Problem 11 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{\alpha}=(\lambda-2) \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\) and \(\vec{\beta}=(4 \lambda-2) \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}\) be two given vectors where vectors \(\vec{a}\) and \(\vec{b}\) are non-collinear. The value of \(\lambda\) for which vectors \(\vec{\alpha}\) and \(\vec{\beta}\) are collinear, is: [Jan. 10, 2019 (II)] (a) \(-4\) (b) \(-3\) (c) 4 (d) 3

Step-by-Step Solution

Answer

The value of \( \lambda \) is -4.

Step 1: Understanding Collinearity Condition

Two vectors \( \vec{\alpha} \) and \( \vec{\beta} \) are collinear if there exists a scalar \( k \) such that \( \vec{\alpha} = k \vec{\beta} \). This means their components must satisfy the relation uniformly.

Step 2: Equating Components Based on Collinearity

Given, \( \vec{\alpha} = (\lambda - 2) \overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}} \) and \( \vec{\beta} = (4 \lambda - 2) \overrightarrow{\mathrm{a}} + 3 \overrightarrow{\mathrm{b}} \). Using the condition of collinearity, \( (\lambda - 2) \) should be proportional to \( (4 \lambda - 2) \), and \( 1 \) should be proportional to \( 3 \).

Step 3: Setting Up Proportions from Components

From the \( \overrightarrow{\mathrm{a}} \) components, we have, \((\lambda - 2) = k(4 \lambda - 2)\). From \( \overrightarrow{\mathrm{b}} \) components, we have \(1 = 3k\). Solve the second proportion to find \( k \): \( k = \frac{1}{3} \).

Step 4: Solving for \( \lambda \)

Substitute \( k = \frac{1}{3} \) into the equation from Step 3: \((\lambda - 2) = \frac{1}{3}(4 \lambda - 2)\). Multiply through by 3 to clear the fraction: \(3(\lambda - 2) = 4\lambda - 2\). Simplifying gives \(3\lambda - 6 = 4\lambda - 2\). Rearranging the terms: \(-6 + 2 = 4\lambda - 3\lambda\). Solving gives \( \lambda = -4 \).

Key Concepts

Vectors and Collinearity

In mathematics, vectors are objects that have both magnitude and direction, represented typically as an arrow, or in component form \( (x, y, z) \). A fundamental concept with vectors is whether they are collinear or not. Vectors are considered collinear if they lie along the same line or are scalar multiples of each other.

This means that if you have two vectors, \( \vec{\alpha} \) and \( \vec{\beta} \), to check if they are collinear, you need to determine if there exists a scalar \( k \) such that \( \vec{\alpha} = k \vec{\beta} \). This condition will hold true if both the direction and magnitude are proportionate across every component of the vector.

• Collinear vectors have a scalar relationship.
• They might not have the same magnitude but do share the same or opposing direction.

Being able to identify and prove the collinearity of vectors is an essential skill in higher mathematics, especially in JEE preparations.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar (a real number), which scales the vector's magnitude without altering its direction.

If you consider a vector \( \vec{v} = (v_1, v_2) \), multiplying by a scalar \( c \) yields a new vector \( (cv_1, cv_2) \). This operation is straightforward, as it uniformly stretches or compresses the vector in space without changing its alignment.

Key features of scalar multiplication include:

• Scalar multiplication does not change the vector's direction.
• If the scalar is positive, the vector's direction remains the same; if it's negative, the direction is reversed.
• The magnitude of the vector changes proportionally to the absolute value of the scalar.

Understanding scalar multiplication is crucial for comprehending relationships and operations involving vectors, particularly in exercises dealing with collinearity.

Component-Wise Comparison

To verify properties like collinearity, component-wise comparison is often used in vector mathematics. When comparing two vectors \( \vec{\alpha} = (\alpha_1, \alpha_2) \) and \( \vec{\beta} = (\beta_1, \beta_2) \), component-wise comparison involves checking if each respective component of the two vectors follows the given condition, such as proportionality.

For example, if \( \vec{\alpha} \) and \( \vec{\beta} \) are to be collinear, then:

• The ratio \( \frac{\alpha_1}{\beta_1} = \frac{\alpha_2}{\beta_2} \) must hold true for some scalar \( k \).
• Each component should reflect the same scalar multiples to ensure the vectors lie on the same straight line.

This comparison helps break down complex vector expressions into manageable algebraic equations that can be solved using standard mathematical principles. It's a useful method in vector analysis.

Solving Linear Equations

Linear equations are equations of the first degree, meaning they include variables raised only to the power of one. Solving these equations is a common task in mathematics, especially in problems involving vectors.

In the context of vectors and collinearity, you often end up with linear equations when you equate the components of two collinear vectors. For example, to solve for \( \lambda \) in the component equation \( (\lambda - 2) = \frac{1}{3}(4\lambda - 2) \), you:

• First clear the fraction by multiplying through by the denominator.
• Simplify the equation to isolate terms involving \( \lambda \).
• Rearrange terms to solve for \( \lambda \).
• Ensue that the solution is clear and logical, as shown in the steps where \( \lambda = -4 \) was derived.

Understanding how to manipulate and solve linear equations is essential for determining unknown scalars and solving broader mathematical problems, especially when dealing with vectors and their orientations.