Problem 65
Question
Approximate the magnitude of each vector and the angle \(\theta, 0^{\circ} \leq \theta<360^{\circ}\), that the vector makes with the positive \(x\)-axis. Round your answers to the nearest tenth. $$\mathbf{W}=\sqrt{5} \mathbf{i}-2 \mathbf{j}$$
Step-by-Step Solution
Verified Answer
Magnitude: 3, Angle: 318.2°
1Step 1: Write the vector in component form
The vector \( \mathbf{W} \) is given as \( \mathbf{W} = \sqrt{5} \mathbf{i} - 2 \mathbf{j} \). This corresponds to the components \( W_x = \sqrt{5} \) and \( W_y = -2 \).
2Step 2: Calculate the magnitude of the vector
The magnitude of the vector \( \mathbf{W} \) is calculated using the formula: \[ |\mathbf{W}| = \sqrt{W_x^2 + W_y^2} = \sqrt{(\sqrt{5})^2 + (-2)^2} = \sqrt{5 + 4} = \sqrt{9} = 3. \] Thus, the magnitude is \( 3 \).
3Step 3: Calculate the angle with the positive x-axis
To find \( \theta \), use the formula \( \theta = \tan^{-1} \left( \frac{W_y}{W_x} \right) \). Substitute the values: \( \theta = \tan^{-1} \left( \frac{-2}{\sqrt{5}} \right) \). Calculate this using a calculator: \( \theta \approx \tan^{-1}(-0.8944) \approx -41.8^{\circ} \).
4Step 4: Adjust the angle to fit the range \(0^{\circ} \leq \theta < 360^{\circ}\)
The angle \( -41.8^{\circ} \) is less than \( 0^{\circ} \) and needs to be adjusted by adding \( 360^{\circ} \). Therefore, \( \theta = 360^{\circ} - 41.8^{\circ} = 318.2^{\circ} \). This angle is now within the specified range.
Key Concepts
Understanding the Magnitude of a VectorCalculating the Angle a Vector Makes with the x-axisTrigonometric Calculations for Vectors
Understanding the Magnitude of a Vector
The magnitude of a vector provides a way to describe its size or length in space. For any vector with components, such as \[ \mathbf{W} = \sqrt{5} \mathbf{i} - 2 \mathbf{j}, \]we can find the magnitude using the formula: \[ |\mathbf{W}| = \sqrt{W_x^2 + W_y^2}, \]where \( W_x \) and \( W_y \) are the components along the x and y axes.
Considering the components:
Considering the components:
- \( W_x = \sqrt{5} \)
- \( W_y = -2 \)
Calculating the Angle a Vector Makes with the x-axis
The angle a vector makes with the positive x-axis can tell us about its direction in the coordinate plane. To find this angle \( \theta \), we need to understand how the vector points relative to the x-axis. This involves using the arctangent function, \( \tan^{-1} \), which can calculate angles based on given ratios.
The formula is: \[ \theta = \tan^{-1}\left(\frac{W_y}{W_x}\right). \]In our case:
The formula is: \[ \theta = \tan^{-1}\left(\frac{W_y}{W_x}\right). \]In our case:
- \( W_x = \sqrt{5} \)
- \( W_y = -2 \)
Trigonometric Calculations for Vectors
Trigonometry plays a vital role in vector analysis, especially when determining angles or directional properties. Trigonometric functions, such as sine, cosine, and tangent, allow us to translate vector components into meaningful angles.
Here's how they come into play:
Properly adjusting angles is crucial when the vector's components change direction (like negative \( y \)-values). Adjustments by adding or subtracting full rotations ensure the angle remains within a standard 0 to 360-degree range. Understanding these trigonometric details assists in comprehensively analyzing vectors and their orientations.
Here's how they come into play:
- **Tangent Function**: \( \frac{\text{opposite}}{\text{adjacent}} \), reflecting the ratio needed to find \( \theta \).
- **Arctangent Function**: \( \tan^{-1} \) is used to find an angle from a tangent ratio.
Properly adjusting angles is crucial when the vector's components change direction (like negative \( y \)-values). Adjustments by adding or subtracting full rotations ensure the angle remains within a standard 0 to 360-degree range. Understanding these trigonometric details assists in comprehensively analyzing vectors and their orientations.
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