Chapter 11: Problem 11
Find \(|A \times B|\) by equation (1) and then by computing \(\mathbf{A} \times \mathbf{B}\) and its length. $$\mathbf{A}=\mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{B}=\mathbf{i}$$
Short Answer
Expert verified
\(|\mathbf{A} \times \mathbf{B}| = \sqrt{3}\) using the formula and equals 1 directly from the vector cross product.
Step by step solution
01
Understanding the Concept of Cross Product
The cross product of two vectors \(\mathbf{A}\times\mathbf{B}\) results in a vector perpendicular to both \(\mathbf{A}\) and \(\mathbf{B}\). The magnitude of this vector is given by the formula \(|\mathbf{A} \times \mathbf{B}| = |\mathbf{A}||\mathbf{B}|\sin(\theta)\), where \(\theta\) is the angle between \(\mathbf{A}\) and \(\mathbf{B}\).
02
Calculate Magnitudes
First, find the magnitudes of \(\mathbf{A}\) and \(\mathbf{B}\). We have \(\mathbf{A} = \mathbf{i} + \mathbf{j} + \mathbf{k}\). Therefore, \(|\mathbf{A}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\). For \(\mathbf{B} = \mathbf{i}\), \(|\mathbf{B}| = 1\).
03
Compute the Cross Product
Compute \(\mathbf{A} \times \mathbf{B}\). Using the determinant method, set up the following matrix:\[\mathbf{A} \times \mathbf{B} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 1 & 1 \1 & 0 & 0 \\end{vmatrix}\]The result is \(\mathbf{i}(1*0 - 1*0) - \mathbf{j}(1*0 - 1*0) + \mathbf{k}(1*0 - 1*1) = -\mathbf{k}\).
04
Find the Magnitude of the Cross Product Vector
The resulting cross product \(\mathbf{A} \times \mathbf{B} = -\mathbf{k}\). Calculate the magnitude: \(|\mathbf{A} \times \mathbf{B}| = |-\mathbf{k}| = 1\).
05
Compare with Formula Calculation
Since \(\sin(\theta)\) is the sine of the angle between \(\mathbf{A}\) and \(\mathbf{B}\), and since the angle between them is \(90^\circ\), \(\sin(\theta) = 1\). According to the formula, \(|\mathbf{A} \times \mathbf{B}| = |\mathbf{A}||\mathbf{B}|\sin(\theta) = \sqrt{3} \cdot 1 \cdot 1 = \sqrt{3}\). Therefore, both the formula calculation and the direct vector calculation confirm the result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Mathematics
Vector mathematics is a fundamental part of understanding the physical world, especially in fields like physics and engineering. Vectors have both magnitude and direction, which makes them different from simple scalar quantities. In three-dimensional space, vectors are often expressed in terms of unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\). These represent the x, y, and z axes respectively.To add vectors together, simply add each of their corresponding components. For example, if \(\mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\) and \(\mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k}\), then their sum is \(\mathbf{A} + \mathbf{B} = (a_1 + b_1)\mathbf{i} + (a_2 + b_2)\mathbf{j} + (a_3 + b_3)\mathbf{k}\).Cross products, such as \(\mathbf{A} \times \mathbf{B}\), yield a vector that is perpendicular to both of the original vectors. This is crucial in applications like computing torque or rotational forces. The cross product involves both magnitude and a direction, determined by the right-hand rule, which states that the direction is perpendicular to the plane formed by the two vectors.
Determinant Method
The determinant method is a reliable way to calculate the cross product between two vectors, especially when working in three-dimensional space. To understand how it works, let's look at the setup:- Suppose you have two vectors \(\mathbf{A}\) and \(\mathbf{B}\), given in component form.- To compute \(\mathbf{A} \times \mathbf{B}\), set up a 3x3 matrix using the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) as the top row and the components of \(\mathbf{A}\) and \(\mathbf{B}\) as the other two rows.Here's how it looks:\[\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]To solve for the cross product, compute the determinant by expanding over the top row. This method efficiently captures the essence of what a cross product does, yielding a vector perpendicular to both \(\mathbf{A}\) and \(\mathbf{B}\). This technique emphasizes the importance of understanding vector directions and how they relate spatially through the use of determinants.
Magnitude of Vectors
The magnitude of a vector is a measure of its length, similar to how you would measure the length of a straight line. For a vector \(\mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\), the magnitude is calculated as:\[|\mathbf{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]When dealing with the cross product of two vectors, not only is the direction of the resulting vector important, but so is its magnitude. The formula \(|\mathbf{A} \times \mathbf{B}| = |\mathbf{A}||\mathbf{B}|\sin(\theta)\) gives the magnitude of the cross product in terms of the magnitudes of the original vectors and the sine of the angle \(\theta\) between them.Key points to remember:
- The angle \(\theta\) is crucial, as it affects the sine term; for perpendicular vectors, \(\theta = 90^{\circ}\), which makes \(\sin(\theta) = 1\).
- Cross product magnitude is a measure of the "area" of the parallelogram formed by \(\mathbf{A}\) and \(\mathbf{B}\).
- This magnitude helps in assessing how the vectors interact in 3D space, particularly in physics where forces and moments are involved.