Problem 81 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

The magnitude of the vector product is approximately 155.95 m².

Step 1: 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, \(\left| \overrightarrow{A} \times \overrightarrow{B} \right|\).

Step 2: 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.

Step 3: 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 \(\cos \theta = \frac{112.0}{12.0 \times 16.0} = \frac{112.0}{192.0} = \frac{7}{12}\).

Step 4: Understanding the Vector Product

The vector product (also known as the cross product) \(\overrightarrow{A} \times \overrightarrow{B}\) has magnitude \(|\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| |\overrightarrow{B}| \sin \theta\).

Step 5: 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 \(\sin \theta = \sqrt{\frac{95}{144}} = \frac{\sqrt{95}}{12}\).

Step 6: 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, \(16.0 \times \sqrt{95}\).

Step 7: Final Calculation

Calculate the magnitude: \(16.0 \times \sqrt{95} \approx 16.0 \times 9.7468 \approx 155.95\, \text{m}^2\).

Key Concepts

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:

• \(\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos \theta\)

Here, \(|\overrightarrow{A}|\) and \(|\overrightarrow{B}|\) represent the magnitudes of the vectors, while \(\theta\) is the angle between them.
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:

• \(|\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| |\overrightarrow{B}| \sin \theta\)

The direction of the resulting vector follows the right-hand rule, meaning if you point the index finger in the direction of \(\overrightarrow{A}\) and the middle finger in the direction of \(\overrightarrow{B}\), the thumb points in the direction of the vector product.
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:

• \(|\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| |\overrightarrow{B}| \sin \theta\)

Its unique characteristic is its resulting vector orthogonal to both \(\overrightarrow{A}\) and \(\overrightarrow{B}\). This is what makes it beneficial in 3D geometry and physics situations, providing both magnitude and direction information simultaneously.

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:

• \(\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos \theta\)

This method of multiplying vectors expresses itself as a scalar number, pointing out the extent to which the vectors point in the same or opposite directions.
It is particularly useful when tackling problems involving projections and energy, where directional components are crucial.