Problem 135 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the plane \(2 \mathrm{ax}-3 \mathrm{ay}+4 \mathrm{az}+6=0\) passes through the midpoint of the line joining the centres of the spheres \(x^{2}+y^{2}+z^{2}+6 x-8 y-2 z=13\) and \(x^{2}+y^{2}+z^{2}-10 x+4 y-2 z=8\) then a equals [2005] (a) \(-1\) (b) 1 (c) \(-2\) (d) 2

Step-by-Step Solution

Answer

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

Step 1: Identify the centers of the spheres

To find the centers of the given spheres, use the formula \((x-h)^2+(y-k)^2+(z-l)^2=r^2\) where \(h, k, l\) is the center and \(r\) is the radius. For the first sphere equation \(x^2+y^2+z^2+6x-8y-2z=13\), complete the square to find the center.

Step 2: Complete squares for the first sphere center

The sphere equation \(x^2+y^2+z^2+6x-8y-2z=13\) can be rewritten by completing the square:- For \(6x\), complete to get \((x+3)^2-9\)- For \(-8y\), complete to get \((y-4)^2-16\)- For \(-2z\), complete to get \((z-1)^2-1\)Adding these in, we have: \((x+3)^2 + (y-4)^2 + (z-1)^2 - 9 - 16 - 1 = 13\). Simplifying gives the center at \((-3, 4, 1)\).

Step 3: Complete squares for the second sphere center

Similarly, for the second sphere equation \(x^2+y^2+z^2-10x+4y-2z=8\), complete the square:- For \(-10x\), complete to get \((x-5)^2-25\)- For \(4y\), complete to get \((y+2)^2-4\)- For \(-2z\), complete to get \((z-1)^2-1\)Resulting in: \((x-5)^2 + (y+2)^2 + (z-1)^2 - 25 - 4 - 1 = 8\). Simplifying gives the center at \((5, -2, 1)\).

Step 4: Find the midpoint of the line joining the centers

Use the midpoint formula to find the midpoint of the line segment joining the centers \((-3, 4, 1)\) and \((5, -2, 1)\). The midpoint is:\(\left(\frac{-3+5}{2}, \frac{4 + (-2)}{2}, \frac{1 + 1}{2}\right) = (1, 1, 1)\).

Step 5: Substitute the midpoint into the plane equation

Substitute the midpoint \((1, 1, 1)\) into the plane equation \(2ax - 3ay + 4az + 6 = 0\), to get:\(2a(1) - 3a(1) + 4a(1) + 6 = 0\).

Step 6: Solve for \(a\)

Substitute the coordinates into the equation to simplify:\(2a - 3a + 4a + 6 = 0\)Combine like terms: \(3a + 6 = 0\)Solve for \(a\):Subtract 6 from both sides to get \(3a = -6\)Divide by 3 to isolate \(a\):\(a = -2\).

Key Concepts

Sphere Midpoint

When dealing with spheres, one common task is finding the midpoint of the line joining two points, specifically the centers of two spheres. To do this, we use the midpoint formula. It is important in problems where we verify whether a point lies on a particular geometrical entity, like a plane. In this exercise, we are given the centers of the spheres obtained through completing the square, which are

• Center of the first sphere:

  \((-3, 4, 1)\)

• Center of the second sphere:

  \((5, -2, 1)\)

To find the midpoint, we apply the midpoint formula

\( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \)

.
For these centers, the calculated midpoint is

\((1, 1, 1)\)

. It's a fundamental concept because it allows us to check if such midpoint satisfies other geometrical conditions, like lying on a given plane.

Plane Equation

The concept of a plane equation is essential in geometry. A plane in 3D space can usually be described by the equation

\(ax + by + cz + d = 0\)

. This equation tells us how each change in location coordinates affects the position relative to the plane.
In this exercise, the given plane equation is

\(2ax - 3ay + 4az + 6 = 0\)

.
This plane passes through the midpoint of the joining line segment formed by two sphere centers, resulting in conditions that allow us to find the unknown parameter \(a\).

• Substitute the midpoint into the plane equation to determine \(a\).

This process showcases how mathematical conditions expressed in coordinate geometry can help determine unknown values in geometric equations.

Completing the Square

The technique known as completing the square is fundamental in simplifying quadratic equations. This technique transforms a quadratic equation into a form that easily shows the vertex or center of a geometric figure like a circle or sphere.
For example, converting an equation such as

\(x^2 + y^2 + z^2 + 6x - 8y - 2z = 13\)

into a completed square form reveals the center of a sphere

• Converted to

  \((x + 3)^2 + (y - 4)^2 + (z - 1)^2 = 39\)

  .

The same process simplifies similar equations revealing their centers like

\((x - 5)^2 + (y + 2)^2 + (z - 1)^2 = 38\)

.Completing the square is like solving a puzzle, as it reconfigures an equation to make the hidden features conspicuous. Such techniques are crucial in identifying essential points in coordinate geometry.

Coordinate Geometry

Coordinate geometry merges algebra and geometry by using coordinates to define geometric shapes. This branch of mathematics allows us to use algebraic equations to elucidate geometric figures, positions, and distances.

• The sphere is defined using

  \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\)

  .
• The plane equation like

  \(ax + by + cz + d = 0\)

  characterizes a plane.

By understanding and utilizing these equations and techniques, such as completing the square, one can precisely detect points, analyze geometric structures, and find the relationships between different figures within a 3D space. When solving problems like the one in this exercise, coordinate geometry provides the toolbox for synthesis and analysis of geometry in an algebraic framework, bridging these two monumental fields.