Problem 64 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three unit vectors, out of which vectors \(\vec{b}\) and \(\vec{c}\) are non-parallel. If \(\alpha\) and \(\beta\) are the angles which vector \(\vec{a}\) makes with vectors \(\vec{b}\) and \(\vec{c}\) respectively and \(\vec{a} \times(\vec{b} \times \vec{c})=\frac{1}{2} \vec{b}\), then \(|\alpha-\beta|\) is equal to : [Jan. 12, 2019 (II)\\} (a) \(30^{\circ}\) (b) \(90^{\circ}\) (c) \(60^{\circ}\) (d) \(45^{\circ}\)

Step-by-Step Solution

Answer

The value of \(|\alpha - \beta|\) is \(30^\circ\).

Step 1: Understand the Given Equation

You are given that \( \vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2} \vec{b} \). Use the vector triple product identity: \( \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \).

Step 2: Applying the Triple Product Identity

Substitute the identity into the given equation: \((\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{1}{2} \vec{b}\). Since \(\vec{b}\) and \(\vec{c}\) are non-parallel, equate the coefficients of \(\vec{b}\) and \(\vec{c}\) separately.

Step 3: Equate the Coefficients of \(\vec{b}\)

From \((\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{1}{2} \vec{b}\), we get \(\vec{a} \cdot \vec{c} = \frac{1}{2}\), representing the cosine of the angle \(\beta\). Therefore, \(\cos \beta = \frac{1}{2}\).

Step 4: Equate the Coefficients of \(\vec{c}\)

Considering \((\vec{a} \cdot \vec{b}) \vec{c} = 0\), this indicates that \(\vec{a} \cdot \vec{b} = 0\), which implies \(\cos \alpha = 0\).

Step 5: Find the Angles \(\alpha\) and \(\beta\)

Given \(\cos \alpha = 0\), \(\alpha\) must be \(90^\circ\). Since \(\cos \beta = \frac{1}{2}\), \(\beta\) is \(60^\circ\) or \(300^\circ\) (in the context of unit vectors, consider \(60^\circ\)).

Step 6: Calculate \(|\alpha - \beta|\)

Calculate \(|\alpha - \beta| = |90^\circ - 60^\circ| = 30^\circ\).

Key Concepts

Vector Triple Product

The vector triple product identity is a vital concept in vector calculus, especially when dealing with cross products. The identity states:

• \( \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \).

This allows us to express the cross product of two vectors in terms of dot products and scalar multiples of vectors. In essence, when you take the cross product of

• a vector \( \vec{a} \) with the cross product of vectors \( \vec{b} \) and \( \vec{c} \), you can break it down into more manageable parts.
• This breakdown involves projecting \( \vec{a} \) onto the vectors \( \vec{b} \) and \( \vec{c} \), justified by the dot product results.

Understanding this identity is crucial because it provides a straightforward way to solve complex vector relations by simplifying them into smaller, easier-to-handle components.

Unit Vectors

Unit vectors are fundamental components in vector algebra. They are vectors with a magnitude of 1 and are primarily used to denote direction.

• Each unit vector is typically represented with a hat symbol, such as \( \hat{i} \), \( \hat{j} \), or \( \hat{k} \) for the standard Cartesian coordinate axes.
• Unit vectors are essential in expressing other vectors. For example, any vector \( \vec{v} \) can be expressed as a scalar multiple of a unit vector.
• In the problem at hand, vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are unit vectors, implying that each of them has a magnitude of 1.

This simplifies calculations and analyses since working with a magnitude value of 1 minimizes the complexity when determining directions and drawing projections on vectors.

Angle Calculation

Calculating angles between vectors is a critical aspect of vector mathematics. The dot product comes handy for this calculation.

• The dot product relationship: \( \vec{a} \cdot \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot \cos(\theta) \), provides a fundamental way to relate angles with vector magnitudes.
• For unit vectors \( \vec{b} \) and \( \vec{c} \), the quantities \( |\vec{b}| \) and \( |\vec{c}| \) are 1, so the dot product further simplifies to just \( \cos(\theta) \).

In this exercise, we found that the angle \( \alpha \) between \( \vec{a} \) and \( \vec{b} \) is 90 degrees because \( \cos \alpha = 0 \). Similarly, \( \cos \beta = \frac{1}{2} \) indicates the angle \( \beta \) is 60 degrees. Understanding these calculations allows one to interpret the relative orientations of the vectors effectively.