Problem 47 - Fundamentals of Physics

Question

Vectors \(\vec{A}\) and \(\vec{B}\) lie in an \(x y\) plane. \(\vec{A}\) has magnitude \(8.00\) and angle \(130^{\circ} ; \vec{B}\) has components \(B_{x}=-7.72\) and \(B_{y}=-9.20\). What are the angles between the negative direction of the \(y\) axis and (a) the direction of \(\vec{A}\), (b) the direction of the product \(\vec{A} \times \vec{B}\), and \((\mathrm{c})\) the direction of \(\vec{A} \times(\vec{B}+3.00 \hat{\mathrm{k}}) ?\)

Step-by-Step Solution

Answer

(a) 140°; (b) 90°; (c) 90°.

Step 1: Determine Angle with Negative Y-Axis for Vector A

Since vector \( \vec{A} \) is at an angle of \( 130^{\circ} \) from the positive x-axis, the angle from the negative y-axis can be calculated by noting that the negative y-axis direction is \( 270^{\circ} \). Thus, the angle between \( \vec{A} \) and the negative y-axis is \( |270^{\circ} - 130^{\circ}| = 140^{\circ} \).

Step 2: Compute the Direction of A x B

The cross product \( \vec{A} \times \vec{B} \) is perpendicular to the plane containing \( \vec{A} \) and \( \vec{B} \). This means the result will either be pointing in the positive or negative z-direction. Since both vectors are in the xy-plane, compute \( \vec{A} \times \vec{B} = A_xB_y - A_yB_x \). Components of \( \vec{A} \) are \( A_x = 8 \cos(130^{\circ}) \) and \( A_y = 8 \sin(130^{\circ}) \). Calculate the cross product to confirm the direction as positive or negative.

Step 3: Calculate Components of Vector A

Vector \( \vec{A}\) has magnitude 8. Using trigonometric formulas, the components are: \( A_x = 8 \cos(130^{\circ}) \approx -5.146 \) and \( A_y = 8 \sin(130^{\circ}) \approx 6.139 \).

Step 4: Evaluate A x B

Using \( A_x = -5.146 \) and \( A_y = 6.139 \), and \( B_x = -7.72, B_y = -9.20 \), calculate:\[ \vec{A} \times \vec{B} = (-5.146) \cdot (-9.20) - (6.139) \cdot (-7.72) = 47.343 + 47.394 \approx 94.737 \].The direction is positive, indicating it points in the positive z-direction.

Step 5: Integrate Additional Vector to B

Add the vector \(3.00 \hat{k}\) to \( \vec{B} \). New vector is \( \vec{B'} = \vec{B} + 3.00 \hat{k}\). Now, calculate \( \vec{A} \times \vec{B'} \), maintaining the same vector \( \vec{A} \) but altering \( \vec{B} \) to account for the new component.

Step 6: Compute A x (B + 3k)

The new cross product \( \vec{A} \times \vec{B'} = \vec{A} \times (\vec{B} + 3.00 \hat{k}) \) involving the vectors and their components will give:\[ (94.737 \hat{k} + (3.00 \times A_x) \hat{j} - (3.00 \times A_y) \hat{i}) \]This simplification leads to:\[ \vec{A} \times \vec{B'} = (94.737 + 15.438) \hat{k} - 18.417 \hat{i} \].The dominant direction is in \( \hat{k} \) axis. Since x-component is non-zero, the vector has a component in y-plane.

Step 7: Determine the Relevant Angles

Between the negative y-direction, and the y-component of the resultant cross product and k-direction (z-axis), manage the resultant vector as primarily in the k-direction. The k-component is dominant, indicating \( \vec{A} \times (\vec{B} + 3.00 \hat{k}) \) is orthogonal to y-axis, so the angle is \(90^{\circ}\).

Key Concepts

Cross Product

When working with vectors in three-dimensional space, understanding the cross product is essential. The cross product, denoted as \( \vec{A} \times \vec{B} \), results in a new vector that is perpendicular to the plane containing the original vectors \( \vec{A} \) and \( \vec{B} \). This orthogonal vector's direction is determined using the right-hand rule, where if you point the index finger of your right hand in direction of \( \vec{A} \) and your middle finger in direction of \( \vec{B} \), your thumb will point in the direction of \( \vec{A} \times \vec{B} \).

To compute the cross product, use the determinant formula involving unit vectors \(\hat{i}, \hat{j}, \hat{k}\), which represent the x, y, and z directions, respectively. For vectors \( \vec{A} = (A_x, A_y, A_z) \) and \( \vec{B} = (B_x, B_y, B_z) \), the cross product is:

• \( \vec{A} \times \vec{B} = (A_yB_z - A_zB_y)\hat{i} - (A_xB_z - A_zB_x)\hat{j} + (A_xB_y - A_yB_x)\hat{k} \)

This calculation not only gives the direction of the resulting vector but also its magnitude and assures that it is orthogonal to both original vectors.

Magnitude and Direction

Understanding the magnitude and direction of a vector is crucial in vector analysis. The magnitude of a vector, often called its length or norm, is a scalar that represents the distance from the vector's tail to its head. For a vector \( \vec{A} \) with components \( (A_x, A_y, A_z) \), the magnitude is given by:

• \( ||\vec{A}|| = \sqrt{A_x^2 + A_y^2 + A_z^2} \)

This formula uses the Pythagorean theorem to compute the length in three-dimensional space.

Direction is typically described using angles, such as the angle relative to a reference axis like the positive x-axis. A shift in angle indicates rotation, changing the direction the vector points in space. To find the direction, you often use trigonometric functions applied to the vector's components. For polar coordinates, the angle \( \theta \) from the positive x-axis may be calculated with:

• \( \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right) \)

This angle can help further understand how vectors interact, especially during cross product calculations.

Angle Calculation

Calculating angles between vectors or between a vector and a specific direction is common in vector analysis. The angle between two vectors \( \vec{A} \) and \( \vec{B} \) is found using their dot product. While the dot product focuses on the projection of one vector onto another rather than cross product, it helps determine this angle through:

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

Where \( \theta \) is the angle between the two vectors.

To isolate \( \theta \), rearrange the formula:

• \( \cos \theta = \frac{\vec{A} \cdot \vec{B}}{||\vec{A}|| \, ||\vec{B}||} \)
• \( \theta = \cos^{-1}\left(\frac{\vec{A} \cdot \vec{B}}{||\vec{A}|| \, ||\vec{B}||}\right) \)

In problems involving specific directions, like angles relative to x or y axes, you may use inversely measured angles, such as \( 270^{\circ} \) for the negative y-axis, as part of establishing angle orientation.

Coordinate Transformation

In vector analysis, transformations allow vectors to adapt between different reference frames or coordinate systems. Commonly, this involves converting between Cartesian coordinates \((x, y, z)\) and other systems like cylindrical or spherical coordinates. Each system provides unique insights into vector behavior, depending on the context.

To transform a vector \( \vec{A} \) from polar to Cartesian coordinates, for example, use the following relations:

• \( A_x = A \cos(\theta) \)
• \( A_y = A \sin(\theta) \)

Where \( A \) is the vector's magnitude and \( \theta \) its angle relative to a reference axis.
Through these transformations, you gain more flexibility in modeling physical systems, such as changing a plane's orientation or aligning vectors with particular axes. Proper knowledge of coordinate transformations is essential in advanced vector computation scenarios, such as 3D modeling or physics simulations.