Problem 101 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \((\lambda, \mu)=\) [2010] (A) \((2,-3)\) (B) \((-2,3)\) (C) \((3,-2)\) (D) \((-3,2)\)

Step-by-Step Solution

Answer

(\lambda, \mu) = (-3, 2) \text{ (Option D)}.

Step 1: Understand Mutual Orthogonality

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

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

Compute the dot product: \(\vec{a} \cdot \vec{b} = (\hat{i} - \hat{j} + 2\hat{k}) \cdot (2\hat{i} + 4\hat{j} + \hat{k}) = 1 \times 2 + (-1) \times 4 + 2 \times 1\). This simplifies to \(2 - 4 + 2 = 0\). Thus, \(\vec{a} \cdot \vec{b} = 0\), confirming mutual orthogonality between \(\vec{a}\) and \(\vec{b}\).

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

Compute the dot product: \(\vec{b} \cdot \vec{c} = (2\hat{i} + 4\hat{j} + \hat{k}) \cdot (\lambda\hat{i} + \hat{j} + \mu\hat{k}) = 2\lambda + 4 \times 1 + 1 \times \mu\). This simplifies to \(2\lambda + 4 + \mu = 0\).

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

Compute the dot product: \(\vec{c} \cdot \vec{a} = (\lambda \hat{i} + \hat{j} + \mu \hat{k}) \cdot (\hat{i} - \hat{j} + 2\hat{k}) = \lambda \times 1 + 1 \times (-1) + \mu \times 2\). This simplifies to \(\lambda - 1 + 2\mu = 0\).

Step 5: Solve the System of Equations

We have two equations: \(2\lambda + \mu = -4\) and \(\lambda + 2\mu = 1\). Solve this system to find \(\lambda\) and \(\mu\). Multiply the second equation by 2: \(2\lambda + 4\mu = 2\). Now subtract the first equation from this result: \(2\lambda + 4\mu - (2\lambda + \mu) = 2 + 4\), resulting in \(3\mu = 6\). Thus \(\mu = 2\). Substitute \(\mu = 2\) into \(\lambda + 2\mu = 1\), yielding \(\lambda + 4 = 1\) so \(\lambda = -3\).

Step 6: Verify the Tuple

The solution is \((\lambda, \mu) = (-3, 2)\). This value satisfies both equations: \(2(-3) + 2 = -4\) and \(-3 + 2\times 2 = 1\). Therefore, the vectors are mutually orthogonal with this choice.

Key Concepts

Mutual Orthogonality

Vectors are said to be mutually orthogonal when there is no overlap in the direction they are pointing, kind of like how two paths that cross at right angles don't share any direction. In the context of vectors, this means the dot product between each pair of vectors is zero.
Consider three vectors, let's name them \(\vec{a}, \vec{b},\) and \(\vec{c}\). For these vectors to be mutually orthogonal:

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

When you calculate each of these dot products and find the result is zero, it confirms that every vector is perpendicular to the others.
This concept is extremely useful, especially in physics and engineering, where figuring out forces or components acting at right angles is essential.

Dot Product

The dot product is one of the fundamental operations you can perform with vectors. It combines two vectors into a single number. This product is heavily used to determine the angle relationship between two vectors.
The formula for the dot product of two vectors \(\vec{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}\) and \(\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}\) is:

• \(\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3\)

If the dot product turns out to be zero, it means that the vectors are orthogonal, or in simpler terms, they make a 90-degree angle with each other.
In our original exercise, the dot product helped us confirm the orthogonality of the vectors \(\vec{a}, \vec{b},\) and \(\vec{c}\), hence identifying the condition of mutual orthogonality.

System of Equations

When solving vector problems, like finding whether vectors are mutually orthogonal, we often end up with a system of equations. A system of equations is simply multiple equations that you need to solve all at once. In our exercise, the mutual orthogonality conditions led us to two equations:

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

The goal is to find values of \(\lambda\) and \(\mu\) that satisfy both equations simultaneously.
There are various methods to solve a system of equations, such as substitution, elimination, or using matrices. We used a step-by-step elimination method which involved:

• Adjusting one equation to eliminate one of the variables
• Solving for the remaining variable
• Back-substituting to find the other variable

After solving, we found \(\lambda = -3\) and \(\mu = 2\). Solution of the system led us to verify the orthogonality of the vectors, confirming their mutual independence in terms of direction.