Problem 54

Question

In Exercises 53–56, let. $$\mathbf{u}=-2 \mathbf{i}+3 \mathbf{j}, \mathbf{v}=6 \mathbf{i}-\mathbf{j}, \mathbf{w}=-3 \mathbf{i}$$ Find each specified vector or scalar. $$3 \mathbf{u}-(4 \mathbf{v}-\mathbf{w})$$

Step-by-Step Solution

Verified
Answer
The specified vector is \(-33\mathbf{i} + 13\mathbf{j}\)
1Step 1: Identify the vectors
From the exercise, the vectors given are \(\mathbf{u}=-2 \mathbf{i}+3 \mathbf{j}, \mathbf{v}=6 \mathbf{i}-\mathbf{j}, \mathbf{w}=-3 \mathbf{i}\).
2Step 2: Distribute the scalar over the vector components
Calculate \(3 \mathbf{u}\). This involves multiplying each component of the vector \(\mathbf{u}\) by 3. So \(3 \mathbf{u} = 3(-2\mathbf{i} + 3\mathbf{j}) = -6\mathbf{i} + 9\mathbf{j}\). Next, calculate \(4 \mathbf{v}\). So \(4 \mathbf{v} = 4(6\mathbf{i} - \mathbf{j}) = 24\mathbf{i} - 4\mathbf{j}\).
3Step 3: Perform the subtraction in the parenthesis
The next operation is to subtract \(\mathbf{w}\) from \(4 \mathbf{v}\). So, \((4 \mathbf{v} - \mathbf{w}) = (24\mathbf{i} - 4\mathbf{j}) - (-3\mathbf{i}) = 27\mathbf{i} - 4\mathbf{j}\).
4Step 4: Final Vector Subtraction
The last operation is to subtract the result from Step 3 from \(3 \mathbf{u}\). So, \(3 \mathbf{u} - (4 \mathbf{v} - \mathbf{w}) = -6\mathbf{i} + 9\mathbf{j} - (27\mathbf{i} - 4\mathbf{j}) = -33\mathbf{i} + 13\mathbf{j}\).

Key Concepts

Vector AdditionVector SubtractionScalar MultiplicationLinear Algebra
Vector Addition
Vector addition is a fundamental operation in linear algebra. It involves adding corresponding components of two vectors. For instance, if you have two vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j}\), the sum \(\mathbf{a} + \mathbf{b}\) is calculated as:
  • Add the \(\mathbf{i}\) components: \(a_1 + b_1\)
  • Add the \(\mathbf{j}\) components: \(a_2 + b_2\)
This results in a new vector \((a_1 + b_1)\mathbf{i} + (a_2 + b_2)\mathbf{j}\).
Understanding how vector addition works assists in visualizing how different forces or directions come together in physics or engineering contexts. It's a straightforward tool for combining various influences acting in different directions.
Vector Subtraction
Vector subtraction is similar to vector addition, but instead of adding the corresponding components, you subtract them. If you have vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j}\), subtracting results in:
  • \(\mathbf{i}\) components: \(a_1 - b_1\)
  • \(\mathbf{j}\) components: \(a_2 - b_2\)
Resulting in the vector \((a_1 - b_1)\mathbf{i} + (a_2 - b_2)\mathbf{j}\).
Subtraction is essentially treated like addition of a negative vector. It's crucial in determining differences between vectors, such as in displacement in physics or changes in certain conditions in engineering.
Scalar Multiplication
Scalar multiplication involves multiplying each component of a vector by a scalar (a constant). Given a vector \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}\) and a scalar \(k\), scalar multiplication produces a new vector by:
  • Multiplying \(a_1\) by \(k\) for the \(\mathbf{i}\) component
  • Multiplying \(a_2\) by \(k\) for the \(\mathbf{j}\) component
Which results in the vector \(k \mathbf{a} = (k \cdot a_1)\mathbf{i} + (k \cdot a_2)\mathbf{j}\).
Scalar multiplication scales the vector's magnitude without changing its direction, making it a fundamental tool in physics to scale forces or velocities.
Linear Algebra
Linear algebra is a branch of mathematics dealing with vectors and operations on vectors, along with other algebraic structures like matrices. It provides foundational tools for modeling multidimensional systems that are prevalent in physical sciences, engineering, and economics.
It focuses on concepts such as:
  • Vector spaces: Sets where vectors can be added and scaled.
  • Transformations: Functions that map vectors to other vectors, often described using matrices.
  • Systems of linear equations: Equations that can be represented and solved using matrix operations.
Understanding linear algebra is crucial for solving practical problems involving multiple variables, including those involving vector operations. It provides techniques for systematic approaches to challenges such as optimizing functions, rotating axes, or finding equilibrium states in dynamic systems.