Problem 108 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three non-zero vectors such that no two of them are collinear and \((\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}\). If \(\theta\) is the angle between vectors \(\vec{b}\) and \(\vec{c}\), then a value of \(\sin \theta\) (A) \(\frac{-\sqrt{2}}{3}\) (B) \(\frac{2}{3}\) (C) \(\frac{-2 \sqrt{3}}{3}\) (D) \(\frac{2 \sqrt{2}}{3}\)

Step-by-Step Solution

Answer

The value of \(\sin \theta\) is \(\frac{2\sqrt{2}}{3}\) (Option D).

Step 1: Understanding the Cross Product Identity

We know that the identity for the cross product of two cross products states:\[(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}\] We can use this identity to solve the given vector equation.

Step 2: Substitute and Rearrange

Given: \[(\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a}\] Using the identity, this becomes:\[(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a}\]We can rearrange it as:\[(\vec{a} \cdot \vec{c})\vec{b} = (\vec{b} \cdot \vec{c} + \frac{1}{3} |\vec{b}| |\vec{c}|) \vec{a}\]

Step 3: Determine Dot Product Relations

Since \( \vec{b} \) and \( \vec{a} \) are not collinear, then for this vector equality to hold the vector coefficients must equate separately:\[\vec{a} \cdot \vec{c} = 0\]which implies \(\vec{c}\) is perpendicular to \(\vec{a}\). Thus:\((\vec{b} \cdot \vec{c}) = -\frac{1}{3} |\vec{b}| |\vec{c}|\).

Step 4: Solve for \(\sin \theta\)

Express \(\vec{b} \cdot \vec{c}\) in terms of the magnitude and angle:\(\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta\).Substituting the expression we found:\[|\vec{b}| |\vec{c}| \cos \theta = -\frac{1}{3} |\vec{b}| |\vec{c}|\]Canceling the magnitudes (assuming non-zero vectors):\[\cos \theta = -\frac{1}{3}\]Since \(\sin^2 \theta + \cos^2 \theta = 1\), solve for \(\sin \theta\):\[\sin^2 \theta = 1 - \left(-\frac{1}{3}\right)^2 = \frac{8}{9}\] \[\sin \theta = \pm \frac{2\sqrt{2}}{3}\].

Step 5: Choose the Correct Option

Given the possible values of \(\sin \theta\) from the choices, we select the positive value \(\frac{2\sqrt{2}}{3}\) as the answer. The negative option \(-\frac{2\sqrt{2}}{3}\) is not listed among the given choices.

Key Concepts

Cross Product

The cross product is a central operation in vector calculus, particularly when dealing with three-dimensional vectors. It generates a new vector that is perpendicular to the plane formed by the original two vectors. If you have two vectors, \( \vec{a} \) and \( \vec{b} \), the cross product, noted as \( \vec{a} \times \vec{b} \), results in a vector defined by the formula:

• \( \vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \ \hat{n}\)

Here, \( |\vec{a}| \) and \( |\vec{b}| \) are the magnitudes of \( \vec{a} \) and \( \vec{b} \) respectively, \( \theta \) is the angle between them, and \( \hat{n} \) is the unit vector perpendicular to the plane.
In the exercise, we also use the triple product identity, which helps in solving expressions involving multiple cross products:

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

This identity facilitates the rewriting of complex cross product equations into a form where scalar dot products and other vector terms appear. Understanding this simplification is crucial for manipulating and solving vector equations effectively.

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors to produce a scalar. This is particularly useful in determining angles between vectors and in projection calculations. Given two vectors \( \vec{a} \) and \( \vec{b} \), the dot product is defined as:

• \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \)

where \( |\vec{a}| \) and \( |\vec{b}| \) are the magnitudes of \( \vec{a} \) and \( \vec{b} \), and \( \theta \) is the angle between the vectors.
In the step-by-step solution provided, the dot product plays a crucial role in finding \( \cos \theta \), which then allows us to determine \( \sin \theta \). Rearranging the equation gives:

• \( \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta = -\frac{1}{3} |\vec{b}| |\vec{c}|\)

This step is vital for connecting the magnitude of the cross product with the angle \( \theta \), ultimately aiding in solving the whole vector equation.

Trigonometry in Vectors

Trigonometry is an indispensable tool in vector mathematics, providing insight into angles and relationships between vectors through sine and cosine functions. In vectors, trigonometric functions help us break down vector components and relate them to their angles.

• Understanding \( \sin \theta \) and \( \cos \theta \) is crucial because it ties into vector magnitudes and directions.
• For instance, the Pythagorean identity helps relate \( \sin \theta \) and \( \cos \theta \), as shown in the solution:\[sin^2 \theta + \cos^2 \theta = 1\]

By using the above identity, when \( \cos \theta \) is known, solving for \( \sin \theta \) becomes straightforward:

• \( \sin^2 \theta = 1 - \left(-\frac{1}{3}\right)^2\)
• Finally, \( \sin \theta = \pm \frac{2\sqrt{2}}{3}\)

This exercise demonstrates how trigonometric principles are applied in vector calculus to find specific angle measurements and how these calculations are crucial for solving vector-related problems. Such principles are enormously beneficial in fields like physics and engineering, ensuring precise and accurate vector analysis.