Problem 2
Question
Find the cross product of \langle 1,0,2\rangle and \(\langle-1,-2,4\rangle .\)
Step-by-Step Solution
Verified Answer
The cross product is \( \langle 4, -6, -2 \rangle \).
1Step 1: Understand the Cross Product Formula
The cross product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be found using the formula:\[ \mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \] where \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \).
2Step 2: Identify Vector Components
Identify the components of vectors \( \mathbf{a} \) and \( \mathbf{b} \):\[ \mathbf{a} = \langle 1, 0, 2 \rangle \text{ and } \mathbf{b} = \langle -1, -2, 4 \rangle \]Thus, \( a_1 = 1, a_2 = 0, a_3 = 2 \) and \( b_1 = -1, b_2 = -2, b_3 = 4 \).
3Step 3: Calculate Each Component of the Cross Product
Use the formula to calculate each component:- First Component: \( a_2b_3 - a_3b_2 = 0\cdot4 - 2\cdot(-2) = 0 + 4 = 4 \).- Second Component: \( a_3b_1 - a_1b_3 = 2\cdot(-1) - 1\cdot4 = -2 - 4 = -6 \).- Third Component: \( a_1b_2 - a_2b_1 = 1\cdot(-2) - 0\cdot(-1) = -2 + 0 = -2 \).
4Step 4: Write the Resulting Cross Product
Combine the calculated components into the resulting vector:\[ \mathbf{a} \times \mathbf{b} = \langle 4, -6, -2 \rangle \]
Key Concepts
Vector MultiplicationVector CalculusThree-Dimensional Vectors
Vector Multiplication
Vector multiplication can be performed using two primary operations: the dot product and the cross product. While the dot product results in a scalar value, the cross product produces another vector. The focus here is on the cross product, which is essential in three-dimensional space.
- The cross product of two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), results in a new vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \).
- The formula for the cross product is \( \mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \).
- This operation is particularly useful in physics and engineering to determine torque, rotational motion, and magnetic forces.
Vector Calculus
Vector calculus is a branch of mathematics focused on differentiation and integration of vector fields. It extends the concepts from traditional calculus to multiple dimensions, making it an essential tool in physics and engineering.
- Vector calculus operations include gradient, divergence, curl, and the Laplacian—each useful for describing various physical phenomena, like fluid flow and electromagnetic fields.
- The cross product is often a part of vector calculus, as it helps in calculating the curl of a vector field, which measures the rotation or swirling strength of the field at a point.
- Understanding vector calculus concepts is pivotal in any field dealing with three-dimensional vectors, as it provides the mathematical foundation for describing changes and interactions within a field.
Three-Dimensional Vectors
Vectors in three-dimensional space have characteristics and properties that make them distinct from their lower-dimensional counterparts. They are used to represent physical quantities with both direction and magnitude, such as forces and velocities.
- A vector is denoted by its components along the x, y, and z axes, written as \( \langle x, y, z \rangle \) in 3D space.
- Operations like addition, scalar multiplication, and the dot and cross products are fundamental manipulations you can perform with vectors.
- The cross product's significance arises due to its perpendicularity to the original vectors, an attribute that enables visualization and computation of spatial orientations and directions.
Other exercises in this chapter
Problem 2
Find an equation for the sphere \(x^{2}+y^{2}+z^{2}=4\) in cylindrical coordinates.
View solution Problem 2
Find an equation of the plane containing (-1,2,-3) and perpendicular to \(\langle 4,5,-1\rangle .\)
View solution Problem 2
Find \(\langle 1,2,0\rangle \cdot\langle 0,0,57\rangle .\)
View solution Problem 2
Draw the vector \(\langle 3,-1,2\rangle\) with its tail at the origin.
View solution