Problem 18 - Contemporary Precalculus

Question

Find \(u+v, v-u,\) and \(2 u-3 v\). $$\mathbf{u}=-4(-\mathbf{i}+\mathbf{j}), \mathbf{v}=-3 \mathbf{i}$$

Step-by-Step Solution

Answer

Question: Given the vectors \(\mathbf{u} = -4(-\mathbf{i}+\mathbf{j})\) and \(\mathbf{v} = -3\mathbf{i}\), compute the following: \(\mathbf{u} + \mathbf{v}\), \(\mathbf{v} - \mathbf{u}\), and \(2\mathbf{u} - 3\mathbf{v}\). Answer: The vector sum \(\mathbf{u} + \mathbf{v} = (1, -4)\), the difference \(\mathbf{v} - \mathbf{u} = (-7, 4)\), and the linear combination \(2\mathbf{u} - 3\mathbf{v} = (17, -8)\).

Step 1: Define the vectors in component form

To perform the operations more easily, we should first write the vectors in component form: $$\mathbf{u}=-4(-\mathbf{i}+\mathbf{j})=(-4)(-1,1)= (4,-4)$$ $$\mathbf{v}=-3 \mathbf{i}=(-3,0)$$

Step 2: Find \(u+v\)

To find the sum of the vectors, we need to add the corresponding components of the vectors: $$\mathbf{u} + \mathbf{v} = (4,-4)+(-3,0) = (4-3, -4+0) = (1,-4)$$

Step 3: Find \(v-u\)

To find the difference of the vectors, we need to subtract the corresponding components of the vectors: $$\mathbf{v} - \mathbf{u} = (-3,0)-(4,-4) = (-3-4, 0-(-4)) = (-7,4)$$

Step 4: Find \(2u - 3v\)

First, we need to multiply the vectors by their respective scalar values and then subtract the results: $$2\mathbf{u} = 2(4,-4) = (8,-8)$$ $$3\mathbf{v} = 3(-3,0) = (-9,0)$$ Now, we can compute \(2u - 3v\): $$2\mathbf{u} - 3\mathbf{v} = (8,-8) - (-9,0) = (8-(-9), -8-0) = (17,-8)$$

Key Concepts

Component Form of Vectors

Understanding vectors in their component form is essential for performing basic vector operations such as addition, subtraction, and scalar multiplication. Visualize a vector as an arrow pointing from one point to another in space. The component form breaks this vector down into its horizontal (x-axis) and vertical (y-axis) movements.

For a two-dimensional vector, the component form is written as \( (x,y) \), where \( x \) is the horizontal movement and \( y \) is the vertical movement from the origin. If we consider vector \( \mathbf{u} \) given as \( -4(-\mathbf{i} + \mathbf{j}) \), translating this into component form results in \( (4,-4) \). The \( -\mathbf{i} \) and \( \mathbf{j} \) denote unit vectors along the x-axis and y-axis, respectively. By converting vectors into their component form before performing operations, we simplify the process and avoid confusion.

Vector Addition and Subtraction

Once vectors are in component form, vector addition and subtraction become straightforward. To add two vectors, simply add their corresponding components. For example, given vectors \( \mathbf{u} = (4,-4) \) and \( \mathbf{v} = (-3,0) \) in their component form, their sum \( \mathbf{u} + \mathbf{v} \) is found by adding the x-components and y-components separately to get \( (1,-4) \).

Subtraction, on the other hand, involves taking the difference of the corresponding components. The vector \( \mathbf{v} - \mathbf{u} \) is found by subtracting the x-component of \( \mathbf{u} \) from the x-component of \( \mathbf{v} \) and the y-component of \( \mathbf{u} \) from the y-component of \( \mathbf{v} \) to yield \( (-7,4) \). Remember that the order in subtraction matters; reversing the vectors would change the result.

Scalar Multiplication of Vectors

Scalar multiplication involves multiplying a vector by a number (a scalar). This operation scales the vector's magnitude without changing its direction when the scalar is positive, or reversing its direction in addition to scaling when the scalar is negative. For instance, if we want to multiply vector \( \mathbf{u} \) by 2, every component of \( \mathbf{u} \) is multiplied by 2. Given \( \mathbf{u} = (4,-4) \) from the example, \( 2\mathbf{u} \) yields \( (8,-8) \).

To find \( 2\mathbf{u} - 3\mathbf{v} \), calculate \( 3\mathbf{v} \) by multiplying each component of \( \mathbf{v} \) by 3, which results in \( (-9,0) \). Subtracting \( 3\mathbf{v} \) from \( 2\mathbf{u} \) involves subtracting their respective components, leading to \( (17,-8) \). Scalar multiplication is a powerful tool in scaling vectors to desired magnitudes for various mathematical and physical applications.