Problem 136 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The intersection of the spheres \(x^{2}+y^{2}+z^{2}+7 x-2 y-z=13\) and \(x^{2}+y^{2}+z^{2}-3 x+3 y+4 z=8\) is the same as the intersection of one of the sphere and the plane [2004] (a) \(2 x-y-z=1\) (b) \(x-2 y-z=1\) (c) \(x-y-2 z=1\) (d) \(x-y-z=1\)

Step-by-Step Solution

Answer

(a) 2x - y - z = 1

Step 1: Rewrite the Equations in Standard Form

First, identify the centers and radii of the given spheres by rewriting each equation in standard form. The general equation for a sphere is \[(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2.\]1st Sphere:\[x^2 + y^2 + z^2 + 7x - 2y - z = 13.\]Complete the square for each variable:\[(x+\frac{7}{2})^2 - \frac{49}{4} + (y-1)^2 - 1 + (z-\frac{1}{2})^2 - \frac{1}{4} = 13.\]Simplifying gives:\[(x+\frac{7}{2})^2 +(y-1)^2 +(z-\frac{1}{2})^2 = \frac{95}{4}.\]2nd Sphere:\[x^2 + y^2 + z^2 - 3x + 3y +4z = 8.\]Complete the square:\[(x-\frac{3}{2})^2 - \frac{9}{4} + (y+\frac{3}{2})^2 - \frac{9}{4} + (z+2)^2 - 4 = 8.\]Simplifying gives:\[(x-\frac{3}{2})^2 +(y+\frac{3}{2})^2 +(z+2)^2 = \frac{65}{4}.\]

Step 2: Determine the Equation of the Intersection

To find the intersection of the two spheres, subtract the second sphere's equation from the first. This elimination yields the equation of the plane that contains their intersection.\[(x+\frac{7}{2})^2 +(y-1)^2 +(z-\frac{1}{2})^2 -(x-\frac{3}{2})^2 -(y+\frac{3}{2})^2 -(z+2)^2 = \frac{95}{4} - \frac{65}{4}.\]Simplifying gives the plane equation:\[10x - 7y - 5z = 10.\]Dividing the entire equation by 10 results in the plane:\[x - \frac{7}{10}y - \frac{1}{2}z = 1.\]

Step 3: Compare the Intersection Plane with Given Options

Now, compare the equation of the intersection plane found in Step 2 to the given options. The correct plane should have coefficients proportional to the equation derived in Step 2. The given options are scaled versions of possible correct answers:(a) \[2x - y - z = 1\](b) \[x - 2y - z = 1\](c) \[x - y - 2z = 1\](d) \[x - y - z = 1\]Notice, option (a) \[2(x - \frac{1}{2}y - \frac{1}{2}z) = 2\] is equivalent to \[x - \frac{1}{2}y - \frac{1}{2}z = 1\] which matches proportionally to the equation from Step 2.

Key Concepts

Intersection of Spheres

Spheres in geometry can intersect in a variety of ways. When two spheres intersect, the result is typically either an empty set, a single point, a circle, or an entire sphere. In this exercise, we are dealing with the intersection of two given spheres. The key is to determine how these spheres meet in space.

The given two spheres can be described by the equations:

• Sphere 1: \(x^2 + y^2 + z^2 + 7x - 2y - z = 13\)
• Sphere 2: \(x^2 + y^2 + z^2 - 3x + 3y + 4z = 8\)

To find their intersection, we subtract one equation from the other, reducing the system to a plane. This plane reveals the line or circle of intersection of the two spheres. Calculating their actual intersection requires knowledge of the positions and radii of each sphere. Once you subtract the equations, you can derive the plane that fits the intersection as shown in the detailed solution.

Sphere Equations

The general equation for a sphere in three-dimensional space is \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\). Here:

• \( (a, b, c) \) are the coordinates of the center of the sphere.
• \( r \) is the radius of the sphere.

To convert a given sphere's equation into its standard form, you must complete the square for each variable. This involves:

• Identifying the linear terms of \(x\), \(y\), and \(z\).
• Completing the square by adjusting constants appropriately and simplifying.

For example, the transformation of the given sphere equation \(x^2 + y^2 + z^2 + 7x - 2y - z = 13\) involves completing the square for each of \(x\), \(y\), and \(z\), rearranging terms, and isolating the constants to derive the standard form with the center and radius defined.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a crucial area of mathematics that uses algebraic equations to describe geometric figures. Here, it allows us to explore the properties and positions of geometric shapes like spheres in space, using Cartesian coordinates \((x, y, z)\).

This branch of mathematics gives us tools to manipulate sphere equations and discover relationships between different geometric entities in 3D space. For instance, when determining the intersection of two spheres, coordinate geometry helps in deriving equations of lines, planes, and points where these shapes meet. The techniques of converting equations into standard forms and then using algebra to find intersections are all part of this essential field that plays a huge role in understanding spatial relationships.

JEE Problem Solving

The JEE Main Mathematics exam includes complex and practical problem-solving exercises such as the one on sphere intersections. Whether you're solving for intersection points, lines, or planes in geometry, the approach used is key to efficiently tackling these questions.

To solve such a problem in an exam setting, you need to:

• Quickly recognize the form of given equations and what transformations are required - completing the square is crucial for spheres.
• Use symmetry and logical deductions to simplify complex algebraic manipulations.
• Understand and apply systematic methods, like eliminating variables or combining equations, to reduce them to simpler forms such as lines or planes.

Practicing problems like these enhances your ability to effectively use basic math principles to solve sophisticated problems, essential for the JEE Main. Remember that clarity, practice, and understanding of fundamental concepts are key to mastering JEE Mathematics questions.