(0,b,0) \), and \( C(0,0,c) \).

Step 4: Calculate the Centroid Coordinates

The centroid \((G)\) of \( \Delta ABC \) with vertices \( (a,0,0), (0,b,0), (0,0,c) \) is given by: \[ G \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right) \].

Step 5: Establish Relationship for the Centroid

Given \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \), substituting the centroid coordinates gives \( \frac{3x}{a} + \frac{3y}{b} + \frac{3z}{c} = 1 \). Simplifying gives \( 3 \left( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} \right) = 1 \).

Step 6: Solve for the Locus of the Centroid

Using step 2 where \( \left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right) = \frac{1}{9} \), the equation for the locus becomes \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 9 \). This is obtained by realizing \( x = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3} \). Substituting gives \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 9 \).

Step 7: Conclusion

After following the calculations and verifying each step, the locus of the centroid of \( \Delta ABC \) is given by \( \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=9 \).

Key Concepts

Centroid of a triangle

A centroid is a crucial point in a triangle, serving as the intersection of its three medians. The median is a line from a vertex to the midpoint of the opposite side. The centroid is often called the triangle's "center of gravity". It is always located inside the triangle, regardless of its shape, making it something that can always be relied upon.

• The coordinates of the centroid, denoted as \(G\), can be calculated as the average of the triangle's vertices coordinates.
• For a triangle with vertices \((x_1, y_1, z_1)\), \((x_2, y_2, z_2)\), \((x_3, y_3, z_3)\), the centroid \(G(x, y, z)\) is determined by:
• \( x = \frac{x_1 + x_2 + x_3}{3}, \quad y = \frac{y_1 + y_2 + y_3}{3}, \quad z = \frac{z_1 + z_2 + z_3}{3} \)

In any triangle, the centroid divides each median into a ratio of 2:1, with the longer segment always extending from the vertex to the centroid. In this exercise, understanding the role of the centroid helps simplify the calculations significantly.

Plane in intercept form

A plane can be represented in its intercept form, which is both simple and neat. The general equation for a plane in intercept form is:

• \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \)
• Here, \(a, b, c\) are the intercepts the plane makes with the x, y, and z-axes respectively.

The intercept form is particularly useful when working with the spatial geometry of planes, making it easier to identify how and where a plane intersects with the axes. In practice, intercepts \(a, b, c\) are the points where the plane touches each of the coordinate axes. They are fundamental when considering triangles formed by these intersections, such as in this exercise.

Locus equation

A locus is a set of points satisfying a certain condition, often forming a geometrical shape or curve. The locus equation is an expression that defines these conditions mathematically.
In the context of this problem, the goal is to find the locus of the centroid of a triangle formed by a plane and the coordinate axes.

• The locus of the centroid refers to all possible positions the centroid can have when you vary the intercepts of the plane.
• From the properties of the plane and the centroid, the locus is expressed as:
• \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 9 \)

This equation results from substituting the relationships between centroid coordinates and the plane intercepts, condensing all possibilities into a single expression. This expression gives a precise description of the set of all centroids as they change with the plane's interactions with space.

Distance of a plane from origin

The distance of a plane from the origin is an important concept in three-dimensional geometry. It provides insight into how planes are positioned relative to the origin point \((0, 0, 0)\).

• The standard formula to find this distance when given an equation \(ax + by + cz = d\) is \(\frac{|d|}{\sqrt{a^2 + b^2 + c^2}}\).
• This formula requires rearranging any given plane equation to a similar format to draw parallels with known distances.

In this exercise, the plane is explicitly stated to be 3 units away from the origin. Using the provided details, calculate how far the plane is from the origin using intercept form constants. Understanding this concept ensures proper setup for solving locus problems involving centroids and three-dimensional space.