Problem 54 - A Complete Resource Book in Mathematics for JEE Main

Question

If the plane \(2 a x-3 a y+4 a z+6=0\) passes through the midpoint of the line joining the centres of the spheres \([2005]\) \(x^{2}+y^{2}+z^{2}+6 x-8 y-2 z=13\) and \(x^{2}+y^{2}+z^{2}-10 x+4 y-2 z=8\), then a equals (A) \(-1\) (B) 1 (C) \(-2\) (D) 2

Step-by-Step Solution

Answer

The value of \(a\) is \(-2\).

Step 1: Identify Centers of Spheres

The equation for a sphere is of the form \[ x^2 + y^2 + z^2 + 2ux + 2vy + 2wz = c \]where \[ (h, k, l) = (-u, -v, -w) \]is the center of the sphere. For the first sphere,\[ x^2 + y^2 + z^2 + 6x - 8y - 2z = 13 \]implies the center \[ (h_1, k_1, l_1) = (-3, 4, 1). \]For the second sphere,\[ x^2 + y^2 + z^2 - 10x + 4y - 2z = 8 \]implies the center \[ (h_2, k_2, l_2) = (5, -2, 1). \]

Step 2: Calculate Midpoint of the Line Segment

The midpoint \[ \left( \frac{h_1 + h_2}{2}, \frac{k_1 + k_2}{2}, \frac{l_1 + l_2}{2} \right) \]of the line joining the centers\[ (h_1, k_1, l_1) = (-3, 4, 1) \]and\[ (h_2, k_2, l_2) = (5, -2, 1) \]is given by:\[ \left( \frac{-3 + 5}{2}, \frac{4 - 2}{2}, \frac{1 + 1}{2} \right) = (1, 1, 1). \]

Step 3: Substitute Midpoint into Plane Equation

The plane is given by\[ 2ax - 3ay + 4az + 6 = 0. \]Substituting the midpoint \[ (1, 1, 1) \]yields:\[ 2a(1) - 3a(1) + 4a(1) + 6 = 0 \]which simplifies to\[ 2a - 3a + 4a + 6 = 0. \]

Step 4: Solve for Parameter \(a\)

Simplify the equation\[ 2a - 3a + 4a + 6 = 0 \]which combines to:\[ 3a + 6 = 0. \]Solving for \(a\), subtract 6 from both sides and divide by 3:\[ 3a = -6. \]\[ a = -2. \]

Key Concepts

Equation of Plane

In coordinate geometry, an equation of a plane is commonly expressed in the form \( Ax + By + Cz + D = 0 \). Here, \( A, B, \) and \( C \) are coefficients that determine the orientation of the plane in the three-dimensional space, while \( D \) is a constant term.
A plane can be visualized as an infinitely extending flat surface, with points \((x, y, z)\) lying on the plane satisfying the plane equation.
Substituting the coordinates of a point into this equation helps to determine if the point lies on the plane or not by checking if the resulting expression equals zero. Finding a plane passing through a specific point requires substituting its coordinates into the plane's equation to derive necessary parameters, just like the exercise where the midpoint coordinates are plugged into the equation \(2ax - 3ay + 4az + 6 = 0\) to solve for \(a\).

Equation of Sphere

The equation of a sphere in three-dimensional geometry is given by the general form \( (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \). Here:

• \((h, k, l)\) represents the center of the sphere
• \( r \) is the radius of the sphere

When expanded, it takes the form \( x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + c = 0 \), where coefficients relate to the sphere's center: \((-u, -v, -w)\).
In the exercise, the center of each sphere was computed by translating this standard equation to its derived format, identifying \(-3, 4, 1\) and \(5, -2, 1\) as the centers of the two given spheres.
Correctly identifying these centers allows for the calculation of further attributes such as the midpoint mentioned in the problem.

Midpoint Formula

The midpoint formula is a vital concept in geometry, used to find the exact center point of a line segment between two points in space. This is calculated using:\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]Where:

• \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) are the coordinates of the endpoints

In simple terms, it finds the average of the corresponding coordinates.
In the given example, it was used to calculate the midpoint between two sphere centers, earning the result \( (1, 1, 1) \) as the midpoint.
This calculated midpoint is paramount in analyzing problems involving coordinate geometry and vastly used in further calculations such as verifying positions or determining relations of points within geometric figures.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, merges algebra and geometry to study figures through equations. It translates geometric shapes into algebraic equations, allowing computations and verifications of properties.
In coordinate geometry:

• Points are represented as coordinates in a plane or space such as \((x, y)\) or \((x, y, z)\).
• Lines, circles, planes, and other shapes are defined through equations capturing their properties like slope, center, or radius.
• Problems utilize these equations to establish relationships between different elements, as seen in the exercise where you determine if a plane passes through a given point by substituting the coordinates.

It simplifies understanding complex problems by providing straightforward manipulation rules. Solving coordinate geometry problems often involves checking the satisfaction of specific equations with given coordinates, ensuring everything aligns, as demonstrated in the exercise by examining the intersection with a plane.