Problem 35 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{a}=2 \hat{i}+\lambda_{1} \hat{j}+3 \hat{k}, \vec{b}=4 \hat{i}+\left(3-\lambda_{2}\right) \hat{j}+6 \hat{k}\) and \(\overrightarrow{\mathrm{c}}=3 \hat{i}+6 \hat{j}+\left(\lambda_{3}-1\right) \hat{k}\) be three vectors such that \(\overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{a}}\) is perpendicular to \(\overrightarrow{\mathrm{c}}\) Then a possible value of \(\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)\) is: [Jan. \(\mathbf{1 0}, \mathbf{2 0 1 9}\) (I)] (a) \((1,3,1)\) (b) \(\left(-\frac{1}{2}, 4,0\right)\) (c) \(\left(\frac{1}{2}, 4,-2\right)\) (d) \((1,5,1)\)

Step-by-Step Solution

Answer

The correct value is \((-\frac{1}{2}, 4, 0)\), option (b).

Step 1: Express Equation for \( \overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{a}} \)

We know \( \overrightarrow{\mathrm{b}} = 4 \hat{i} + (3 - \lambda_2) \hat{j} + 6 \hat{k} \) and \( \overrightarrow{\mathrm{a}} = 2 \hat{i} + \lambda_1 \hat{j} + 3 \hat{k} \). Since \( \overrightarrow{\mathrm{b}} = 2 \overrightarrow{\mathrm{a}} \), equate components to find relationships:\[ 4 = 2 \times 2 \]\[ (3 - \lambda_2) = 2 \lambda_1 \]\[ 6 = 2 \times 3 \]From these, it is clear the equations for the \( \hat{j} \) component is: \( 3 - \lambda_2 = 2 \lambda_1 \).

Step 2: Express Perpendicular Condition \( \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}} = 0 \)

Since \( \overrightarrow{\mathrm{a}} \) is perpendicular to \( \overrightarrow{\mathrm{c}} \), their dot product is zero.We have \( \overrightarrow{\mathrm{c}} = 3 \hat{i} + 6 \hat{j} + (\lambda_3 - 1) \hat{k} \).Calculate the dot product:\[ \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}} = (2)(3) + (\lambda_1)(6) + (3)(\lambda_3 - 1) = 0 \]Simplifying, we get:\[ 6 + 6\lambda_1 + 3\lambda_3 - 3 = 0 \]\[ 3 + 6\lambda_1 + 3\lambda_3 = 0 \]\[ 1 + 2\lambda_1 + \lambda_3 = 0 \].

Step 3: Solve the System of Equations

We now have a system of two equations:1. \( 3 - \lambda_2 = 2 \lambda_1 \)2. \( 1 + 2\lambda_1 + \lambda_3 = 0 \)Let's try the given options one by one and check if they satisfy both equations.

Step 4: Check Option (a) \((1,3,1)\)

Substitute \((\lambda_1, \lambda_2, \lambda_3) = (1, 3, 1)\) into both equations:For equation 1:\[ 3 - 3 = 2 \times 1 \rightarrow 0 = 2 \quad (Not \, true) \].This option does not satisfy the first equation, so it's incorrect.

Step 5: Check Option (b) \(\left(-\frac{1}{2}, 4, 0\right)\)

Substitute \((\lambda_1, \lambda_2, \lambda_3) = \left(-\frac{1}{2}, 4, 0\right)\):For equation 1:\[ 3 - 4 = 2\left(-\frac{1}{2}\right) \rightarrow -1 = -1 \quad (True) \].For equation 2:\[ 1 + 2\left(-\frac{1}{2}\right) + 0 = 0 \rightarrow 0 = 0 \quad (True) \].This option satisfies both equations.

Step 6: Verification of Other Options

Verify other options (c) and (d) aren't correct:- Option (c) \(\left(\frac{1}{2}, 4, -2\right)\): Fails the check for equation 2.- Option (d) \((1,5,1)\): Fails the check for equation 1.Thus, option (b) is the only one satisfying both equations.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra. It combines two vectors to produce a scalar. Think of it as a way to measure how much one vector goes in the direction of another.

For vectors \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \), the dot product is calculated as:

• \( \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \)

The result is a single number, not a vector, which tells us about the angle between the vectors. If this product equals zero, it means the vectors are perpendicular. This is crucial for understanding vector relationships and geometrical interpretations.

Perpendicular Vectors

Vectors are perpendicular when they form a 90-degree angle with each other. In vector algebra, this relationship can be confirmed using the dot product.

If the dot product of two vectors is zero, \( \vec{a} \cdot \vec{b} = 0 \), then the vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular.

• This orthogonality condition is often utilized to find unknowns in vector problems.
• Understanding perpendicular vectors is important for many applications, including physics and engineering, where directions play a critical role.

In our exercise, knowing that \( \vec{a} \) is perpendicular to \( \vec{c} \) helped us establish a condition to find the values of the variables.

System of Equations

A system of equations involves solving for multiple unknowns at the same time. These equations often arise naturally in problems involving vectors, especially when dealing with conditions like parallelism or perpendicularity.

To solve a system:

• Identify all the equations you have based on the problem conditions. These could represent component equality, perpendicularity via dot products, etc.
• Substitute the values of known components and solve for the unknowns in a step-by-step manner.

In our exercise, we derived two equations:

• From component comparison \( 3 - \lambda_2 = 2\lambda_1 \).
• From the dot product condition \( 1 + 2\lambda_1 + \lambda_3 = 0 \).

This system allowed us to test the possible values for \( (\lambda_1, \lambda_2, \lambda_3) \) presented in the options.

Vector Components

Vector components are the building blocks of vectors, each representing a portion along one of the coordinate axes. In three-dimensional space, we commonly talk about components along the \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) axes.

A vector \( \vec{v} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k} \) has:

• \( v_1 \) as the component in the direction of the \( \hat{i} \) axis (x-axis).
• \( v_2 \) as the component in the direction of the \( \hat{j} \) axis (y-axis).
• \( v_3 \) as the component in the direction of the \( \hat{k} \) axis (z-axis).

Understanding vector components is crucial because they allow us to analyze and manipulate vectors mathematically. They enable us to apply operations like addition, dot products, and solving equations, which is what we used in this exercise to equate and solve for unknown variables using vector relationships.