Problem 25
Question
Find the indicated scalar or vector without using \((5),(13)\), or \((15)\). $$ (\mathbf{i}+\mathbf{j}) \times(\mathbf{i}+5 \mathbf{k}) $$
Step-by-Step Solution
Verified Answer
The resulting vector is \(5\mathbf{i} - 5\mathbf{j} - \mathbf{k}\).
1Step 1: Expand the vectors
First, write out the vectors explicitly in terms of their components. \( \mathbf{i} + \mathbf{j} \) can be represented as \([1, 1, 0]\) and \( \mathbf{i} + 5 \mathbf{k} \) as \([1, 0, 5]\).
2Step 2: Use the cross product formula
For vectors \(\mathbf{a} = [a_1, a_2, a_3]\) and \(\mathbf{b} = [b_1, b_2, b_3]\), the cross product \(\mathbf{a} \times \mathbf{b} = [a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1]\). We will now apply this formula.
3Step 3: Calculate each component of the cross product
Using the formula:- The i-th component: \(1 \cdot 5 - 0 \cdot 0 = 5\)- The j-th component: \(0 \cdot 1 - 1 \cdot 5 = -5\)- The k-th component: \(1 \cdot 0 - 1 \cdot 1 = -1\)
4Step 4: Form the resulting vector
Combine the components calculated in Step 3 to form the resulting vector: \(\mathbf{v} = [5, -5, -1]\).
Key Concepts
Vector OperationsComponent MethodThree-Dimensional Vectors
Vector Operations
Vector operations are essential tools in physics and engineering that help us describe and manipulate quantities that have both magnitude and direction. In these disciplines, vectors are used to represent quantities like force, velocity, and acceleration. One key operation involving vectors is the cross product, which occurs between two three-dimensional vectors.
- The **cross product** of two vectors results in a third vector that is perpendicular to the plane formed by the original vectors.
- This operation is not just a simple multiplication; it has its unique set of rules and results.
- Unlike the dot product, which results in a scalar, the cross product always results in another vector.
Component Method
The component method is a powerful strategy in vector mathematics for simplifying and solving complex vector equations. Breaking down vectors into their components is essential because it allows operations, such as addition, subtraction, and cross products, to be carried out more easily.
- Vectors can be broken down into components along the x, y, and z axes in three-dimensional space. For instance, a vector \( \mathbf{a} \) might be represented as \([a_1, a_2, a_3]\).
- Using the component form, complex calculations, such as cross products, become simple arithmetic.
- In the cross product operation, each component of the resulting vector is calculated using a specific combination of the components of the original vectors.
Three-Dimensional Vectors
Three-dimensional vectors are crucial in modeling and solving problems that arise in three-dimensional space, such as those encountered in physics and engineering. These vectors provide a more comprehensive understanding of spatial relationships.
- A three-dimensional vector is expressed using three components: usually denoted as the x, y, and z components. These correspond to the vector's projection on the Cartesian coordinate axes.
- Unlike two-dimensional vectors, three-dimensional vectors enable representation and analysis of systems involving depth, not just length and width.
- Operations such as cross and dot products rely heavily on the properties of three-dimensional vectors to find solutions that accurately represent real-world phenomena.
Other exercises in this chapter
Problem 25
In Problems \(25-28\), determine whether the given vectors are linearly independent or linearly dependent. $$ \langle 4,-8\rangle,(-6,12) \text { in } R^{2} $$
View solution Problem 25
In Problems \(25-28\), find the direction cosines and direction angles of the given vector. $$ \mathbf{a}=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k} $$
View solution Problem 25
The given three points form a triangle. Determine which triangles are isosceles and which are right triangles. $$ (0,0,0),(3,6,-6),(2,1,2) $$
View solution Problem 25
Find a unit vector (a) in the same direction as \(\mathbf{a}\), and (b) in the opposite direction of \(\mathbf{a}\). \(\mathbf{a}=\langle 2,2\rangle\)
View solution