Problem 54 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

3 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 \([\mathbf{2 0 0 5}]\) \(x^{2}+y^{2}+z^{2}+6 x-8 y-2 z=13\) 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: Determine the Centers of the Spheres

The center of a sphere given by the equation \(x^{2}+y^{2}+z^{2}+2gx+2fy+2hz=c\) is \((-g, -f, -h)\). **First sphere:**Equation: \(x^{2}+y^{2}+z^{2}+6x-8y-2z=13\).Center: \((-3, 4, 1)\).**Second sphere:**Equation: \(x^{2}+y^{2}+z^{2}-10x+4y-2z=8\).Center: \((5, -2, 1)\).

Step 2: Find the Midpoint of the Line Segment

To find the midpoint of the line joining two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), use the formula:\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)\).Substituting the centers we found:\(\left(\frac{-3+5}{2}, \frac{4+(-2)}{2}, \frac{1+1}{2}\right)\).This gives midpoint \((1, 1, 1)\).

Step 3: Substitute the Midpoint into the Plane Equation

The equation of the plane is \(2ax - 3ay + 4az + 6 = 0\).Substitute the midpoint \((1, 1, 1)\) into the plane's equation:\(2a(1) - 3a(1) + 4a(1) + 6 = 0\).This simplifies to \(3a + 6 = 0\).

Step 4: Solve for 'a'

From the equation \(3a + 6 = 0\), solve for \(a\):Subtract 6 from both sides:\(3a = -6\).Divide both sides by 3:\(a = -2\).

Key Concepts

Sphere Equations

A sphere in coordinate geometry is defined by an equation of the form \[ x^2 + y^2 + z^2 + 2gx + 2fy + 2hz = c \], where \(g\), \(f\), \(h\) are constants, and \(c\) is a real number. The key feature of this equation is that it helps to identify the center and radius of the sphere.

• The center of the sphere can be found by using the coefficients of \(x\), \(y\), and \(z\). For an equation like the one above, the center is \((-g, -f, -h)\).
• The radius of the sphere is determined by substituting the center back into the equation and solving for the radius. However, in this problem, we are primarily interested in the center for finding the midpoint.

In our exercise, we have two sphere equations:- For the first sphere, the equation is \[ x^2 + y^2 + z^2 + 6x - 8y - 2z = 13 \], giving the center as \((-3, 4, 1)\).- For the second sphere, the equation is \[ x^2 + y^2 + z^2 - 10x + 4y - 2z = 8 \], giving the center as \((5, -2, 1)\).These centers are crucial as they form the endpoints of a line segment whose midpoint we need to determine for the subsequent part of the solution.

Midpoint Formula

The midpoint of a line segment joining two points in 3-dimensional space is calculated using the midpoint formula.This formula is \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)\]where \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) are coordinates of the two points.

• This formula averages the x, y, and z coordinates of the two points to find a center point, or midpoint.
• The midpoint is important as it often acts as a point of symmetry and helps in various geometric calculations.

In this exercise, the calculated midpoint of the line segment between centers \((-3, 4, 1)\) and \((5, -2, 1)\) is\((1, 1, 1)\).This midpoint will be used in the plane equation to find the value of \(a\).

Plane Equation

A plane in coordinate geometry is often written in the form \[ ax + by + cz + d = 0 \],where \(a\), \(b\), and \(c\) are direction ratios of the normal to the plane.

• This equation is significant because it can represent any flat surface in 3-dimensional space.
• To find if a point lies on the plane, substitute the point's coordinates into the equation. If the equation holds true, the point is on the plane.

For our scenario, the equation is \[ 2ax - 3ay + 4az + 6 = 0 \]. By substituting the midpoint \((1, 1, 1)\),we set up the equation \[ 2a(1) - 3a(1) + 4a(1) + 6 = 0 \],which simplifies to \[ 3a + 6 = 0 \].By solving this, one discovers that \( a = -2 \).This calculation shows the relation between the midpoint of the spheres' centers and the plane.