Problem 77 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The plane which bisects the line segment joining the points \((-3,-3,4)\) and \((3,7,6)\) at right angles, passes through which one of the following points? [Jan. 10, 2019 (II)] (a) \((-2,3,5)\) (b) \((4,-1,7)\) (c) \((2,1,3)\) (d) \((4,1,-2)\)

Step-by-Step Solution

Answer

Point (4,1,-2) satisfies the plane equation.

Step 1: Find the Midpoint

The midpoint of a line segment can be found using the formula \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \). For points \((-3,-3,4)\) and \((3,7,6)\), the midpoint is \( M = \left( \frac{-3 + 3}{2}, \frac{-3 + 7}{2}, \frac{4 + 6}{2} \right) = (0, 2, 5) \).

Step 2: Determine the Direction Ratios

The direction ratios of the line segment are computed as differences between corresponding coordinates: \( 3 - (-3), 7 - (-3), 6 - 4 \), resulting in \( 6, 10, 2 \).

Step 3: Find the Equation of the Plane

The plane bisects the segment at right angles, so its normal vector is \( (6, 10, 2) \). The general equation of a plane is \( ax + by + cz = d \). Here, \( a = 6, b = 10, c = 2 \), and substituting the midpoint \((0, 2, 5)\), we find \( 6(0) + 10(2) + 2(5) = d \) gives \( d = 30 \). Therefore, the equation is \( 6x + 10y + 2z = 30 \).

Step 4: Check Which Point Satisfies the Plane Equation

Substitute each option into the plane equation \( 6x + 10y + 2z = 30 \): - For \((-2,3,5)\): \( 6(-2) + 10(3) + 2(5) = -12 + 30 + 10 = 28 \).- For \((4,-1,7)\): \( 6(4) + 10(-1) + 2(7) = 24 - 10 + 14 = 28 \).- For \((2,1,3)\): \( 6(2) + 10(1) + 2(3) = 12 + 10 + 6 = 28 \).- For \((4,1,-2)\): \( 6(4) + 10(1) + 2(-2) = 24 + 10 - 4 = 30 \).This satisfies the equation.

Key Concepts

Midpoint Formula

The concept of a midpoint in coordinate geometry is essential for finding the center point of a line segment in space. To find the midpoint of a line segment joining two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \), we use the formula \[M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right).\]

• It averages the respective coordinates of the endpoints.
• This midpoint represents a point that is equidistant from both endpoints along each axis.

For example, given two points \( (-3,-3,4) \) and \( (3,7,6) \), the midpoint calculation involves:

• Calculating \( \frac{-3 + 3}{2} = 0 \) for the x-coordinate.
• Calculating \( \frac{-3 + 7}{2} = 2 \) for the y-coordinate.
• Calculating \( \frac{4 + 6}{2} = 5 \) for the z-coordinate.

Thus, the midpoint is \( (0, 2, 5) \). This is often the first step in problems involving lines and planes as it sets the stage for further calculations.

Direction Ratios

Direction ratios are a fundamental component in understanding the direction of a vector or line in coordinate space. These ratios are the differences between the coordinates of two points. Consider a line segment connecting \( (x_1, y_1, z_1) \) to \( (x_2, y_2, z_2) \):

• The direction ratios are given by \( x_2 - x_1, \ \, y_2 - y_1, \ \, z_2 - z_1\).
• These values essentially describe the slope or inclination of the line in three-dimensional space.

For the segment between \( (-3,-3,4) \) and \( (3,7,6) \), the direction ratios are:

• For x-direction: \( 3 - (-3) = 6 \).
• For y-direction: \( 7 - (-3) = 10 \).
• For z-direction: \( 6 - 4 = 2 \).

These direction ratios, \(6, 10, 2\), indicate how the line rises or falls along each axis and also serve as normal vectors for planes that are perpendicular to the line.

Equation of a Plane

The equation of a plane in three-dimensional space is fundamental in coordinate geometry. To define a plane, we use a general equation \( ax + by + cz = d \), where:

• The coefficients \( a, b, \) and \( c \) are the components of the normal vector to the plane. This vector is perpendicular to any line lying on the plane.
• The constant \( d \) is calculated using a point \( (x_0, y_0, z_0) \) on the plane, leading to \( ax_0 + by_0 + cz_0 = d \).

In our problem, the normal vector is determined by the direction ratios \( (6, 10, 2) \) from the line segment. We can use the midpoint \( (0, 2, 5) \) as a reference point. Substituting into the equation gives:

• \[ 6(0) + 10(2) + 2(5) = 20 + 10 = 30 \]
• Thus, our plane equation becomes \( 6x + 10y + 2z = 30 \).

To find which given point lies on the plane, substitute each point into the plane equation and verify equality. This method is very effective in 3D geometry tasks involving planes and points.