Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 1: Problem 81
Vector \(\overrightarrow{A}\) has magnitude 12.0 m, and vector \(\overrightarrow{B}\) has magnitude 16.0 m. The scalar product \(\overrightarrow{A}\) \(\cdot\) \(\overrightarrow{B}\) is 112.0 m\(^2\). What is the magnitude of the vector product between these two vectors?
Short Answer
Expert verified
The magnitude of the vector product is approximately 155.95 m虏.
Step by step solution
01
Understanding the Given Information
We have two vectors: \(\overrightarrow{A}\) with magnitude 12.0 m and \(\overrightarrow{B}\) with magnitude 16.0 m. We are given that their scalar product \(\overrightarrow{A} \cdot \overrightarrow{B} = 112.0\, \text{m}^2\). We need to find the magnitude of their vector product, \(\left| \overrightarrow{A} \times \overrightarrow{B} \right|\).
02
Scalar Product Definition
The scalar product (also known as the dot product) of two vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) is given by the formula \(\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos \theta\), where \(\theta\) is the angle between the two vectors.
03
Calculating the Cosine of the Angle
Using the formula for the scalar product, we set \(112.0 = 12.0 \times 16.0 \times \cos \theta\). Solving for \(\cos \theta\), we have \(\cos \theta = \frac{112.0}{12.0 \times 16.0} = \frac{112.0}{192.0} = \frac{7}{12}\).
04
Understanding the Vector Product
The vector product (also known as the cross product) \(\overrightarrow{A} \times \overrightarrow{B}\) has magnitude \(|\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| |\overrightarrow{B}| \sin \theta\).
05
Calculating the Sine of the Angle
Since \(\sin^2 \theta + \cos^2 \theta = 1\), we have \(\sin^2 \theta = 1 - \left(\frac{7}{12}\right)^2\). Therefore, \(\sin^2 \theta = 1 - \frac{49}{144} = \frac{95}{144}\), and \(\sin \theta = \sqrt{\frac{95}{144}} = \frac{\sqrt{95}}{12}\).
06
Finding the Magnitude of the Vector Product
Using the formula for the magnitude of the vector product, we get \(|\overrightarrow{A} \times \overrightarrow{B}| = 12.0 \times 16.0 \times \frac{\sqrt{95}}{12}\). Simplifying, \(16.0 \times \sqrt{95}\).
07
Final Calculation
Calculate the magnitude: \(16.0 \times \sqrt{95} \approx 16.0 \times 9.7468 \approx 155.95\, \text{m}^2\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Scalar Product
The scalar product, also known as the dot product, is a fundamental concept in vector calculus. This operation takes two vectors and results in a scalar quantity. It is calculated by multiplying the magnitudes of the two vectors with the cosine of the angle between them. For vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\), the formula is:
This product is useful in determining how much of one vector "goes in the direction" of the other. For example, it is often used when computing work done or projecting one vector onto another.
- \(\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos \theta\)
This product is useful in determining how much of one vector "goes in the direction" of the other. For example, it is often used when computing work done or projecting one vector onto another.
Vector Product
The vector product, or cross product, is quite different from the scalar product. Instead of a scalar, it results in a new vector that is orthogonal to the plane containing the original vectors. The magnitude of the vector product is given by:
This product is especially significant in physics, for example, in torque and angular momentum calculations.
- \(|\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| |\overrightarrow{B}| \sin \theta\)
This product is especially significant in physics, for example, in torque and angular momentum calculations.
Cross Product
The cross product is specifically a vector operation that combines two vectors in three-dimensional space to create another vector. This operation is significant because it retains the rotational properties of the input vectors relative to each other.
The magnitude of the cross product, similar to the vector product, is determined using:
The magnitude of the cross product, similar to the vector product, is determined using:
- \(|\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| |\overrightarrow{B}| \sin \theta\)
Dot Product
The dot product is an essential mathematical tool in both physics and engineering, providing valuable insight into the relationship between two vectors. While often used interchangeably with the scalar product, it specifically focuses on how the vectors align with each other. The formula is:
It is particularly useful when tackling problems involving projections and energy, where directional components are crucial.
- \(\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos \theta\)
It is particularly useful when tackling problems involving projections and energy, where directional components are crucial.
