Problem 102 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the points \((1,1, \lambda)\) and \((-3,0,1)\) are equidistant from the plane, \(3 x+4 y-12 z+13=0\), then \(\lambda\) satisfies the equation: \(\quad\) [Online April 10, 2015] (a) \(3 x^{2}+10 x-13=0\) (b) \(3 x^{2}-10 x+21=0\) (c) \(3 x^{2}-10 x+7=0\) (d) \(3 x^{2}+10 x-7=0\)

Step-by-Step Solution

Answer

\(\lambda\) satisfies the equation \(3x^2 - 10x + 7 = 0\).

Step 1: Understanding the Problem

We need to find the value of \(\lambda\) such that the points \((1,1, \lambda)\) and \((-3,0,1)\) are equidistant from the plane \(3x + 4y - 12z + 13 = 0\). This means setting the perpendicular distance from each point to the plane equal to each other, and solving for \(\lambda\).

Step 2: Formula for Distance from a Point to a Plane

The distance \(D\) from a point \((x_0, y_0, z_0)\) to a plane \(ax + by + cz + d = 0\) is given by:\[ D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

Step 3: Finding Distance of Point (1, 1, \(\lambda\)) to the Plane

Substitute the point \((1,1,\lambda)\) into the distance formula:\[ D_1 = \frac{|3(1) + 4(1) - 12(\lambda) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|17 - 12\lambda|}{13} \]

Step 4: Finding Distance of Point (-3, 0, 1) to the Plane

Substitute the point \((-3,0,1)\) into the distance formula:\[ D_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|13 - 21|}{13} = \frac{8}{13} \]

Step 5: Setting the Distances Equal

Because the points are equidistant from the plane, we set \(D_1 = D_2\):\[ \frac{|17 - 12\lambda|}{13} = \frac{8}{13} \]

Key Concepts

Equidistant Points

When discussing equidistant points, it's all about measuring the distance between points and a reference plane, and making sure these distances are the same. In our exercise, two points are evaluated to check if they are equidistant from a plane. The coordinates given in our exercise are

• The point \( (1,1, \lambda) \)
• The point \((-3, 0, 1)\)

To achieve equidistance:

• You calculate the perpendicular distance of each point to the plane.
• The result should be that both distances are equal when the right value of \(\lambda\) is used.

Equidistant points assure symmetry or equal spacing between a set of points and a reference, which is helpful in solving geometrical problems. Understanding the concept of equidistance is key in coordinate geometry, enabling precise distance calculations from given points to a plane.

Perpendicular Distance Formula

To find the distance from a specific point to a plane, we use the Perpendicular Distance Formula. It involves substituting the point's coordinates into the equation of the plane, and then computing the result. Here's how the formula works:Consider the general equation of a plane: \( ax + by + cz + d = 0 \).For a point \( (x_0, y_0, z_0) \), the formula for calculating the perpendicular distance \( D \) is:\[ D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

• The numerator consists of evaluating the plane equation using \( (x_0, y_0, z_0) \).
• The denominator is the square root of the sum of squares of the coefficients from the plane equation.

This formula provides the shortest distance from the point to the plane, by dropping a perpendicular line from the point to the plane. Using this formula, you can find the distance from any point to a specified plane accurately.

Coordinate Geometry

Coordinate Geometry involves studying geometry using a coordinate system, and it's handy for solving problems involving distances and relationships in a geometrical space. In three-dimensional coordinate geometry:

• The position of each point is described using three coordinates: \( (x, y, z) \).
• Geometrical problems can be visualized in 3D space, making computations easier when finding distances or locating points relative to planes or other surfaces.

In our exercise, the plane's equation \( 3x + 4y - 12z + 13 = 0 \) and the given points are all part of this three-dimensional coordinate system. By utilizing the related formulas, we solve for vector distances and equidistant conditions. Coordinate geometry thus transforms visual concepts into solvable numerical equations, bridging purely visual geometry with algebraic methods and calculations.