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(II) What is the angle \(\theta\) between two vectors \(\vec{\mathbf{A}}\) and \(\vec{\mathbf{B}}\) , if \(|\vec{\mathbf{A}} \times \vec{\mathbf{B}}|=\vec{\mathbf{A}} \cdot \vec{\mathbf{B}} ?\)

Short Answer

Expert verified
\(\theta = 45^\circ\) or \(\frac{\pi}{4}\) radians.

Step by step solution

01

Understanding the Problem

We are given two vectors, \(\vec{\mathbf{A}}\) and \(\vec{\mathbf{B}}\), and need to find the angle \(\theta\) between them given \(|\vec{\mathbf{A}} \times \vec{\mathbf{B}}| = \vec{\mathbf{A}} \cdot \vec{\mathbf{B}}\). This condition involves both the cross product and the dot product of the vectors.
02

Using the Cross Product Formula

The magnitude of the cross product \(|\vec{\mathbf{A}} \times \vec{\mathbf{B}}|\) is given by the formula \(|\vec{\mathbf{A}}||\vec{\mathbf{B}}|\sin\theta\).
03

Using the Dot Product Formula

The dot product \(\vec{\mathbf{A}} \cdot \vec{\mathbf{B}}\) is given by the formula \(|\vec{\mathbf{A}}||\vec{\mathbf{B}}|\cos\theta\).
04

Equating and Simplifying the Problem

Since \(|\vec{\mathbf{A}} \times \vec{\mathbf{B}}| = \vec{\mathbf{A}} \cdot \vec{\mathbf{B}}\), we substitute the formulas: \(|\vec{\mathbf{A}}||\vec{\mathbf{B}}|\sin\theta = |\vec{\mathbf{A}}||\vec{\mathbf{B}}|\cos\theta\).
05

Canceling and Solving for \(\theta\)

Cancel \(|\vec{\mathbf{A}}||\vec{\mathbf{B}}|\) from both sides (assuming neither is zero), resulting in \(\sin\theta = \cos\theta\). Dividing both sides by \(\cos\theta\) gives \(\tan\theta = 1\).
06

Finding the Angle

The equation \(\tan\theta = 1\) implies that \(\theta\) is \(45^\circ\) or \(\frac{\pi}{4}\) radians, since tangent has a period of \(\pi\) and \(\tan 45^\circ = 1\).

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Key Concepts

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

Dot Product
The dot product, also known as the scalar product, is a way to multiply two vectors to result in a scalar or a single number. It's one of the fundamentals of vector mathematics.When calculating the dot product of two vectors, say \( \vec{\mathbf{A}} \) and \( \vec{\mathbf{B}} \), you follow the formula:\[ \vec{\mathbf{A}} \cdot \vec{\mathbf{B}} = |\vec{\mathbf{A}}||\vec{\mathbf{B}}|\cos\theta \]Here, \(|\vec{\mathbf{A}}|\) and \(|\vec{\mathbf{B}}|\) are the magnitudes (or lengths) of vectors \(\vec{\mathbf{A}}\) and \(\vec{\mathbf{B}}\), while \(\theta\) is the angle between them.
The resulting dot product value can tell us several things:
  • If the dot product is positive, the angle between the vectors is acute (less than 90°).
  • If it's zero, the vectors are perpendicular.
  • If negative, the angle is obtuse (greater than 90°).
Understanding how to use and interpret the dot product can provide insights into the direction and relationship between two vectors.
Cross Product
The cross product, also known as the vector product, differs from the dot product as it results in a vector rather than a scalar. It's calculated using two vectors and results in a third vector that is perpendicular to the plane of the first two vectors.The formula for the magnitude of the cross product of vectors \( \vec{\mathbf{A}} \) and \( \vec{\mathbf{B}} \) is:\[ |\vec{\mathbf{A}} \times \vec{\mathbf{B}}| = |\vec{\mathbf{A}}||\vec{\mathbf{B}}|\sin\theta \]Here, \(|\vec{\mathbf{A}}|\) and \(|\vec{\mathbf{B}}|\) are still the magnitudes of the vectors, while \(\theta\) is the angle between them.
Some key points about the cross product are:
  • The resulting vector's direction is determined by the right-hand rule.
  • The magnitude of the cross product gives the area of the parallelogram formed by the two vectors.
  • If the vectors are parallel, the cross product is zero because \(\sin 0 = 0\).
These characteristics make the cross product a handy tool in physics and engineering, especially in scenarios involving rotational forces and torques.
Angle Between Vectors
Finding the angle between two vectors involves using both the dot product and the cross product.If we have two vectors \( \vec{\mathbf{A}} \) and \( \vec{\mathbf{B}} \), we can find the angle \( \theta \) between them through the equations for their dot and cross products:
  • Dot Product: \( \vec{\mathbf{A}} \cdot \vec{\mathbf{B}} = |\vec{\mathbf{A}}||\vec{\mathbf{B}}|\cos\theta \)
  • Cross Product: \( |\vec{\mathbf{A}} \times \vec{\mathbf{B}}| = |\vec{\mathbf{A}}||\vec{\mathbf{B}}|\sin\theta \)
In scenarios where \( |\vec{\mathbf{A}} \times \vec{\mathbf{B}}| = \vec{\mathbf{A}} \cdot \vec{\mathbf{B}} \), it implies that \( \sin\theta = \cos\theta \).
Solving this gives \( \tan\theta = 1 \), which means \( \theta = 45^\circ \) or \( \frac{\pi}{4} \) radians.
This relationship and calculation showcase how vector operations can help us understand geometric and spatial relationships between different directions in space.

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Most popular questions from this chapter

A particle of mass \(1.00 \mathrm{~kg}\) is moving with velocity \(\overrightarrow{\mathbf{v}}=(7.0 \hat{\mathbf{i}}+6.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} . \quad(a)\) Find the angular momentum \(\overrightarrow{\mathbf{L}}\) relative to the origin when the particle is at \(\overrightarrow{\mathbf{r}}=(2.0 \hat{\mathbf{j}}+4.0 \hat{\mathbf{k}}) \mathrm{m} .(b)\) At position \(\overrightarrow{\mathbf{r}}\) a force of \(\overrightarrow{\mathbf{F}}=4.0 \mathrm{Ni} \hat{\mathbf{i}}\) is applied to the particle. Find the torque relative to the origin.

(II) A person of mass 75 \(\mathrm{kg}\) stands at the center of a rotating merry-go-round platform of radius 3.0 \(\mathrm{m}\) and moment of inertia 920 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) The platform rotates without friction with angular velocity 0.95 \(\mathrm{rad} / \mathrm{s}\) . The person walks radially to the edge of the platform. (a) Calculate the angular velocity when the person reaches the cdge. (b) Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.

(II) A uniform disk turns at 3.7 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk, Fig. 31. They then turn together around the spindle with their centers superposed. What is the angular frequency in \(\mathrm{rev} / \mathrm{s}\) of the combination?

For what directions of velocity would the Coriolis effect on an object moving at the Earth's equator be zero?

A particle of mass \(m\) uniformly accelerates as it moves counterclockwise along the circumference of a circle of radius \(R\) : $$ \overrightarrow{\mathbf{r}}=\hat{\mathbf{i}} R \cos \theta+\hat{\mathbf{j}} R \sin \theta $$ with \(\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2},\) where the constants \(\omega_{0}\) and \(\alpha\) are the initial angular velocity and angular acceleration, respectively. Determine the object's tangential acceleration \(\overrightarrow{\mathbf{a}}_{\tan }\) and determine the torque acting on the object using \((a) \vec{\tau}=\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{F}}\) (b) \(\vec{\tau}=I \vec{\alpha}\)

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