Problem 94 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

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

Step-by-Step Solution

Answer

\(\sin \theta = \frac{2 \sqrt{2}}{3}\) is the answer, which corresponds to option (a).

Step 1: Write Vector Triple Product Formula

The vector triple product identity states that for three vectors \(\bar{x}, \bar{y}, \bar{z}\), we have \((\bar{x} \times \bar{y}) \times \bar{z} = (\bar{z} \cdot \bar{y})\bar{x} - (\bar{z} \cdot \bar{x})\bar{y}\). We shall use this identity to simplify the given problem.

Step 2: Apply Triple Product Identity

Given \((\bar{a} \times \bar{b}) \times \bar{c} = \frac{1}{3} |\bar{b}| |\bar{c}| \bar{a}\), use the identity from Step 1: \(((\bar{a} \times \bar{b}) \times \bar{c}) = (\bar{c} \cdot \bar{b})\bar{a} - (\bar{c} \cdot \bar{a})\bar{b}\). Since this is equal to \(\frac{1}{3} |\bar{b}| |\bar{c}| \bar{a}\), it implies that \((\bar{c} \cdot \bar{b})\bar{a} - (\bar{c} \cdot \bar{a})\bar{b} = \frac{1}{3} |\bar{b}| |\bar{c}| \bar{a}\).

Step 3: Compare Coefficients of \(\bar{a}\)

Equating the coefficients of \(\bar{a}\) from both sides, we have \(\bar{c} \cdot \bar{b} = \frac{1}{3} |\bar{b}| |\bar{c}|\).

Step 4: Use Dot Product Formula

The dot product \(\bar{c} \cdot \bar{b} = |\bar{c}| |\bar{b}| \cos \theta\). Thus, we have \(|\bar{c}| |\bar{b}| \cos \theta = \frac{1}{3} |\bar{b}| |\bar{c}|\).

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

Cancel out \(|\bar{b}| |\bar{c}|\) from both sides to get \(\cos \theta = \frac{1}{3}\). Now that we know \(\cos\theta = \frac{1}{3}\), use the sine identity \(\sin^2 \theta + \cos^2 \theta = 1\) to find \(\sin \theta\).

Step 6: Calculate \(\sin \theta\)

Using \(\sin^2 \theta + \cos^2 \theta = 1\), substitute \(\cos^2 \theta = (\frac{1}{3})^2 = \frac{1}{9}\). Then \(\sin^2 \theta = 1 - \frac{1}{9} = \frac{8}{9}\). Thus, \(\sin \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\).

Step 7: Conclusion

Therefore, the value of \(\sin \theta\) is \(\frac{2\sqrt{2}}{3}\). This corresponds to option (a).

Key Concepts

Dot Product

Vectors can be multiplied in a variety of ways, and one of these methods is the dot product. The dot product of two vectors \(\bar{u}\) and \(\bar{v}\) is a scalar, and is calculated as \(\bar{u} \cdot \bar{v} = |\bar{u}| |\bar{v}| \cos \theta\), where \(|\bar{u}|\) and \(|\bar{v}|\) are the magnitudes of the vectors, and \(\theta\) is the angle between them.

The dot product is vital because it gives valuable insights into the directional relationship between the vectors. When the dot product is zero, the vectors are orthogonal, meaning they are at a right angle. In our problem, recognizing the role of the dot product allowed us to equate \(\cos \theta\) with a given value derived from coefficients, specifically, \(\cos \theta = \frac{1}{3}\). This step was instrumental in solving for \(\sin \theta\) later on.

Sine and Cosine Identities

Trigonometric identities often play a crucial role in vector problems. The fundamental Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) was utilized in this exercise to solve for \(\sin \theta\) once \(\cos \theta\) was known.

Here, with \(\cos \theta = \frac{1}{3}\), you can square this value to get \(\cos^2 \theta = \frac{1}{9}\). With this squared value, apply the identity to get:

• \(\sin^2 \theta = 1 - \frac{1}{9} = \frac{8}{9}\)

From this, it follows that \(\sin \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\).

These identities simplify the complexities of trigonometric relationships, allowing us to easily transition from cosine to sine and ensure a more straightforward calculation.

Cross Product

The cross product creates a new vector that is perpendicular to the plane formed by the original two vectors. The magnitude of the cross product \(\bar{u} \times \bar{v}\) is \(|\bar{u}| |\bar{v}| \sin \theta\). This magnitude tells us how much \(\bar{u}\) and \(\bar{v}\) are moving in separate directions, making it ideal for calculating areas and determining scalar triple products.

In our initial exercise, the expression \((\bar{a} \times \bar{b}) \times \bar{c}=\frac{1}{3}|\bar{b}| |\bar{c}| \bar{a}\) involves cross products. The vector triple product identity, \((\bar{x} \times \bar{y}) \times \bar{z} = (\bar{z} \cdot \bar{y})\bar{x} - (\bar{z} \cdot \bar{x})\bar{y}\), helped us simplify this triple cross product sequence, which returns the problem back to dot products.

Understanding this identity is crucial as it peels back layers of vector operations, revealing foundational scalar relationships underpinning the bulk calculations involving dot and cross products. In solving the problem, pulling the relationship back to the simple dot product allowed for direct computation of \(\theta\).