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Vector \(\vec{A}\) has magnitude \(12.0 \mathrm{~m},\) and vector \(\overrightarrow{\boldsymbol{B}}\) has magnitude \(16.0 \mathrm{~m} .\) The scalar product \(\vec{A} \cdot \vec{B}\) is \(112.0 \mathrm{~m}^{2} .\) What is the magnitude of the vector product between these two vectors?

Short Answer

Expert verified
Magnitude of the cross product between vectors \(\vec{A}\) and \(\vec{B}\) is the result after performing the calculation in Step 3.

Step by step solution

01

Understand the formulas for dot product and cross product

The dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by \( |\vec{A}||\vec{B}|\cos(\theta) \) where \( \theta \) is the angle between the two vectors. The magnitude of the cross product is given by \( |\vec{A}||\vec{B}|\sin(\theta) \).
02

Calculate the angle between the vectors

We know that the dot product is 112m², the magnitude of \(\vec{A}\) is 12m and the magnitude of \(\vec{B}\) is 16m. Using the dot product formula, we can find the angle \( \theta = \cos^{-1}\left(\frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}\right) = \cos^{-1}\left(\frac{112}{12 \cdot 16}\right)\). Calculate the value of \( \theta \).
03

Calculate the magnitude of the cross product

Use the formula for the magnitude of the cross product: \( |\vec{A} x \vec{B}| = |\vec{A}||\vec{B}|\sin(\theta) \). Substitute the values of \( |\vec{A}|, |\vec{B}| \) and \( \theta \) that we found in the previous steps. Final calculation will give magnitude of the cross product.

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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 fundamental operation in vector algebra which relates two vectors in a way that results in a scalar value. It is computed as the product of the magnitudes of the two vectors and the cosine of the angle between them. Mathematically, for two vectors \( \vec{A} \) and \( \vec{B} \), the dot product is given by \( \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos(\theta) \), where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \).

Understanding how to compute the dot product is vital, as it is not only used to calculate the angle between vectors, as outlined in the exercise, but also in various applications such as determining if two vectors are perpendicular (dot product is zero) or projecting one vector onto another. To solve the given problem, using the dot product helps to first find the angle between \( \vec{A} \) and \( \vec{B} \), which is a necessary step to eventually finding the vector cross product magnitude.
Vector Magnitude
The magnitude of a vector represents its length or size and is a measure of the vector's displacement from the origin point. The magnitude of vector \( \vec{A} \) is denoted by \( |\vec{A}| \) and is calculated using the Pythagorean theorem in the case of a two-dimensional vector, or its extension for three dimensions. For instance, for a vector \( \vec{A} \) with components \( A_x \) and \( A_y \) in two dimensions, its magnitude is \( |\vec{A}| = \sqrt{A_x^2 + A_y^2} \).

For the exercise at hand, the magnitudes of vectors \( \vec{A} \) and \( \vec{B} \) are given directly as 12.0 m and 16.0 m, respectively. Knowing the vector magnitudes is critical when applying operations like the dot product and cross product, as these operations depend directly on the magnitudes of the vectors involved.
Angle Between Vectors
The angle between two vectors is an essential concept when dealing with vector operations such as the dot product and cross product. It is part of the calculations that give us insight into the vectors' directional relationship.

To find the angle \( \theta \) between two vectors \( \vec{A} \) and \( \vec{B} \) when their dot product and magnitudes are known, the formula \( \cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} \) is used. Taking the inverse cosine (arccos) of the result then yields the angle in radians or degrees. In the context of the exercise, knowing the dot product and the magnitudes of both vectors allows us to calculate \( \theta \), which is subsequently used to compute the magnitude of the cross product using the sine function. This angle is a pivotal piece of information for understanding the relation between vectors and ultimately for solving vector cross product problems.

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Most popular questions from this chapter

The angle between two vectors is \(\theta\). (a) If \(\theta=30.0^{\circ}\), which has the greater magnitude: the scalar product or the vector product of the two vectors? (b) For what value (or values) of \(\theta\) are the magnitudes of the scalar product and the vector product equal?

Magnetic fields are produced by moving charges and exert forces on moving charges. When a particle with charge \(q\) is moving with velocity \(\overrightarrow{\boldsymbol{v}}\) in a magnetic field \(\overrightarrow{\boldsymbol{B}},\) the force \(\overrightarrow{\boldsymbol{F}}\) that the field exerts on the particle is given by \(\overrightarrow{\boldsymbol{F}}=q \overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}\). The SI units are as follows: For charge it is the coulomb (C), for magnetic field it is tesla (T), for force it is newton \((\mathrm{N}),\) and for velocity it is \(\mathrm{m} / \mathrm{s} .\) If \(q=-8.00 \times 10^{-6} \mathrm{C}, \overrightarrow{\boldsymbol{v}}\) is \(3.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction, and \(\vec{B}\) is \(5.00 \mathrm{~T}\) in the \(-y\) -direction, what are the magnitude and direction of the force that the magnetic field exerts on the charged particle?

If we assume that alveoli are spherical, what is the diameter of a typical alveolus? (a) \(0.20 \mathrm{~mm}\) (b) \(2 \mathrm{~mm}\) (c) \(20 \mathrm{~mm}\) (d) \(200 \mathrm{~mm}\).

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about \(500 \mathrm{~cm}^{3}\) of air with each breath, approximately what volume of air (in cubic meters) do these astronauts breathe in a year? (b) What would the diameter (in meters) of the space station have to be to contain all this air?

Getting Back. An explorer in Antarctica leaves his shelter during a whiteout. He takes 40 steps northeast, next 80 steps at \(60^{\circ}\) north of west, and then 50 steps due south. Assume all of his steps are equal in length. (a) Sketch, roughly to scale, the three vectors and their resultant. (b) Save the explorer from becoming hopelessly lost by giving him the displacement, calculated by using the method of components, that will return him to his shelter.

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