Problem 38 - Fundamentals of Physics

Question

For the following three vectors, what is \(3 \vec{C} \cdot(2 \vec{A} \times \vec{B}) ?\) $$\begin{aligned} &\vec{A}=2.00 \hat{\mathrm{i}}+3.00 \hat{\mathrm{j}}-4.00 \hat{\mathrm{k}} \\ &\vec{B}=-3.00 \hat{\mathrm{i}}+4.00 \hat{\mathrm{j}}+2.00 \hat{\mathrm{k}} \quad \vec{C}=7.00 \hat{\mathrm{i}}-8.00 \hat{\mathrm{j}}\end{aligned}$$

Step-by-Step Solution

Answer

The solution to the exercise is 156.

Step 1: Multiply Vector A by 2

First, we need to calculate \(2\vec{A}\). Multiply each component of \(\vec{A}\) by 2: \[ 2 \vec{A} = 2(2.00 \hat{\mathrm{i}} + 3.00 \hat{\mathrm{j}} - 4.00 \hat{\mathrm{k}}) = 4.00\hat{\mathrm{i}} + 6.00\hat{\mathrm{j}} - 8.00\hat{\mathrm{k}} \].

Step 2: Calculate the Cross Product \(2 \vec{A} \times \vec{B}\)

Use the cross product formula for vectors \(\vec{u} \times \vec{v} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix}\). Therefore: \[2 \vec{A} \times \vec{B} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 4.00 & 6.00 & -8.00 \ -3.00 & 4.00 & 2.00 \end{vmatrix} \].

Step 3: Solve the Determinant for Cross Product

Calculate the determinant from the above matrix: \(2 \vec{A} \times \vec{B} = \hat{\mathrm{i}} (6.00 \cdot 2.00 - (-8.00) \cdot 4.00) - \hat{\mathrm{j}} (4.00 \cdot 2.00 - (-8.00) \cdot -3.00) + \hat{\mathrm{k}} (4.00 \cdot 4.00 - 6.00 \cdot -3.00)\). Simplifying gives: \[ 2 \vec{A} \times \vec{B} = \hat{\mathrm{i}} (12 + 32) - \hat{\mathrm{j}} (8 + 24) + \hat{\mathrm{k}} (16 + 18)\] \[ = 44\hat{\mathrm{i}} - 32\hat{\mathrm{j}} + 34\hat{\mathrm{k}} \].

Step 4: Multiply Vector C by 3

Next, multiply each component of \(\vec{C}\) by 3: \[ 3 \vec{C} = 3(7.00 \hat{\mathrm{i}} - 8.00 \hat{\mathrm{j}}) = 21.00 \hat{\mathrm{i}} - 24.00 \hat{\mathrm{j}} \].

Step 5: Calculate the Dot Product \(3 \vec{C} \cdot (2\vec{A} \times \vec{B})\)

Use the dot product formula \(\vec{m} \cdot \vec{n} = m_1 n_1 + m_2 n_2 + m_3 n_3\). For vectors \(3 \vec{C} = 21.00 \hat{\mathrm{i}} - 24.00 \hat{\mathrm{j}} \) and \(2 \vec{A} \times \vec{B} = 44\hat{\mathrm{i}} - 32\hat{\mathrm{j}} + 34\hat{\mathrm{k}}\): \[ 3 \vec{C} \cdot (2\vec{A} \times \vec{B}) = 21.00 \times 44 - 24.00 \times 32 + 0 \times 34 \]. This simplifies to: \[ = 924 - 768 \] \[ = 156 \].

Key Concepts

Cross Product

In vector calculus, the cross product is a binary operation on two vectors in three-dimensional space. It results in a third vector that is perpendicular to the plane containing the original vectors, making it a very useful tool in physics and engineering.
To calculate the cross product of \(\vec{u} = a_1 \hat\textbf{i} + a_2 \hat\textbf{j} + a_3 \hat\textbf{k}\) and \(\vec{v} = b_1 \hat\textbf{i} + b_2 \hat\textbf{j} + b_3 \hat\textbf{k}\), use the determinant of a matrix:

• Place the unit vectors \(\hat\textbf{i}, \hat\textbf{j}, \) and \(\hat\textbf{k}\) in the first row.
• Place the components of \(\vec{u}\) and \(\vec{v}\) in the second and third rows, respectively.

For the given vectors in the problem, \(2 \vec{A} \) and \ \vec{B}\, the determinant method provides a step-by-step calculation that ultimately gives the resulting vector. The resulting vector, \(2 \vec{A} \times \vec{B} = 44 \hat\textbf{i} - 32 \hat\textbf{j} + 34 \hat\textbf{k}\), describes the spatial relationship between the initial vectors, emphasizing its importance in finding forces and rotations.

Dot Product

The dot product, or scalar product, is another fundamental operation in vector calculus. It differs significantly from the cross product in that it results in a scalar, a single real number, rather than a vector. This operation measures the extent to which two vectors point in the same direction.
The dot product is calculated by multiplying the corresponding components of two vectors and summing these products:

• For vectors \( \vec{m} = m_1 \hat\textbf{i} + m_2 \hat\textbf{j} + m_3 \hat\textbf{k} \) and \( \vec{n} = n_1 \hat\textbf{i} + n_2 \hat\textbf{j} + n_3 \hat\textbf{k} \), the dot product is \( \vec{m} \cdot \vec{n} = m_1n_1 + m_2n_2 + m_3n_3 \).

In the exercise, the dot product \( 3 \vec{C} \cdot (2 \vec{A} \times \vec{B}) =156\) results from multiplying the components of \( 3 \vec{C} \) with those from \( 2 \vec{A} \times \vec{B} \) and summing them. This value helps evaluate specific physical entities like work or energy when assessing forces.

Vector Operations

Vector operations, such as addition, subtraction, and scalar multiplication, are the building blocks of more complex vector calculus problems. Understanding these operations is essential for solving problems involving vectors in physics and engineering.
First, let's break down scalar multiplication, an operation used in this exercise. This involves multiplying each component of a vector by a scalar. For example, multiplying \( \vec{C} \) by 3, transforms each component like so: \(3(7.00 \hat{\text{\textbf{i}}} - 8.00 \hat{\text{\textbf{j}}}) = 21.00 \hat{\text{\textbf{i}}} - 24.00 \hat{\text{\textbf{j}}}\).
These simple operations allow for quick manipulation of vectors to achieve desired results, preparing them for use in dot and cross products, as shown previously in this exercise. Mastery of basic vector operations is key for understanding how more complex physical phenomena are described.