Problem 6 - A Complete Resource Book in Mathematics for JEE Main

Question

Given a cube \(A B C D A_{1} B_{1} C_{1} D_{1}\) with lower base \(A B C D\), upper base \(A_{1} B_{1} C_{1} D_{1}\) and the lateral edges \(A A_{1}, B B_{1}\), \(C C_{1}\) and \(D D_{1} ; M\) and \(M_{1}\) are the centres of the faces \(A B C D\) and \(A_{1} B_{1} C_{1} D_{1}\) respectively. \(O\) is a point on line \(M M_{1}\), such that \(O A+O B+O C+O D=O M_{1}\), then \(O M=\lambda O M_{1}\) if \(\lambda=\) (A) \(\frac{1}{4}\) (B) \(\frac{1}{2}\) (C) \(\frac{1}{6}\) (D) \(\frac{1}{8}\)

Step-by-Step Solution

Answer

The value of \( \lambda \) is \( \frac{1}{2} \) (Option B).

Step 1: Understanding the Problem

We are given a cube with certain geometrical entities: bases, lateral edges, and centers of bases. M is the center of the lower base ABCD, M_1 is the center of the upper base A_1B_1C_1D_1, and O is on the line MM_1 such that OA + OB + OC + OD = OM_1. We need to find the ratio OM to OM_1.

Step 2: Set up Coordinate System

Place the cube in a coordinate system where A(0,0,0), B(a,0,0), C(a,a,0), D(0,a,0), A_1(0,0,a), B_1(a,0,a), C_1(a,a,a), D_1(0,a,a) with a being the side of the cube. M is the midpoint of ABCD at (a/2, a/2, 0) and M_1 is the midpoint of A_1B_1C_1D_1 at (a/2, a/2, a).

Step 3: Coordinates of Point O

Express point O on line MM_1 parametrically as O(a/2, a/2, z) where z is a parameter between 0 and a, since it lies on the line joining (a/2, a/2, 0) and (a/2, a/2, a).

Step 4: Calculating OA, OB, OC, OD and OM_1

Calculate distances using the distance formula: \( OA = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2 + z^2} \), \( OB = \sqrt{(\frac{a}{2} - a)^2 + (\frac{a}{2})^2 + z^2} \), \( OC = \sqrt{(\frac{a}{2} - a)^2 + (\frac{a}{2} - a)^2 + z^2} \), \( OD = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2} - a)^2 + z^2} \), and \( OM_1 = a - z \).

Step 5: Solve Equation OA + OB + OC + OD = OM_1

Simplify and use symmetry: Since all four distances OA, OB, OC, and OD are equal due to symmetry, express them in terms of one variable to solve the equation obtained in Step 4, to find relation between \( z \) and \( a \).

Step 6: Determine \( \lambda \) using Ratio

Calculate distances \( OM = z \) and \( OM_1 = a - z \), use these to get \( \frac{OM}{OM_1} = \lambda \). Solve for \( \lambda \).

Step 7: Choose the Correct Option

Using the value found for \( \lambda \), choose the matching option from (A) \( \frac{1}{4} \), (B) \( \frac{1}{2} \), (C) \( \frac{1}{6} \), (D) \( \frac{1}{8} \).

Key Concepts

Distance Formula

Coordinate geometry helps us find the distances between points in a geometric space. The distance formula is a key component in determining these distances. For two points given by coordinates

• Point 1: \(x_1, y_1, z_1\)
• Point 2: \(x_2, y_2, z_2\)

the distance \(d\) between them is calculated as \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \).

In the exercise, we used the distance formula to calculate distances like \(OA\), which has the formula \( \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2 + z^2} \). This gives a tangible sense of measurement in the spatial setting of the cube.

Using the distance formula allows us to understand how far one point is from another, which is crucial in solving problems like this one involving geometric structures and their properties.

Parametric Equations

Geometrically, parametric equations allow us to express the positions of points on a line or curve in terms of a parameter. This is extremely useful when dealing with points that move along a line.

In the context of our cube exercise, point \(O\) is expressed parametrically. Since \(O\) is on the line connecting \(M\) and \(M_1\), \(O\)'s position can be defined by \(O(\frac{a}{2}, \frac{a}{2}, z)\).

Here, \(z\) serves as the parameter that varies from the start to the end of the line, precisely from \(z = 0\) to \(z = a\). This allows us to explore different positions of \(O\) along the line and find the particular value of \(z\) that satisfies given conditions like \(OA + OB + OC + OD = OM_1\).

Parametric equations give us flexibility and control when analyzing geometrical locations, facilitating the exploration of structures within coordinate geometry.

Geometric Symmetry

Symmetry is a significant concept in geometry that simplifies complex problems by identifying repetitive or equal parts. It involves understanding similar patterns within a shape or figure.

In our exercise, we leverage the geometric symmetry of the cube and its center points. Since points \(A, B, C, \) and \(D\) are symmetrically placed around \(M\), the center, the distances \(OA, OB, OC, \) and \(OD\) become equal due to this symmetry.

Recognizing this symmetry allows us to express these distances with a single variable, making the equations easier to handle. This utilization of symmetry helps to streamline solutions and verify correct calculations.

By embracing geometric symmetry, we simplify our investigation within coordinate geometry, effectively addressing challenges presented by complex shapes like cubes.