Problem 138 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The shortest distance from the plane \(12 x+4 y+3 z=327\) to the sphere \(x^{2}+y^{2}+z^{2}+4 x-2 y-6 z=155\) is [2003] (a) 39 (b) 26 (c) \(11 \frac{4}{13}\) (d) 13 .

Step-by-Step Solution

Answer

The shortest distance is \(11 \frac{4}{13}\).

Step 1: Identify the sphere's center and radius

The equation of the sphere is given by \[x^2 + y^2 + z^2 + 4x - 2y - 6z = 155\]First, we need to complete the square for the terms containing \(x\), \(y\), and \(z\):For \(x\):\(x^2 + 4x = (x+2)^2 - 4\)For \(y\):\(y^2 - 2y = (y-1)^2 - 1\)For \(z\):\(z^2 - 6z = (z-3)^2 - 9\)Substituting back, we have:\[(x+2)^2 + (y-1)^2 + (z-3)^2 - 4 - 1 - 9 = 155\]Simplifying gives:\[(x+2)^2 + (y-1)^2 + (z-3)^2 = 169\]Thus, the center of the sphere is \((-2, 1, 3)\) and its radius is \(\sqrt{169} = 13\).

Step 2: Calculate the distance from the center of the sphere to the plane

The plane equation is:\[12x + 4y + 3z = 327\]The distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\) is calculated by:\[d = \frac{|Ax_1 + By_1 + Cz_1 + D'|}{\sqrt{A^2 + B^2 + C^2}}\]where \(D' = -327\). Substituting the center of the sphere \((-2, 1, 3)\), we find:\[A = 12, \quad B = 4, \quad C = 3, \quad D' = -327\]\[d = \frac{|12(-2) + 4(1) + 3(3) + 327|}{\sqrt{12^2 + 4^2 + 3^2}}\]\[d = \frac{|-24 + 4 + 9 + 327|}{\sqrt{169}} = \frac{316}{13} = 24.31\]

Step 3: Calculate the shortest distance from the sphere to the plane

The shortest distance from the sphere to the plane is the distance from the center of the sphere to the plane, minus the radius of the sphere. From Step 2, we found that distance to be \(24.31\). The radius from Step 1 is 13.Therefore, the shortest distance from the sphere to the plane is:\[24.31 - 13 = 11.31\]This matches the option \(11 \frac{4}{13}\) after converting \(11.31\) to a fraction.

Key Concepts

Sphere and Plane Distance

Understanding the distance between a sphere and a plane is crucial for solving many geometric problems in coordinate space. A straightforward way to find this shortest distance is to determine the perpendicular distance from the sphere's center to the plane.

• First, calculate the straight-line (perpendicular) distance from the center of the sphere to the plane.
• The formula used for this calculation involves taking the absolute value of the plane's equation when substituted with the sphere's center coordinates.
• Finally, subtract the sphere's radius from the distance obtained because the closest point on the sphere will be one radius length away from the center point on the plane.

By approaching the problem this way, we account for both the center-to-plane distance and the size of the sphere itself.

Coordinate Geometry

Coordinate geometry, or analytic geometry, allows us to convert geometrical problems into algebraic equations. This transformation enables algebraic manipulation to solve complex problems about distance, intersections, and angles in space.

• Each point or shape in space can be defined with coordinates, making it easier to apply mathematical operations.
• Spheres and planes can be represented in specific equations that describe their positions and dimensions within three-dimensional space.
• By solving these equations, you can find the relationships and distances between geometrical elements like spheres and planes.

This analytical approach simplifies the understanding and solving of spatial relationships by allowing numerical and algebraic strategies to be applied.

Equation of a Sphere

The equation of a sphere in standard form is \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\]where \((h, k, l)\) is the center of the sphere, and\(r\)is the radius.
To convert any given sphere equation to this standard form, you might need to complete the square for each variable.

• Identify the coefficients of the terms: these help locate the center of the sphere.
• Complete the square for each variable (x, y, z) to simplify the equation.
• Once simplified, read off the center and the radius from the equation directly.

Completing the square lets you see the standard form of the equation, making it easier to work with and apply in solving geometry problems.

Shortest Distance Calculation

Calculating the shortest distance in coordinate geometry often involves defining specific points or lines that simplify the problem. For calculating the distance from a point (like a sphere's center) to a plane, a particular formula helps:
\[d = \frac{|Ax_1 + By_1 + Cz_1 + D'|}{\sqrt{A^2 + B^2 + C^2}}\]This formula gives the perpendicular distance from a point to a plane.

• Insert the coordinates of the point into the plane's equation to find the numerator.
• The denominator is the magnitude of the plane's normal vector.
• The shortest distance from the sphere to the plane involves subtracting the sphere's radius from this calculated distance.

This straightforward calculation is essential when determining how objects in three-dimensional space relate to each other, ensuring accuracy in spatial planning and analysis.