Problem 32 - Trigonometry

Question

Find the magnitude of each of the following vectors. $$\mathbf{U}=15 \mathbf{i}-8 \mathbf{j}$$

Step-by-Step Solution

Answer

The magnitude of vector \(\mathbf{U}\) is 17.

Step 1: Understand the Vector Components

Recognize that the vector \({U} = 15 \mathbf{i} - 8 \mathbf{j}\) is presented in terms of its unit vectors \(\mathbf{i}\) and \(\mathbf{j}\), with 15 being the x-component and -8 being the y-component.

Step 2: Recall the Magnitude Formula

The magnitude of a vector \(\mathbf{V} = a\mathbf{i} + b\mathbf{j}\) is calculated using the formula: \[\|\mathbf{V}\| = \sqrt{a^2 + b^2}\]Here, identify \(a = 15\) and \(b = -8\) for the vector \(\mathbf{U}\).

Step 3: Calculate the Squared Components

Square each component:- Calculate \(15^2 = 225\)- Calculate \((-8)^2 = 64\)

Step 4: Sum the Squared Components

Add the squared components together:\(225 + 64 = 289\)

Step 5: Compute the Square Root

Find the square root of the sum:\(\sqrt{289} = 17\)

Step 6: State the Magnitude

The magnitude of vector \(\mathbf{U}\) is 17.

Key Concepts

Vector Components

Vectors in mathematics are described by their components. These components are the values that specify the vector's direction and magnitude in a coordinate plane. In the given vector \(\mathbf{U} = 15 \mathbf{i} - 8 \mathbf{j}\), the vector is expressed using the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\). Unit vectors are the standard basis vectors that represent one unit in the direction of the x and y axes respectively.

• The coefficient of \(\mathbf{i}\) (here 15) is called the x-component.
• The coefficient of \(\mathbf{j}\) (here -8) is known as the y-component.

Understanding these components is essential. They help in organizing the vector along the coordinate axes, providing a clear direction and magnitude.

Unit Vectors

Unit vectors are fundamental in vector mathematics. They denote direction but have a magnitude of 1. In two-dimensional space:

• \(\mathbf{i}\) represents the unit vector along the x-axis: \(\langle 1, 0 \rangle\)
• \(\mathbf{j}\) indicates the unit vector along the y-axis: \(\langle 0, 1 \rangle\)

When constructing vectors like \(\mathbf{U}\), these unit vectors are scaled by their respective components, here 15 and -8, to give the vector its direction and length. This use of unit vectors simplifies finding products, such as dot and cross products, and assists in forming easily interpretable vector expressions.

Magnitude Formula

The magnitude of a vector quantifies its length regardless of its direction. Calculating magnitude involves the well-known Pythagorean theorem. It tells us that the magnitude of vector \(\mathbf{V} = a\mathbf{i} + b\mathbf{j}\) is determined by the formula: \[\|\mathbf{V}\| = \sqrt{a^2 + b^2}\] For our vector \(\mathbf{U}\), substituting \(a = 15\) and \(b = -8\), we use:

• \(15^2 = 225\)
• \((-8)^2 = 64\)

This formula is necessary for determining vector length, essential in physics and geometry for understanding directional quantities and constructing further vector analysis.

Square Root Calculation

The final step in finding the vector magnitude involves computing the square root of the sum of squared components. From the formula, \[a^2 + b^2 = 225 + 64 = 289\] It follows that the magnitude of \(\mathbf{U}\) is:

• \(\sqrt{289} = 17\)

Calculating the square root helps provide the exact length of the vector, translating the directional arrows into a scalar measurement. This step is crucial for transitioning from abstract vector components to concrete numerical interpretations of vector magnitude.