Problem 84 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(\bar{a}, \bar{b}\) and \(\bar{c}\) be non-zero vectors such that \((\bar{a} \times \bar{b}) \times \bar{c}=-|\bar{b}| \bar{c} \mid \bar{a} . \quad\) If \(\theta\) is the acute angle between the vectors \(\bar{b}\) and \(\bar{c}\) then \(\sin \theta\) equals [2004] (A) \(\frac{1}{3}\) (B) \(\frac{\sqrt{2}}{3}\) (C) \(\frac{2}{3}\) (D) \(\frac{2 \sqrt{2}}{3}\)

Step-by-Step Solution

Answer

The value of \(\sin \theta\) is (A) \(\frac{1}{3}\).

Step 1: Understand the Triple Cross Product

The identity for triple cross product is \((\bar{a} \times \bar{b}) \times \bar{c} = (\bar{a} \cdot \bar{c})\bar{b} - (\bar{b} \cdot \bar{c})\bar{a}\). We need to apply this identity to simplify the given expression \((\bar{a} \times \bar{b}) \times \bar{c} = -|\bar{b}| \bar{c} |\bar{a}|\).

Step 2: Apply the Identity

Applying the identity, we have \((\bar{a} \cdot \bar{c})\bar{b} - (\bar{b} \cdot \bar{c})\bar{a} = -|\bar{b}| \bar{c} |\bar{a}|\). This means \((\bar{a} \cdot \bar{c})\bar{b} - (\bar{b} \cdot \bar{c})\bar{a} = -|\bar{b}||\bar{c}||\bar{a}| .\cos\theta\), aligned with \(\bar{c}\).

Step 3: Analyze the Resulting Expression

Since \((\bar{a} \cdot \bar{c})\bar{b} - (\bar{b} \cdot \bar{c})\bar{a}\) must balance with \(-|\bar{b}|\bar{c}|\bar{a}|\), it indicates that the vectors \((\bar{a} \cdot \bar{c})\bar{b}\) and \( - (\bar{b} \cdot \bar{c})\bar{a}\) become zero vectors. Simplifying, we find the components of these vectors are parallel to \(\bar{c}\), thus \((\bar{b} \cdot \bar{c}) = |\bar{b}||\bar{c}|\cos\theta\).

Step 4: Relate Dot and Cross Product

Using \((\bar{b} \cdot \bar{c}) = |\bar{b}||\bar{c}|\cos\theta\) and \( |\bar{b}|\bar{c}|\bar{a}| = |\bar{b}|^2|\bar{c}| |\bar{a}|\sin\theta\), these simplify and provide a solution when equating coefficients. Solving the resultant trigonometric relationship gives us \(\sin\theta\).

Step 5: Solve for \(\sin\theta\)

For this particular setup, we equate magnitudes of both sides and solve the resultant equation that equates the sine side. Ultimately, solving yields \(\sin\theta = \frac{1}{3}\).

Key Concepts

Triple Cross Product

The triple cross product is a significant concept in vector algebra. It involves the cross product of three vectors, represented as \((\bar{a} \times \bar{b}) \times \bar{c}\). The triple cross product can be computed using the vector identity:\[ (\bar{a} \times \bar{b}) \times \bar{c} = (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{b} \cdot \bar{c}) \bar{a} \]This identity essentially breaks down the triple cross product into a combination of dot products and scalar multiplications. By using this identity, complex vector expressions can be simplified.

• It transforms a potentially complicated vector product into terms that often allow easy evaluation using known quantities.
• The expression \((\bar{a} \cdot \bar{c}) \bar{b} - (\bar{b} \cdot \bar{c}) \bar{a}\) is frequently easier to interpret and calculate.

Understanding how to apply this identity can make solving vector algebra problems more manageable and offer deeper insights into vector interactions.

Dot Product

The dot product is a fundamental operation in vector algebra involving two vectors, which results in a scalar. Given two vectors \( \bar{a} \) and \( \bar{b} \), their dot product is calculated as:\[ \bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos \theta \]Here, \(|\bar{a}|\) and \(|\bar{b}|\) are the magnitudes of the vectors, and \(\theta\) is the angle between them.

• The dot product provides a measure of how much one vector goes in the direction of another vector.
• It is particularly useful for projecting vectors onto another and determining parallelism.
• If the dot product is zero, the vectors are perpendicular.

This arithmetic operation helps simplify vector expressions and directly influences both geometric interpretations and problem-solving strategies in linear algebra.

Magnitude of Vectors

The magnitude of a vector, often termed its length or norm, provides a scalar value representing the size of the vector. For any vector \(\bar{v}= (v_1, v_2, v_3)\), its magnitude is calculated as:\[ |\bar{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]The magnitude reflects the Euclidean distance of the point represented by the vector from the origin.

• Knowing the magnitude is essential for tasks like normalizing vectors, which requires dividing the vector by its magnitude to get a unit vector.
• It also plays a crucial role in the computation of dot products and cross products.
• The magnitude is always a non-negative quantity.

When dealing with vectors, understanding how to compute and apply magnitudes can significantly aid in the analysis and resolution of vector-related problems.