Problem 101 - A Complete Resource Book in Mathematics for JEE Main

Question

If the vectors \(\vec{a}=\hat{i}-\hat{j}+2 \hat{k}, \hat{b}=2 \hat{i}+4 \hat{j}+\hat{k}\) and \(\vec{c}=\lambda \hat{i}+\hat{j}\) \(+\mu \hat{k}\) are mutually orthogonal, then the tuple \((\lambda, \mu)=\) \([\mathbf{2 0 1 0}]\) (A) \((2,-3)\) (B) \((-2,3)\) (C) \((3,-2)\) (D) \((-3,2)\)

Step-by-Step Solution

Answer

The tuple \((\lambda, \mu)\) is \((-3, 2)\).

Step 1: Understand Orthogonality

Two vectors are orthogonal if their dot product is zero. Therefore, for vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) to be mutually orthogonal, the conditions \(\vec{a} \cdot \vec{b} = 0\), \(\vec{a} \cdot \vec{c} = 0\), and \(\vec{b} \cdot \vec{c} = 0\) must be satisfied.

Step 2: Dot Product of \(\vec{a}\) and \(\vec{b}\)

Calculate \(\vec{a} \cdot \vec{b}\) using the components of the vectors. Given \(\vec{a} = \hat{i} - \hat{j} + 2\hat{k}\) and \(\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}\), the dot product is:\[ (1)(2) + (-1)(4) + (2)(1) = 2 - 4 + 2 = 0 \]Thus, \(\vec{a} \cdot \vec{b} = 0\) verifies that \(\vec{a}\) and \(\vec{b}\) are orthogonal.

Step 3: Dot Product of \(\vec{a}\) and \(\vec{c}\)

Calculate \(\vec{a} \cdot \vec{c}\). Given \(\vec{c} = \lambda \hat{i} + \hat{j} + \mu \hat{k}\), the dot product is:\[ (1)\lambda + (-1)(1) + 2(\mu) = \lambda - 1 + 2\mu = 0 \]This simplifies to \(\lambda + 2\mu = 1\).

Step 4: Dot Product of \(\vec{b}\) and \(\vec{c}\)

Calculate \(\vec{b} \cdot \vec{c}\). The dot product is: \[ (2)\lambda + (4)(1) + (1)(\mu) = 2\lambda + 4 + \mu = 0 \]This simplifies to \(2\lambda + \mu = -4\).

Step 5: Solve the system of equations

You now have a system of two equations:1. \(\lambda + 2\mu = 1\)2. \(2\lambda + \mu = -4\)Multiply the first equation by 2: \[2\lambda + 4\mu = 2\]Subtract the second equation from this equation:\[(2\lambda + 4\mu) - (2\lambda + \mu) = 2 - (-4)\]\[3\mu = 6\]\[\mu = 2\]Plug \(\mu = 2\) into the first equation:\[\lambda + 2(2) = 1\]\[\lambda + 4 = 1\]\[\lambda = -3\]

Step 6: Verify the solution

Substitute \(\lambda = -3\) and \(\mu = 2\) back into both equations:1. \((-3) + 2(2) = -3 + 4 = 1\) \(\Rightarrow\) Satisfied.2. \(2(-3) + 2 = -6 + 2 = -4\) \(\Rightarrow\) Satisfied.Both equations are satisfied, confirming that \((\lambda, \mu) = (-3, 2)\).

Key Concepts

Understanding the Dot Product

The dot product is a crucial concept in understanding the relationship between vectors. It helps us determine whether two vectors are orthogonal by calculating their dot product. To compute the dot product of two vectors \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), simply multiply each of their respective components and sum up the results:

• \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \).

If the result is zero, the vectors are orthogonal. This helps us in problems involving mutual orthogonality, where we need several dot products to be zero. In our exercise, calculating the dot products \( \vec{a} \cdot \vec{b} \), \( \vec{a} \cdot \vec{c} \), and \( \vec{b} \cdot \vec{c} \) confirmed pairwise orthogonality among the vectors.

Solving the System of Equations

A system of equations arises when we have multiple conditions that need to be satisfied simultaneously. Solving these systems is essential for determining unknowns—like \( \lambda \) and \( \mu \) in an orthogonal vector problem. The dot product's conditions guides us to form equations for these unknowns. For example:

• From \( \vec{a} \cdot \vec{c} = 0 \), we formed \( \lambda + 2\mu = 1 \).
• From \( \vec{b} \cdot \vec{c} = 0 \), we formed \( 2\lambda + \mu = -4 \).

Solving the system involves algebraic manipulation. You can use methods like substitution or elimination to find the values of \( \lambda \) and \( \mu \). In this exercise, we used elimination to first find \( \mu = 2 \) and then substituted it back to find \( \lambda = -3 \). Consistency across the original equations verified these values.

Understanding Mutual Orthogonality

Mutual orthogonality is when a set of vectors are all orthogonal to each other. This means their pairwise dot products result in zero. This property is essential when dealing with problems that involve three or more vectors, such as in our exercise.For vectors \( \vec{a}, \vec{b}, \) and \( \vec{c} \) to be mutually orthogonal:

• \( \vec{a} \cdot \vec{b} = 0 \), confirming \( \vec{a} \) and \( \vec{b} \) are orthogonal.
• \( \vec{a} \cdot \vec{c} = 0 \), confirming \( \vec{a} \) and \( \vec{c} \) are orthogonal.
• \( \vec{b} \cdot \vec{c} = 0 \), confirming \( \vec{b} \) and \( \vec{c} \) are orthogonal.

This concept helps determine vector configurations in multi-dimensional space, ensuring vector independence and simplifying complex problems into manageable systems of equations. By understanding mutual orthogonality, solving such vector-related exercises becomes much clearer.