Chapter 7: Problem 2
find \(\mathbf{a} \times \mathbf{b}\). $$ a=2 \mathrm{i}+\mathrm{j}, \mathbf{b}=4 \mathbf{i}-\mathbf{k} $$
Short Answer
Expert verified
The cross product is \(-\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\).
Step by step solution
01
Write the Vectors in Component Form
To solve for the cross product \(\mathbf{a} \times \mathbf{b}\), start by writing the vectors \(\mathbf{a}\) and \(\mathbf{b}\) in component form: \(\mathbf{a} = 2\mathbf{i} + 1\mathbf{j} + 0\mathbf{k}\) and \(\mathbf{b} = 4\mathbf{i} + 0\mathbf{j} - 1\mathbf{k}\).
02
Formula for Cross Product
Recall the formula for the cross product of two vectors \(\mathbf{a} \times \mathbf{b}\) given by:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 1 & 0 \ 4 & 0 & -1 \end{vmatrix}\]This is a determinant that will give us the vector orthogonal to \(\mathbf{a}\) and \(\mathbf{b}\).
03
Compute the Determinant
Calculate the determinant:\[\mathbf{c} = \mathbf{i}(1 \cdot (-1) - 0 \cdot 0) - \mathbf{j}(2 \cdot (-1) - 0 \cdot 4) + \mathbf{k}(2 \cdot 0 - 4 \cdot 1)\]Simplify each component:\[\mathbf{c} = \mathbf{i}(-1) + \mathbf{j}(2) - \mathbf{k}(4)\]Thus, \(\mathbf{c} = -\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\).
04
Write the Resulting Vector
The vector resulting from the cross product \(\mathbf{a} \times \mathbf{b}\) is \(-\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\). This vector is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Vector Operations
Vector operations, such as addition, subtraction, and the cross product, play a crucial role in physics and engineering. While addition and subtraction are relatively straightforward, the cross product can be a bit more complex. The cross product of two vectors, denoted as \( \mathbf{a} \times \mathbf{b} \), results in another vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). This is particularly useful in physics for finding torque or rotational motion when the direction is involved.
- Addition: Combine vectors by adding their corresponding components.
- Subtraction: Subtract vectors by subtracting their corresponding components.
- Cross Product: Use it to find a vector perpendicular to two given vectors.
Calculating the Determinant for Cross Product
The determinant provides a systematic way to find areas, volumes, and in the case of cross products, vector magnitudes in 3D spaces. When computing a cross product, you use a specific determinant formula that captures the perpendicular vector required. This involves setting up a matrix with vectors as rows and solving the determinant. The formula used for two vectors \(\mathbf{a}\) and \(\mathbf{b}\), as shown in the original problem, involves placing the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) in the first row. The components of \(\mathbf{a}\) and \(\mathbf{b}\) fill the next two rows. From here, you calculate the determinant to find the cross product:\[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \2 & 1 & 0 \4 & 0 & -1 \end{vmatrix} \]By solving this step-by-step, each sub-determinant provides a component of the resulting vector. This process exemplifies the power of determinants in vector operations, turning matrix calculations into geometric insights.
Properties of Perpendicular Vectors
Perpendicular vectors have a dot product of zero, indicating orthogonality. In geometry, two vectors are perpendicular if they meet at a right angle. This right angle carries significant implications in 3D modeling, physics, and other fields requiring spatial awareness. When you compute the cross product of two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), the resulting vector \(\mathbf{c}\) is perpendicular to both. This property is useful because it provides a standard way to find a direction orthogonal to a plane.
- The dot product of \(\mathbf{c}\) with \(\mathbf{a}\) and \(\mathbf{b}\) should be zero, confirming \(\mathbf{c}\) is perpendicular.
- Applications include calculating normals to surfaces, which are essential in graphics and physics simulations.