Problem 84 - University Physics with Modern Physics

Question

Two vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) have magnitudes \(A\) = 3.00 and \(B\) = 3.00. Their vector product is \(\overrightarrow{A}\) \\(\times\\) \(\overrightarrow{B}$$^{\circ}\) = \(-\)5.00\(\hat{k}\) + 2.00\(\hat{\imath}\). What is the angle between \(\overrightarrow{A}\) and \(\overrightarrow{B}\)?

Step-by-Step Solution

Answer

The angle between the vectors is approximately \(36.9^{\circ}\).

Step 1: Recall the Cross Product Formula

The cross product of two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \) is given by the formula \( \overrightarrow{A} \times \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \sin(\theta) \hat{n} \), where \( \theta \) is the angle between the vectors and \( \hat{n} \) is the unit vector perpendicular to the plane of \( \overrightarrow{A} \) and \( \overrightarrow{B} \).

Step 2: Determine the Magnitude of the Cross Product

From the given vector product \( \overrightarrow{A} \times \overrightarrow{B} = -5.00\hat{k} + 2.00\hat{\imath} \), calculate the magnitude using the Pythagorean theorem: \[ |\overrightarrow{A} \times \overrightarrow{B}| = \sqrt{(-5.00)^2 + (2.00)^2} = \sqrt{25 + 4} = \sqrt{29} \approx 5.39 \]

Step 3: Use the Magnitude to Find the Angle

We know that \( |\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| |\overrightarrow{B}| \sin(\theta) \). Substitute the values: \[ 5.39 = 3.00 \times 3.00 \times \sin(\theta) \] \[ \sin(\theta) = \frac{5.39}{9.00} \approx 0.599 \]

Step 4: Solve for the Angle

Calculate the angle \( \theta \) by taking the inverse sine (\( \sin^{-1} \)) of \( 0.599 \): \[ \theta = \sin^{-1}(0.599) \approx 36.9^{\circ} \]

Key Concepts

Cross Product

In vector mathematics, the cross product, also known as the vector product, is a crucial operation that results in another vector. When you take the cross product of two vectors, the result is a vector that is perpendicular to both original vectors. This new vector's direction is determined by the right-hand rule, where the thumb points in the direction of the cross product if the fingers are curled from the first vector to the second.

A key aspect of the cross product is its formula: \( \overrightarrow{A} \times \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \sin(\theta) \hat{n} \). Here, \( |\overrightarrow{A}| \) and \( |\overrightarrow{B}| \) are the magnitudes of vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \), \( \theta \) is the angle between them, and \( \hat{n} \) denotes the unit vector perpendicular to the plane created by the two original vectors.

• The cross product is not defined in two dimensions, making it unique to three-dimensional vector mathematics.
• Unlike the dot product, the cross product is not commutative. So, \( \overrightarrow{A} \times \overrightarrow{B} eq \overrightarrow{B} \times \overrightarrow{A} \).
• The cross product operation results in maximum magnitude when the vectors are orthogonal (\( \theta = 90^{\circ} \)).

Angle Between Vectors

The angle between two vectors can reveal much about their relationship in a geometric space. To find this angle, especially when dealing with a cross product, we rely on the sine function. The relationship is given by \( \sin(\theta) \), derived from the cross product's magnitude.

Knowing information about the cross product's magnitude is essential in calculating the angle. The magnitude of the cross product \( |\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| |\overrightarrow{B}| \sin(\theta) \). Solving for \( \sin(\theta) \) involves dividing the cross product's magnitude by the product of the magnitudes of the original vectors. Whether vectors are parallel (\( \theta = 0^{\circ} \) or \( 180^{\circ} \)) or orthogonal (\( \theta = 90^{\circ} \)) can be deduced by the calculated angle.

• For parallel vectors, \( \sin(\theta) = 0 \), leading to a zero cross product magnitude.
• Compute the angle using the inverse sine function, \( \theta = \sin^{-1}\left(\frac{|\overrightarrow{A} \times \overrightarrow{B}|}{|\overrightarrow{A}| |\overrightarrow{B}|}\right) \).
• Different angles can help explain phenomena in physics, such as torque or the orientation of magnetic fields.

Magnitude of Vectors

The magnitude of a vector is like the vector's "length". It provides a measure of how much physical quantity exists, irrespective of its direction. In mathematical terms, the magnitude of a vector \( \overrightarrow{A} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} \) is calculated using the formula \( |\overrightarrow{A}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

This calculation is foundational in understanding vector quantities in physics and engineering, as it allows for assessing the size of forces, velocities, and other vector quantities. When solving problems involving the cross product, the magnitude helps in determining how vectors interact spatially.

• The magnitude helps in determining the scalars needed to utilize vectors in calculations.
• Units of the vector's magnitude will depend on the physical context, such as meters for distance or newtons for force.
• If a vector's components are expressed in unit vectors \( \hat{i}, \hat{j}, \hat{k} \), calculating the magnitude gives insights into its real-world size.