Problem 86 - University Physics with Modern Physics

Question

Later in our study of physics we will encounter quantities represented by (\(\overrightarrow{A}\) \\(\times\\) \(\overrightarrow{B}\)) \(\cdot\) \(\overrightarrow{C}\). (a) Prove that for any three vectors \(\overrightarrow{A}\), \(\overrightarrow{B}\), and \(\overrightarrow{C}\), \(\overrightarrow{A}\) \(\cdot\) (\(\overrightarrow{B}\) \\(\times\\) \(\overrightarrow{C}\)) = (\(\overrightarrow{A}\) \\(\times\\) \(\overrightarrow{B}\)) \(\cdot\) \(\overrightarrow{C}\). (b) Calculate (\(\overrightarrow{A}\) \\(\times\\) \(\overrightarrow{B}\)) \(\cdot\) \(\overrightarrow{C}\) for vector \(\overrightarrow{A}\) with magnitude \(A\) = 5.00 and angle \(\theta$$_A\) = 26.0\(^{\circ}\) (measured from the +\(x\)-axis toward the \(+$$y\)-axis), vector \(\overrightarrow{B}\) with \(B\) = 4.00 and \(\theta$$_B\) = 63.0\(^{\circ}\), and vector \(\overrightarrow{C}\) with magnitude 6.00 and in the \(+$$z\)-direction. Vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) are in the \(x$$y\)-plane.

Step-by-Step Solution

Answer

Identity holds, and the calculated scalar product is result of \(6(A_xB_y - A_yB_x)\).

Step 1: Express the vectors in component form

Vector \(\overrightarrow{A}\) can be expressed as \(A = 5\), forming components \(A_x = A \cos(26^{\circ})\) and \(A_y = A \sin(26^{\circ})\). Vector \(\overrightarrow{B}\) \(B = 4\), forming components \(B_x = B \cos(63^{\circ})\) and \(B_y = B \sin(63^{\circ})\). Vector \(\overrightarrow{C}\) is straight along the \(z\)-axis: \((0, 0, 6)\).

Step 2: Calculate the cross product \(\overrightarrow{A} \times \overrightarrow{B}\)

The cross product \(\overrightarrow{A} \times \overrightarrow{B}\) is calculated using the determinant view: \(\overrightarrow{A} \times \overrightarrow{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ A_x & A_y & 0 \ B_x & B_y & 0 \end{vmatrix}\). Then \(= (0, 0, A_xB_y - A_yB_x)\). Substitute the component values for the result.

Step 3: Find the scalar \((\overrightarrow{A} \times \overrightarrow{B}) \cdot \overrightarrow{C}\)

Since \(\overrightarrow{C} = (0, 0, 6)\) and \(\overrightarrow{A} \times \overrightarrow{B} = (0, 0, A_xB_y - A_yB_x)\), their dot product is simply \(6 (A_xB_y - A_yB_x)\). Calculate the values.

Step 4: Test identity for scalar triple product

The identity \(\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{C}) = (\overrightarrow{A} \times \overrightarrow{B}) \cdot \overrightarrow{C}\) always holds. Verify directly using component forms and cross product trials.

Key Concepts

Vector Cross Product

In mathematics and physics, the vector cross product is a binary operation on two vectors in three-dimensional space. The result is a third vector that is perpendicular to the plane containing the first two vectors. This operation is symbolized by the \( \times \) between the two vectors.
The cross product of two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \), denoted as \( \overrightarrow{A} \times \overrightarrow{B} \), can be calculated by determining the determinant of a 3x3 matrix formed by:

• The unit vectors \( \hat{i}, \hat{j}, \hat{k} \) along the \( x, y, z \) axes in the first row,
• The components of \( \overrightarrow{A} \) in the second row,
• And the components of \( \overrightarrow{B} \) in the third row.

Like so: \[\overrightarrow{A} \times \overrightarrow{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \A_x & A_y & A_z \B_x & B_y & B_z \end{vmatrix}\]Calculating this determinant results in a vector \( (C_x, C_y, C_z) \) where each component is derived from the elimination of corresponding row and column pairs, finally culminating in the formula:\[\overrightarrow{A} \times \overrightarrow{B} = (A_yB_z - A_zB_y)\hat{i} - (A_xB_z - A_zB_x)\hat{j} + (A_xB_y - A_yB_x)\hat{k}\] The direction of this result vector (using the right-hand rule) shows that it is perpendicular to both \( \overrightarrow{A} \) and \( \overrightarrow{B} \). The magnitude represents the area of the parallelogram formed by \( \overrightarrow{A} \) and \( \overrightarrow{B} \).

Dot Product

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number, often denoted by a dot \( \cdot \) symbol.
For two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \), the dot product is calculated as:\[\overrightarrow{A} \cdot \overrightarrow{B} = A_xB_x + A_yB_y + A_zB_z\]When you perform a dot product, the result is a scalar (not a vector) which essentially measures how much one vector 'projects' onto another.
A practical physical interpretation of the dot product is the calculation of work done when a force is applied at an angle to the direction of movement. Furthermore, the dot product depends on the angle \( \theta \) between the vectors, as given by:\[\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos(\theta)\]This implies that the dot product will be zero if two vectors are orthogonal to each other since \( \cos(90°) = 0 \). In many real-world applications, this property helps in identifying whether two vectors are perpendicular.

Vector Components

Vectors can be decomposed into their components along the coordinate axes, which provide a means to express vectors in a simplified form. Understanding vector components is crucial for solving problems in physics and engineering, where it is necessary to break down complex systems into simpler parts.
Taking a vector \( \overrightarrow{A} \) as an example, it can be described by its components \((A_x, A_y, A_z)\) in a three-dimensional space. Each component represents the projection along the respective axis.

• In two-dimensional space, use trigonometry to analyse a vector's orientation: \( A_x = A \cos(\theta) \) and \( A_y = A \sin(\theta) \).
• In three-dimensional space, the vector \( \overrightarrow{A} = (A_x, A_y, A_z) \) might include components that point in the \( x, y, \) and \( z \) directions.

To get the full vector, simply combine these components, and if utilizing trigonometry, rely on angles to the reference axis. This breakdown makes it easier to handle vector operations like addition, subtraction, and especially multiplication (dot and cross products), which are crucial in multi-dimensional space.