Problem 49
Question
In Problems 45-50, give a proof of the indicated property for two-dimensional vectors. Use \(\mathbf{u}=\left\langle u_{1}, u_{2}\right\rangle, \mathbf{v}=\left\langle v_{1}, v_{2}\right\rangle\), and \(\mathbf{w}=\left\langle w_{1}, w_{2}\right\rangle\). $$ \mathbf{0} \cdot \mathbf{u}=0 $$
Step-by-Step Solution
Verified Answer
The dot product of the zero vector with any vector is zero: \( \mathbf{0} \cdot \mathbf{u} = 0 \).
1Step 1: Understanding the Problem
We are tasked to prove that the dot product of the zero vector \( \mathbf{0} \) with any vector \( \mathbf{u} \) in two dimensions results in zero: \( \mathbf{0} \cdot \mathbf{u} = 0 \). Let's break this down using the properties of vectors.
2Step 2: Defining the Zero Vector
The zero vector in two-dimensional space is defined as \( \mathbf{0} = \langle 0, 0 \rangle \). It has no direction and its magnitude is zero.
3Step 3: Defining the Dot Product
For two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is defined as: \[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2. \]
4Step 4: Applying Dot Product to Zero Vector and \(\mathbf{u}\)
Let the vector \( \mathbf{u} = \langle u_1, u_2 \rangle \). The dot product \( \mathbf{0} \cdot \mathbf{u} \) is calculated as: \[ \mathbf{0} \cdot \mathbf{u} = \langle 0, 0 \rangle \cdot \langle u_1, u_2 \rangle = 0 \cdot u_1 + 0 \cdot u_2. \] This simplifies to \( 0 + 0 \), which is equal to 0.
5Step 5: Conclusion
We have shown that the dot product of the zero vector with any vector \( \mathbf{u} \) results in zero by definition of the dot product and the properties of multiplication with zero. Therefore, the property \( \mathbf{0} \cdot \mathbf{u} = 0 \) is proven.
Key Concepts
Dot ProductZero VectorVector ProofTwo-Dimensional Vectors
Dot Product
The dot product is a fundamental operation in vector calculus. It is the result of multiplying two vectors, resulting not in a vector but in a scalar quantity. For two-dimensional vectors, the dot product of \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) is calculated as:
- \( a_1 \times b_1 \)
- plus \( a_2 \times b_2 \)
Zero Vector
The zero vector is a unique vector in mathematics and physics. It is represented by \( \mathbf{0} \) or \( \langle 0, 0 \rangle \) in two-dimensional space. This vector:
- Has no magnitude
- And no direction
Vector Proof
A vector proof involves using vector properties and operations to demonstrate a particular truth. In the exercise, we are tasked to prove the property:\[ \mathbf{0} \cdot \mathbf{u} = 0 \]Proofs in vector calculus often rely on breaking down operations to fundamental definitions. This particular proof employs the definition of the zero vector and the basic properties of multiplication. By carefully applying these definitions, we can confidently show that any two-dimensional vector, when `dotted` with the zero vector, always equals zero. This is a simple yet powerful demonstration of how arithmetic properties extend into the realm of vectors.
Two-Dimensional Vectors
Two-dimensional vectors are represented in the form \( \langle u_1, u_2 \rangle \), where \( u_1 \) and \( u_2 \) are components along the \( x \)- and \( y \)-axes, respectively. These vectors:
- Can be visualized as arrows in a plane
- Have both magnitude and direction
Other exercises in this chapter
Problem 48
In Problems 45-50, give a proof of the indicated property for two-dimensional vectors. Use \(\mathbf{u}=\left\langle u_{1}, u_{2}\right\rangle, \mathbf{v}=\left
View solution Problem 48
Find the tangential and normal components \(\left(a_{T}\right.\) and \(a_{N}\) ) of the acceleration vector at \(t\). Then evaluate at \(t=t_{1}\). \(x=t, y=t^{
View solution Problem 49
Find the tangential and normal components \(\left(a_{T}\right.\) and \(a_{N}\) ) of the acceleration vector at \(t\). Then evaluate at \(t=t_{1}\). \(x=e^{-t},
View solution Problem 50
In Problems 45-50, give a proof of the indicated property for two-dimensional vectors. Use \(\mathbf{u}=\left\langle u_{1}, u_{2}\right\rangle, \mathbf{v}=\left
View solution