Chapter 6: Problem 1

If \(\mathbf{A}\) and \(\mathbf{B}\) are unit vectors with an angle \(\theta\) between them, and \(\mathbf{C}\) is a unit vector perpendicular to both \(\mathbf{A}\) and \(\mathbf{B},\) evaluate \([(\mathbf{A} \times \mathbf{B}) \times(\mathbf{B} \times \mathbf{C})] \times(\mathbf{C} \times \mathbf{A})\).

Short Answer

Expert verified

(\textbf{A} \bullet \textbf{B}) \textbf{A}.

Step by step solution

Understand the Vectors

Given \(\textbf{A}\), \(\textbf{B}\), and \(\textbf{C}\) are unit vectors. The angle between \(\textbf{A}\) and \(\textbf{B}\) is \(\theta\). Additionally, \(\textbf{C}\) is perpendicular to both \(\textbf{A}\) and \(\textbf{B}\).

Apply Vector Identities

Utilize the vector triple product identity: \[ \textbf{X} \times (\textbf{Y} \times \textbf{Z}) = (\textbf{X} \bullet \textbf{Z}) \textbf{Y} - (\textbf{X} \bullet \textbf{Y}) \textbf{Z}. \] Apply this identity on \[ \textbf{A} \times \textbf{B} \times (\textbf{B} \times \textbf{C}), \] and note that \(\textbf{B} \times \textbf{C} \) is a vector in the direction of \(\textbf{A}\) since \(\textbf{C}\) is perpendicular to \(\textbf{A}\) and \(\textbf{B}\).

Simplify Intermediate Expression

Since \(\textbf{A}\) and \(\textbf{C}\) are perpendicular, \( \textbf{A} \bullet \textbf{C} = 0 \). Thus, applying the identity: \[ (\textbf{A} \times \textbf{B}) \times (\textbf{B} \times \textbf{C}) = (\textbf{A} \bullet \textbf{C}) \textbf{B} - (\textbf{A} \bullet \textbf{B}) \textbf{C}. \] Given \( \textbf{A} \bullet \textbf{C} = 0 \), the intermediate result reduces to: \[ - (\textbf{A} \bullet \textbf{B}) \textbf{C}. \]

Evaluate Final Cross Product

Now simplify \( \text{Final cross product} \) as: \[ [ - (\textbf{A} \bullet \textbf{B}) \textbf{C}] \times (\textbf{C} \times \textbf{A}). \] Use the identity again: \[ \textbf{C} \times (\textbf{C} \times \textbf{A}) = \textbf{C} (\textbf{C} \bullet \textbf{A}) - \textbf{A} (\textbf{C} \bullet \textbf{C}). \] Since \( \textbf{C} \bullet \textbf{A} = 0 \) and \( \textbf{C} \bullet \textbf{C} = 1, \) it reduces to: \[ \textbf{C} \times (\textbf{C} \times \textbf{A}) = - \textbf{A}. \]

Combine Simplifications

Given that \(- (\textbf{A} \bullet \textbf{B}) \textbf{C} \) is a scalar times a vector, combine this with the final cross product result of \(- \textbf{A}\), giving: \[ - (\textbf{A} \bullet \textbf{B}) (\textbf{C} \times \textbf{A}) = (\textbf{A} \bullet \textbf{B}) \textbf{A}. \]

Evaluate the Final Expression

Finally, considering \( \textbf{A} \bullet \textbf{B} \) as a scalar, the final answer simplifies to: \[ (\textbf{A} \bullet \textbf{B}) \textbf{A}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Vectors

A unit vector is a vector that has a magnitude of exactly 1. They are used to specify directions without specifying magnitude. When dealing with unit vectors in vector calculus, it's essential to note:

The notation commonly used for unit vectors is with a hat symbol, for example, \( \textbf{A} \hat{} \).
They maintain their direction, while their magnitude remains constant at 1.
They are fundamental in simplifying and understanding vector operations like the dot product and cross product.

In this exercise, vectors \( \textbf{A} \), \( \textbf{B} \), and \( \textbf{C} \) are all unit vectors. Remember, when you take the dot product of any unit vector with itself, you get 1, and for perpendicular unit vectors, the dot product is 0.

Cross Product

The cross product of two vectors results in a vector that is perpendicular to both original vectors. This vector’s direction is determined by the right-hand rule. The magnitude of the cross product vector is equal to the area of the parallelogram formed by the original vectors.

Key points to remember about cross products:

The cross product operation is denoted by \( \textbf{A} \times \textbf{B} \).
The result of the cross product is a vector.
The formula for the cross product in three dimensions is given by: \( \textbf{A} \times \textbf{B} = | \textbf{A} || \textbf{B} | \text{sin} \theta \textbf{n} \), where \( \textbf{n} \) is a unit vector perpendicular to both \( \textbf{A} \) and \( \textbf{B} \).

In the given exercise, applying the cross product to vectors \( \textbf{A} \) and \( \textbf{B} \), and analyzing their respective orthogonal unit vector \( \textbf{C} \), helps in determining the complex expression using vector identities.

Vector Identities

Vector identities are mathematical tools that simplify the manipulation of vector equations. One such identity used in the given exercise is the vector triple product identity:

The identity for the vector triple product is: \( \textbf{X} \times (\textbf{Y} \times \textbf{Z}) = (\textbf{X} \bullet \textbf{Z}) \textbf{Y} - (\textbf{X} \bullet \textbf{Y}) \textbf{Z} \).
It helps to reduce complex cross product operations by transforming them into simpler dot product terms.
The triple product identity is used twice in the exercise to simplify intermediate expressions.

In the context of the problem, understanding and correctly applying these identities is crucial for reaching the solution efficiently.

Dot Product

The dot product of two vectors results in a scalar. It is a measure of how much one vector extends in the direction of another.

Essential aspects of the dot product:

The dot product is denoted by \( \textbf{A} \bullet \textbf{B} \).
It can be calculated using the formula: \( \textbf{A} \bullet \textbf{B} = | \textbf{A} || \textbf{B} | \text{cos} \theta \), where \( \theta \) is the angle between vectors \( \textbf{A} \) and \( \textbf{B} \).
The result of the dot product is a scalar value, not a vector.

In the given exercise, the dot products \( \textbf{A} \bullet \textbf{C} = 0 \), because vectors \( \textbf{A} \) and \( \textbf{C} \) are perpendicular. Understanding the properties and outcomes of the dot product is key to simplifying the given vector expression.