L 02 Find {(2a + B) X (3ã - 2b)} Solution With Vector Algebra

In this comprehensive article, we delve into the intricate problem of finding the value of the expression {(2a + b) x (3ã - 2b)} given the conditions (a x b)² + (a.b)² = 9 and that a is a unit vector with the angle between a and b being π/4. This problem seamlessly blends vector algebra concepts, including cross products, dot products, and the properties of unit vectors, making it a valuable exercise for students and enthusiasts alike. We will meticulously break down each step, providing clear explanations and insightful observations along the way. This exploration will not only reveal the solution but also enhance your understanding of vector algebra principles.

Problem Statement

Given that (a x b)² + (a.b)² = 9, a is a unit vector, and the angle between a and b is π/4, find the value of {(2a + b) x (3ã - 2b)}.

Solution Approach

To solve this problem, we will follow a structured approach:

1. Expand the Expression: We begin by expanding the expression {(2a + b) x (3ã - 2b)} using the distributive property of the cross product.
2. Utilize Vector Identities: We will leverage key vector identities, such as a x a = 0, a x b = -b x a, and the magnitude relationship |a x b| = |a||b|sin(θ), where θ is the angle between a and b.
3. Apply Given Conditions: We will incorporate the given conditions, including (a x b)² + (a.b)² = 9, that a is a unit vector (|a| = 1), and the angle between a and b is π/4.
4. Simplify and Solve: Through careful simplification and substitution, we will arrive at the final solution.

Step-by-Step Solution

1. Expanding the Expression

We start by expanding the expression {(2a + b) x (3ã - 2b)} using the distributive property of the cross product:

(2a + b) x (3ã - 2b) = (2a x 3ã) + (2a x -2b) + (b x 3ã) + (b x -2b)

2. Applying Vector Identities

Now, we apply the following vector identities:

• a x a = 0: The cross product of a vector with itself is zero.
• a x b = -b x a: The cross product is anti-commutative.
• |a x b| = |a||b|sin(θ): The magnitude of the cross product is the product of the magnitudes and the sine of the angle between the vectors.

Using these identities, we can simplify the expanded expression:

(2a x 3ã) + (2a x -2b) + (b x 3ã) + (b x -2b) = 6(a x a) - 4(a x b) + 3(b x a) - 2(b x b)

Since a x a = 0 and b x b = 0, and using b x a = - (a x b), we get:

- 4(a x b) + 3(b x a) = -4(a x b) - 3(a x b) = -7(a x b)

3. Utilizing Given Conditions

We are given that (a x b)² + (a.b)² = 9. We also know that |a| = 1 (since a is a unit vector) and the angle between a and b is π/4.

We can express the magnitudes of the cross product and dot product as follows:

• |a x b| = |a||b|sin(π/4) = 1 * |b| * (1/√2) = |b|/√2
• a.b = |a||b|cos(π/4) = 1 * |b| * (1/√2) = |b|/√2

Now, let's rewrite the given equation (a x b)² + (a.b)² = 9 in terms of magnitudes:

(|a x b|)² + (a.b)² = (|b|/√2)² + (|b|/√2)² = |b|²/2 + |b|²/2 = |b|² = 9

Therefore, |b|² = 9, which implies |b| = 3.

4. Finding the Magnitude of (a x b)

Now we know |b| = 3, so we can find the magnitude of a x b:

|a x b| = |b|/√2 = 3/√2

5. Determining the Final Result

We found that (2a + b) x (3ã - 2b) = -7(a x b). Now we need to find the magnitude of this vector:

|(2a + b) x (3ã - 2b)| = |-7(a x b)| = 7|a x b| = 7 * (3/√2) = 21/√2

To rationalize the denominator, we multiply the numerator and denominator by √2:

(21/√2) * (√2/√2) = (21√2)/2

So, the magnitude of the vector (2a + b) x (3ã - 2b) is (21√2)/2.

Conclusion

In this detailed exploration, we successfully determined the value of {(2a + b) x (3ã - 2b)} given the conditions (a x b)² + (a.b)² = 9, a is a unit vector, and the angle between a and b is π/4. The solution involved expanding the expression, applying vector identities, utilizing the given conditions, and simplifying to arrive at the final answer of (21√2)/2. This problem highlights the importance of a strong foundation in vector algebra principles and the ability to apply them in problem-solving scenarios. Understanding the properties of cross products, dot products, and unit vectors is crucial for tackling such challenges. By systematically breaking down the problem and applying the relevant concepts, we were able to navigate the complexities and arrive at a clear and concise solution.

Key Takeaways

• Vector Identities: Mastering vector identities, such as a x a = 0 and a x b = -b x a, is essential for simplifying vector expressions.
• Magnitude Relationships: Understanding the relationships between the magnitudes of cross products and dot products, such as |a x b| = |a||b|sin(θ) and a.b = |a||b|cos(θ), is crucial for solving problems involving angles between vectors.
• Unit Vectors: Recognizing that the magnitude of a unit vector is 1 (|a| = 1) simplifies calculations and allows for easier manipulation of vector equations.
• Systematic Approach: A structured approach, involving expanding expressions, applying identities, utilizing given conditions, and simplifying, is key to solving complex vector algebra problems.

By internalizing these key takeaways, you can enhance your problem-solving skills in vector algebra and tackle similar challenges with confidence. This exploration not only provides a solution to the specific problem but also equips you with the tools and knowledge to excel in related mathematical domains.

Further Exploration

To further enhance your understanding of vector algebra, consider exploring the following topics:

• Applications of Cross and Dot Products: Investigate the applications of cross and dot products in physics, engineering, and computer graphics.
• Vector Equations of Lines and Planes: Learn how to represent lines and planes using vector equations.
• Linear Transformations: Explore the concept of linear transformations and their representation using matrices.
• Eigenvalues and Eigenvectors: Delve into the world of eigenvalues and eigenvectors and their applications in various fields.

By expanding your knowledge in these areas, you can gain a deeper appreciation for the power and versatility of vector algebra in solving real-world problems. Remember, mathematics is a journey of continuous learning and exploration, and the more you delve into its depths, the more rewarding the experience becomes. So, keep practicing, keep exploring, and keep pushing the boundaries of your mathematical understanding.