/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 \(\mathbf{X}\) is an unknown vec... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\mathbf{X}\) is an unknown vector satisfying the following relations involving the known vectors \(\mathbf{A}\) and \(\mathbf{B}\) and the scalar \(\phi\) $$\mathbf{A} \times \mathbf{X}=\mathbf{B}, \quad \mathbf{A} \cdot \mathbf{X}=\phi$$ Express \(\mathbf{X}\) in terms of \(\mathbf{A}, \mathbf{B}, \phi,\) and the magnitude of \(\mathbf{A}\)

Short Answer

Expert verified
The expression for the vector \(\mathbf{X}\) in terms of the known vectors \(\mathbf{A}\) and \(\mathbf{B}\), scalar \(\phi\), and the magnitude of \(\mathbf{A}\) is: \(\mathbf{X} = \frac{\mathbf{B}}{\|\mathbf{A}\|^2}\mathbf{A} + \frac{\phi \|\mathbf{A}\|^2 - \mathbf{A} \cdot \mathbf{B}}{\|\mathbf{A}\|^4} (\mathbf{A} \times (\mathbf{A} \times \mathbf{B}))\)

Step by step solution

01

Write down the given equations

We are given these equations: \[ \begin{cases} \mathbf{A} \times \mathbf{X} = \mathbf{B}\\ \mathbf{A} \cdot \mathbf{X} = \phi \end{cases} \]
02

Take dot product of the first equation with \(\mathbf{A}\)

Taking the dot product of both sides of the first equation with \(\mathbf{A}\), we get: \(\mathbf{A} \cdot (\mathbf{A} \times \mathbf{X}) = \mathbf{A} \cdot \mathbf{B}\)
03

Use the vector triple product identity

We apply the vector triple product identity on the left side of the equation. The vector triple product identity states: \(\mathbf{A} \cdot (\mathbf{B} \times \mathbf{X}) = \mathbf{B} \cdot (\mathbf{A} \times \mathbf{X})\). Hence, we obtain: \(\mathbf{X} \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\mathbf{A} \times \mathbf{X})\) Now, we substitute the given equation \(\mathbf{A} \times \mathbf{X} = \mathbf{B}\) into the vector triple product identity, and we get: \(\mathbf{X} \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot \mathbf{B}\)
04

Rewrite \(\mathbf{X}\) using the given equations

We express \(\mathbf{X}\) in terms of the given equations as follows: \(\mathbf{X} = \frac{\mathbf{B}}{\|\mathbf{A}\|^2}\mathbf{A} + \frac{\phi \|\mathbf{A}\|^2 - \mathbf{A} \cdot \mathbf{B}}{\|\mathbf{A}\|^4} (\mathbf{A} \times (\mathbf{A} \times \mathbf{B}))\) Here, \(\|\mathbf{A}\|\) represents the magnitude of vector \(\mathbf{A}\).
05

Final expression for \(\mathbf{X}\)

The expression for the vector \(\mathbf{X}\) in terms of the known vectors \(\mathbf{A}\) and \(\mathbf{B}\), scalar \(\phi\), and the magnitude of \(\mathbf{A}\) is as follows: \(\mathbf{X} = \frac{\mathbf{B}}{\|\mathbf{A}\|^2}\mathbf{A} + \frac{\phi \|\mathbf{A}\|^2 - \mathbf{A} \cdot \mathbf{B}}{\|\mathbf{A}\|^4} (\mathbf{A} \times (\mathbf{A} \times \mathbf{B}))\)

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Key Concepts

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

Cross Product
The cross product, denoted as \( \mathbf{A} \times \mathbf{X} \), is an operation that finds a vector perpendicular to both \( \mathbf{A} \) and \( \mathbf{X} \). It produces a vector, rather than a scalar, and its direction is given by the right-hand rule. If you point your index finger in the direction of vector \( \mathbf{A} \) and your middle finger in the direction of vector \( \mathbf{X} \), your thumb will point in the direction of the cross product. The magnitude of this vector is given by \( \|\mathbf{A} \| \cdot \| \mathbf{X} \| \cdot \sin \theta \), where \( \theta \) is the angle between \( \mathbf{A} \) and \( \mathbf{X} \). The cross product is particularly useful in physics and engineering where it describes rotational effects and is used in calculations involving torque, rotational motion, and electromagnetic forces.
  • It is essential in determining the perpendicularity of directions.
  • Non-zero cross products mean the vectors are not parallel.
  • Cross product results in a vector orthogonal to the original two vectors.
Dot Product
The dot product, noted as \( \mathbf{A} \cdot \mathbf{X} \), results in a scalar. It measures the extent to which two vectors point in the same direction. To compute it, multiply the components of the vectors and sum the results: \[ \mathbf{A} \cdot \mathbf{X} = A_1 X_1 + A_2 X_2 + A_3 X_3 \]This operation is commutative, meaning \( \mathbf{A} \cdot \mathbf{X} = \mathbf{X} \cdot \mathbf{A} \). The geometrical interpretation is that it relates to the cosine of the angle between \( \mathbf{A} \) and \( \mathbf{X} \). Specifically, \( \mathbf{A} \cdot \mathbf{X} = \| \mathbf{A} \| \cdot \| \mathbf{X} \| \cdot \cos \theta \), where \( \theta \) is the angle between them. This concept is widely used in various applications:
  • It tells how much one vector projects onto another.
  • It is used in determining angles between vectors in space.
  • A zero dot product means the vectors are orthogonal.
Vector Identity
A vector identity is a mathematical equation that holds for all vectors under particular operations. In the context of the given exercise, the vector triple product identity is used: \[ \mathbf{A} \times (\mathbf{B} \times \mathbf{X}) = (\mathbf{A} \cdot \mathbf{X}) \mathbf{B} - (\mathbf{A} \cdot \mathbf{B}) \mathbf{X} \] Using identities like this helps simplify complex vector expressions, transform them, or find unknowns. The vector triple product, specifically, is highly useful. It allows us to rewrite a vector as a linear combination of other vectors. In the original solution, it helps express \( \mathbf{X} \) in terms of \( \mathbf{A}, \mathbf{B}, \) and \( \phi \). This identity can be powerful:
  • It allows manipulating terms in vector equations.
  • It can express a vector in terms of other known vectors.
  • It simplifies the calculation in physics and engineering problems.

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