Problem 26 - A Complete Resource Book in Mathematics for JEE Main

Question

If the \(p\) th, \(q\) th and \(r\) th terms of a G. P. are positive numbers \(a, b\) and \(c\), respectively, then the angle between the vectors \(i l_{n} a+j l_{n} b+k l_{n} c\) and \(i(q-r)+j(r-p)+k\) \((p-q)\) is (A) \(\frac{\pi}{3}\) (B) \(\frac{\pi}{6}\) (C) \(\frac{\pi}{2}\) (D) none of these

Step-by-Step Solution

Answer

The angle between the vectors is \(\frac{\pi}{2}\), option (C).

Step 1: Understanding the Problem

We know that for a geometric progression (G.P.), the terms are related by a common ratio. Given the problem, the \(p\) th, \(q\) th, and \(r\) th terms are \(a, b,\) and \(c\) respectively, which can be expressed in terms of the first term \(A\) and common ratio \(R\) as: \(a = AR^{p-1}, b = AR^{q-1}, c = AR^{r-1}\). The vectors are constructed using the logarithms of these terms and specific coefficients.

Step 2: Calculate the Components of Vectors

The first vector is given by \(\mathbf{v_1} = i \ln{a} + j \ln{b} + k \ln{c}\). Substituting the terms in the G.P., we get: \(\ln a = \ln A + (p-1)\ln R\), \(\ln b = \ln A + (q-1)\ln R\), \(\ln c = \ln A + (r-1)\ln R\). Thus, \(\mathbf{v_1} = i((p-1)\ln R) + j((q-1)\ln R) + k((r-1)\ln R)\). Simplifying, \(\mathbf{v_1} = \ln R (i(p-1) + j(q-1) + k(r-1))\).

Step 3: Simplify the Second Vector

The second vector \(\mathbf{v_2}\) is given as \(i(q-r) + j(r-p) + k(p-q)\). This is a direct expression from the problem statement, and we'll use these components in calculating the dot product of the two vectors related to the angle between them.

Step 4: Compute Dot Product of the Vectors

To find the angle, compute the dot product: \(\mathbf{v_1} \cdot \mathbf{v_2} = \ln R \left[i(p-1) + j(q-1) + k(r-1)\right]\cdot [i(q-r) + j(r-p) + k(p-q)]\). This results in \(\ln R [(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)]\).

Step 5: Identify Condition for Perpendicular Vectors

For vectors to be perpendicular, their dot product must equal zero. Simplifying the expression, each of the terms \((p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)\) sums to zero, confirming \(\mathbf{v_1} \cdot \mathbf{v_2} = 0\).

Step 6: Conclude the Angle between Vectors

Since the dot product equals zero, the angle between the vectors must be \(\frac{\pi}{2}\). Therefore, the answer is (C) \(\frac{\pi}{2}\).

Key Concepts

Geometric Progression in Vectors

In the context of vectors, a geometric progression (G.P.) refers to a sequence of terms where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In vector terms for this exercise, we have the vectors defined by the G.P. sequence's terms.For example, if the terms of the G.P. are denoted by \(a\), \(b\), and \(c\), these can be expressed in terms of a first term \(A\) and a common ratio \(R\) as follows:

• \(a = AR^{p-1}\)
• \(b = AR^{q-1}\)
• \(c = AR^{r-1}\)

This formulation indicates that each term maintains a consistent relationship through multiplication by \(R\). Understanding how these terms relate through their logarithms is pivotal when dealing with vector operations. In this exercise, the components of the first vector are derived using the logarithm of each term to create specific vector elements.The first vector \(\mathbf{v_1}\) is constructed using these logarithmic values, transforming the problem of geometric progression into the vector domain, linking algebraic properties with geometric interpretations.

Dot Product of Vectors

The dot product is a crucial operation when handling vectors, especially when determining angles or projections. The dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is calculated as:\[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \]where \(u_1, u_2, u_3\) and \(v_1, v_2, v_3\) are the components of the vectors \(\mathbf{u}\) and \(\mathbf{v}\), respectively.In this exercise, the dot product is used to find the angle between two vectors, \(\mathbf{v_1}\) and \(\mathbf{v_2}\), derived from logarithmic expressions of the terms in a geometric progression and specific coefficients, respectively.The crucial aspect is calculating this dot product as:\[\ln R \left[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)\right]\]The result of this dot product directly informs us of the angle between the vectors. If the dot product simplifies to zero, it elucidates a particular relational property between these vectors, essential in the ensuing sections. The simplicity of multiplying and adding the vector components demonstrates the power of dot products in analyzing vector relationships in geometric spaces.

Perpendicular Vectors Condition

Two vectors are said to be perpendicular or orthogonal if their dot product equals zero. This condition is pivotal when evaluating the angle between vectors.In the context of this exercise, if the dot product \(\mathbf{v_1} \cdot \mathbf{v_2}\) is zero, it implies the vectors \(\mathbf{v_1}\) and \(\mathbf{v_2}\) form a right angle with each other, confirming orthogonality. This orthogonality translates to the angle between the vectors being \(\frac{\pi}{2}\) radians or 90 degrees.The expression:\[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q) = 0\]yields such a condition, revealing that the vectors don't just intersect but do so perpendicularly. Understanding this property is crucial when dealing with vector spaces, providing insights into geometric configurations and the specific orientations of vectors relative to each other. Hence, in this exercise, recognizing the orthogonality condition simplifies the solution significantly, demonstrating the blend of algebraic manipulation and geometric insight.