Problem 41 - Trigonometry

Question

For each pair of vectors, find \(\mathbf{U}+\mathbf{V}, \mathbf{U}-\mathbf{V}\), and \(2 \mathbf{U}-3 \mathbf{V}\). $$\mathbf{U}=\langle-5,0\rangle, \mathbf{V}=\langle 0,1\rangle$$

Step-by-Step Solution

Answer

\( \mathbf{U} + \mathbf{V} = \langle -5, 1 \rangle \); \( \mathbf{U} - \mathbf{V} = \langle -5, -1 \rangle \); \( 2\mathbf{U} - 3\mathbf{V} = \langle -10, -3 \rangle \)."

Step 1: Calculate \( \mathbf{U} + \mathbf{V} \)

To find \( \mathbf{U} + \mathbf{V} \), we need to add the corresponding components of \( \mathbf{U} \) and \( \mathbf{V} \). Given \( \mathbf{U} = \langle -5, 0 \rangle \) and \( \mathbf{V} = \langle 0, 1 \rangle \): \[ \mathbf{U} + \mathbf{V} = \langle -5 + 0, 0 + 1 \rangle = \langle -5, 1 \rangle \].

Step 2: Calculate \( \mathbf{U} - \mathbf{V} \)

To find \( \mathbf{U} - \mathbf{V} \), subtract the components of \( \mathbf{V} \) from \( \mathbf{U} \):\[ \mathbf{U} - \mathbf{V} = \langle -5 - 0, 0 - 1 \rangle = \langle -5, -1 \rangle \].

Step 3: Calculate \( 2\mathbf{U} - 3\mathbf{V} \)

For this, first multiply \( \mathbf{U} \) by 2 and \( \mathbf{V} \) by 3:\[ 2\mathbf{U} = 2\langle -5, 0 \rangle = \langle -10, 0 \rangle \]\[ 3\mathbf{V} = 3\langle 0, 1 \rangle = \langle 0, 3 \rangle \]Now subtract the vectors:\[ 2\mathbf{U} - 3\mathbf{V} = \langle -10 - 0, 0 - 3 \rangle = \langle -10, -3 \rangle \].

Key Concepts

Vector Operations

Vector operations are fundamental actions that can be performed on vectors, such as addition and subtraction. These operations allow us to combine and manipulate vectors in various ways, helping us analyze and understand different physical and mathematical scenarios.

Vectors, which are mathematical tools used to represent quantities with both magnitude and direction, can be added together by summing their respective components. To demonstrate, consider two vectors, \( \mathbf{U} = \langle -5, 0 \rangle \) and \( \mathbf{V} = \langle 0, 1 \rangle \). When you add these two vectors, you simply add their respective components:

• \( \mathbf{U} + \mathbf{V} = \langle -5 + 0, 0 + 1 \rangle = \langle -5, 1 \rangle \).

Subtraction of vectors follows a similar approach but instead involves subtracting the corresponding components:

• \( \mathbf{U} - \mathbf{V} = \langle -5 - 0, 0 - 1 \rangle = \langle -5, -1 \rangle \).

These simple operations are used across physics, engineering, and computer graphics, enabling the manipulation of quantities such as force, velocity, and more.

Scalar Multiplication of Vectors

Scalar multiplication involves multiplying a vector by a scalar (a constant number), which scales the magnitude of the vector while keeping its direction intact. This is a key operation in vector mathematics, allowing us to adjust the length of a vector.

When a vector \( \mathbf{U} = \langle -5, 0 \rangle \) is multiplied by a scalar such as 2, each component of the vector is multiplied by this scalar value:

• \( 2\mathbf{U} = 2 \langle -5, 0 \rangle = \langle -10, 0 \rangle \).

Similarly, for another vector \( \mathbf{V} = \langle 0, 1 \rangle \) and a scalar of 3:

• \( 3\mathbf{V} = 3 \langle 0, 1 \rangle = \langle 0, 3 \rangle \).

Understanding scalar multiplication is crucial as it leads to other operations, such as vector projection and scaling transformations. This fundamental process swaps real-world scenarios, such as doubling a force or changing speed, into a calculable format.

Algebra of Vectors

The algebra of vectors involves the set of rules governing the manipulation of vectors, such as addition, subtraction, scalar multiplication, and dot products. These algebraic operations are essential in solving equations involving vectors and understanding geometry in higher dimensions.

In practical scenarios, these operations allow us to evaluate expressions combining multiple vector operations. For example, a complex expression, like \( 2\mathbf{U} - 3\mathbf{V} \), can be broken down into manageable steps using algebraic rules.

• Start by performing scalar multiplication on each vector:
• \( 2\mathbf{U} = \langle -10, 0 \rangle \)
• \( 3\mathbf{V} = \langle 0, 3 \rangle \)
• The next step is subtracting the resulting vectors:
• \( 2\mathbf{U} - 3\mathbf{V} = \langle -10, 0 \rangle - \langle 0, 3 \rangle = \langle -10, -3 \rangle \)

These components exemplify how vector algebra works in step-by-step problem-solving, providing a solid foundation for tackling more complex mathematical challenges involving vectors.