Problem 48 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the vectors \(\vec{a}=\hat{i}-\hat{j}+2 \hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}+\hat{\hat{k}}\) and \(\vec{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}\) are mutually orthogonal, then \((\lambda, \mu)=\) \(\begin{array}{ll}\text { (a) }(2,-3) & \text { (b) }(-2,3)\end{array}\) [2010] (c) \((3,-2)\) (d) \((-3,2)\)

Step-by-Step Solution

Answer

The correct option is (d) \((-3, 2)\).

Step 1: Understand Orthogonality

Vectors are mutually orthogonal if the dot product of each pair of vectors is zero. Therefore, we have to ensure that \( \vec{a} \cdot \vec{b} = 0 \), \( \vec{a} \cdot \vec{c} = 0 \), and \( \vec{b} \cdot \vec{c} = 0 \).

Step 2: Calculate \( \vec{a} \cdot \vec{b} \)

The dot product \( \vec{a} \cdot \vec{b} = (1)(2) + (-1)(4) + (2)(1) = 2 - 4 + 2 = 0 \). This confirms \( \vec{a} \) and \( \vec{b} \) are orthogonal.

Step 3: Calculate \( \vec{a} \cdot \vec{c} \)

Calculate using the formula for dot product: \( \vec{a} \cdot \vec{c} = (1)(\lambda) + (-1)(1) + (2)(\mu) = \lambda - 1 + 2\mu = 0 \). We will use this equation later.

Step 4: Calculate \( \vec{b} \cdot \vec{c} \)

Calculate this dot product: \( \vec{b} \cdot \vec{c} = (2)(\lambda) + (4)(1) + (1)(\mu) = 2\lambda + 4 + \mu = 0 \). This is the second equation needed.

Step 5: Solve the System of Equations

Solve the two equations: \( \lambda - 1 + 2\mu = 0 \) and \( 2\lambda + 4 + \mu = 0 \).- From the first equation: \( \lambda = 1 - 2\mu \)- Substitute into the second equation: \( 2(1 - 2\mu) + 4 + \mu = 0 \)- Simplify into: \( 2 - 4\mu + \mu + 4 = 0 \)- Combine equations: \( 6 - 3\mu = 0 \)- Solve for \( \mu \): \( \mu = 2 \)- Solve for \( \lambda \): \( \lambda = 1 - 2(2) = -3 \).

Step 6: Determine Solution Combination

The calculated values are \((\lambda, \mu) = (-3, 2)\). Match this to the available options, which corresponds to option (d).

Key Concepts

Dot Product

The dot product is a mathematical operation used in vector calculus that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is essential for calculating orthogonality in vectors. In simple terms, if you have two vectors, say \( \vec{u} = u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k} \) and \( \vec{v} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k} \), their dot product is calculated as:

• \( \vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \)

The result of a dot product is a scalar. It helps to determine if two vectors are orthogonal, i.e., perpendicular to each other.
If the dot product of two vectors is zero, it means the vectors are orthogonal. In our exercise, the calculations of the dot products \( \vec{a} \cdot \vec{b} \), \( \vec{a} \cdot \vec{c} \), and \( \vec{b} \cdot \vec{c} \) help identify the solution by confirming the orthogonality.

Mutually Orthogonal Vectors

Vectors are mutually orthogonal when every pair of vectors in the set is orthogonal to each other. In the context of the given exercise, this means that the dot product of each pair of the vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) must equal zero.

• \( \vec{a} \cdot \vec{b} = 0 \)
• \( \vec{a} \cdot \vec{c} = 0 \)
• \( \vec{b} \cdot \vec{c} = 0 \)

This condition ensures that all vectors form a right angle with each other in their dimensional space.
Understanding this concept is crucial for solving problems involving vector spaces and linear algebra, as it helps to determine relations between vectors and solve vector equations effectively. Orthogonal vectors are independent of each other, which is an important property in many areas, including physics and engineering.

System of Linear Equations

A system of linear equations involves finding the exact values of variables that satisfy all the equations simultaneously. In this exercise, after calculating dot products to establish orthogonality, two linear equations emerge:

• \( \lambda - 1 + 2\mu = 0 \)
• \( 2\lambda + 4 + \mu = 0 \)

Solving these equations requires combining them to eliminate variables step by step.
Here's how it's done:
- From the first equation, express \( \lambda \) in terms of \( \mu \):
\( \lambda = 1 - 2\mu \)- Substitute into the second equation: \( 2(1 - 2\mu) + 4 + \mu = 0 \)- Simplify to find \( \mu \): \( 6 - 3\mu = 0 \)- Solve for \( \mu \): \( \mu = 2 \)- Substitute back to find \( \lambda \): \( \lambda = 1 - 2(2) = -3 \)The calculated values \( (\lambda, \mu) = (-3, 2) \) solve the system, showing how these techniques fit together like pieces of a puzzle.