Problem 54
Question
Let \(\boldsymbol{u}=\langle a, b\rangle\) and \(\boldsymbol{v}=\langle c, d\rangle,\) and let \(r\) and s be scalars. Prove that the stated property holds by calculating the vector on each side of the equal sign. $$\mathbf{v}+(-\mathbf{v})=\mathbf{0}$$
Step-by-Step Solution
Verified Answer
Question: Prove that the property 𝐯 + (−𝐯) = 𝐨 (vector o) holds for a given vector 𝑣 = ⟨c, d⟩.
Answer: By adding vector 𝑣 and its negation, we found that (𝐯 + (-𝐯)) = ⟨0, 0⟩, which is the zero vector. This proves that the property 𝐯 + (−𝐯) = 𝐨 holds true for the given vector.
1Step 1: Identify the given vectors
We are given the vectors 𝑢 = ⟨a, b⟩, 𝑣 = ⟨c, d⟩, r and s as scalars. We will only use 𝑣 = ⟨c, d⟩ in this exercise since the property we're proving only involves it.
2Step 2: Find the negation of vector 𝑣
To find the negation of vector 𝑣, we need to negate each of its components:
(-𝐯) = ⟨-c, -d⟩
3Step 3: Add vector 𝑣 and its negation
Now we need to add vector 𝑣 and its negation to see if it results in the zero vector. To add two vectors, we need to add their corresponding components:
(𝐯 + (-𝐯)) = ⟨c + (-c), d + (-d)⟩
4Step 4: Simplify the addition
We can simplify the addition as:
⟨c - c, d - d⟩ = ⟨0, 0⟩
5Step 5: Conclude the statement
Since the result of adding vector 𝑣 and its negation is ⟨0, 0⟩, which is the zero vector, we can conclude that the property:
(𝐯 + (-𝐯)) = 𝐨
has been successfully proven.
Key Concepts
Vector NegationZero VectorScalar Multiplication
Vector Negation
Vector negation is an operation that takes a vector and produces another vector pointing in the opposite direction. In essence, negating a vector \textbf{v} means creating a vector \textbf{-v} such that each component of the original vector is multiplied by -1. When you have a vector, say \textbf{v} = \(\langle c, d \rangle\), its negation is simply \(\langle -c, -d \rangle\). This operation is crucial in the world of vector mathematics as it allows us to effectively 'subtract' vectors.
Understanding vector negation is fundamental when working with vector addition and subtraction. It helps to visualize vector negation in terms of movement: if one step forward is represented by the vector \textbf{v}, then one step backward is represented by \textbf{-v}. When you add a vector to its negation, \(\mathbf{v} + (-\mathbf{v})\), you effectively end up with no movement at all, which leads us into our next concept, the zero vector.
Understanding vector negation is fundamental when working with vector addition and subtraction. It helps to visualize vector negation in terms of movement: if one step forward is represented by the vector \textbf{v}, then one step backward is represented by \textbf{-v}. When you add a vector to its negation, \(\mathbf{v} + (-\mathbf{v})\), you effectively end up with no movement at all, which leads us into our next concept, the zero vector.
Zero Vector
The zero vector, often denoted as \(\mathbf{0}\) or simply 0, acts as the identity element in vector addition. It can be imagined as a vector with no direction or magnitude, essentially a point in space or the 'origin' in a coordinate system. For any vector \textbf{v}, when you add it to the zero vector, you get the vector \textbf{v} back: \(\mathbf{v} + \mathbf{0} = \mathbf{v}\). Similarly, when you add a vector to its negation, as shown in our exercise, you get the zero vector: \(\mathbf{v}+(-\mathbf{v}) = \mathbf{0}\).
The concept of the zero vector is a cornerstone in vector spaces since it represents the absence of any vector quantity. In physics, it might represent a state of no velocity or no force applied. The property that adding a vector to its negation results in the zero vector demonstrates the existence of an additive inverse for every vector, fulfilling one of the key requirements for a mathematical structure to be a vector space.
The concept of the zero vector is a cornerstone in vector spaces since it represents the absence of any vector quantity. In physics, it might represent a state of no velocity or no force applied. The property that adding a vector to its negation results in the zero vector demonstrates the existence of an additive inverse for every vector, fulfilling one of the key requirements for a mathematical structure to be a vector space.
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar (a real number), which changes the magnitude of the vector, and possibly its direction if the scalar is negative. When you multiply a vector \textbf{v} by a scalar r, each component of the vector gets multiplied by r, resulting in a new vector \(r\mathbf{v}\). If \textbf{v} is \(\langle c, d \rangle\) and r is a real number, then scalar multiplication gives us \(r\mathbf{v} = \langle rc, rd \rangle\).
Through scalar multiplication, you can stretch or compress vectors, or even reverse their direction. This operation is a fundamental building block in vector algebra and has applications in areas such as physics, where it might represent scaling a velocity or force. It's also integral to the process of creating linear combinations of vectors, which in turn are key in understanding the span and linear independence, concepts that are essential to the study of vector spaces and linear algebra.
Through scalar multiplication, you can stretch or compress vectors, or even reverse their direction. This operation is a fundamental building block in vector algebra and has applications in areas such as physics, where it might represent scaling a velocity or force. It's also integral to the process of creating linear combinations of vectors, which in turn are key in understanding the span and linear independence, concepts that are essential to the study of vector spaces and linear algebra.
Other exercises in this chapter
Problem 53
In Exercises \(53-64,\) perform the indicated multiplication or division. Express your answer in both polar form \(r(\cos \theta+i \sin \theta)\) and rectangula
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A child pulls a wagon along a level sidewalk by exerting a force of 18 pounds on the wagon handle, which makes an angle of \(25^{\circ}\) with the horizontal. H
View solution Problem 54
In Exercises \(53-64,\) perform the indicated multiplication or division. Express your answer in both polar form \(r(\cos \theta+i \sin \theta)\) and rectangula
View solution Problem 55
Let \(\boldsymbol{u}=\langle a, b\rangle\) and \(\boldsymbol{v}=\langle c, d\rangle,\) and let \(r\) and s be scalars. Prove that the stated property holds by c
View solution