Problem 58 - A Graphical Approach to Precalculus with Limits

Question

Find the magnitude and direction angle (to the nearest tenth) for each vector. Give the measure of the direction angle as an angle in \(\left[0,360^{\circ}\right)\). $$\langle- 7,24\rangle$$

Step-by-Step Solution

Answer

Magnitude is 25, direction angle is 106.3°.

Step 1: Understand the Vector Components

The given vector is \(\langle -7, 24 \rangle\), where \(-7\) is the horizontal component and \(24\) is the vertical component.

Step 2: Calculate the Magnitude

Use the formula for the magnitude of a vector \(\langle a, b \rangle\), which is \(\sqrt{a^2 + b^2}\). For our vector, substituting the values gives us \(\sqrt{(-7)^2 + 24^2}\).

Step 3: Compute the Magnitude

Calculate \((-7)^2 = 49\) and \(24^2 = 576\). Add these results to get \(49 + 576 = 625\). Take the square root: \(\sqrt{625} = 25\). Thus, the magnitude of the vector is 25.

Step 4: Calculate the Direction Angle

Use the formula for the direction angle \(\theta\), which is \(\tan^{-1}\left(\frac{b}{a}\right)\). For our vector, this is \(\tan^{-1}\left(\frac{24}{-7}\right)\).

Step 5: Compute the Initial Angle

Compute \(\theta = \tan^{-1}(-\frac{24}{7})\), which gives an angle of approximately \(-73.7^{\circ}\).

Step 6: Adjust for Standard Position

Since the vector is in the second quadrant (negative x, positive y), add \(180^{\circ}\) to the initial angle to find the angle in standard position: \(-73.7^{\circ} + 180^{\circ} = 106.3^{\circ}\).

Step 7: Finalize the Direction

Verify that the angle falls within the required range \([0, 360^{\circ})\). In this case, \(106.3^{\circ}\) is correct and within bounds.

Key Concepts

Vector Components

Vectors are essentially quantities that have both magnitude and direction. They are often represented in a two-dimensional plane with horizontal (x-axis) and vertical (y-axis) components. The given vector is \(\langle -7, 24 \rangle\). Here,

• \(-7\) is the horizontal component, indicating movement to the left on the x-axis.
• \(24\) is the vertical component, indicating upward movement on the y-axis.

These components help us understand the vector's orientation and influence the calculations for magnitude and direction. They form the building blocks necessary to compute other attributes of the vector. Understanding vector components is crucial for applying vector operations effectively.

Magnitude Calculation

The magnitude of a vector describes its length and is calculated using the Pythagorean theorem. For a vector \(\langle a, b \rangle\), the magnitude \(M\) can be found with the formula:\[M = \sqrt{a^2 + b^2}\]For the vector \(\langle -7, 24 \rangle\):

• Calculate \((-7)^2 = 49\)
• Calculate \(24^2 = 576\)
• Add the results: \(49 + 576 = 625\)

Finally, take the square root of 625, which gives a magnitude of \(25\). The vector's magnitude is essentially its distance from the origin (0,0), indicating how "long" the vector is in the plane.

Direction Angle

Vectors not only have magnitude but also direction, described by the direction angle \(\theta\). The direction angle measures how the vector deviates from the positive x-axis. For calculation purposes, we use the arctangent of the ratio of vector components. Initially, the angle \(\theta\) is found using:\[\theta = \tan^{-1}\left(\frac{b}{a}\right)\]Given our vector \(\langle -7, 24 \rangle\):

• \(\theta = \tan^{-1}\left(\frac{24}{-7}\right)\)

This gives approximately \(-73.7^{\circ}\), a measure usually adjusted based on the vector's position in one of the four quadrants of the Cartesian plane. In our scenario, the vector is in the second quadrant, necessitating the addition of \(180^{\circ}\) for proper adjustment, resulting in a direction angle of \(106.3^{\circ}\). The direction gives us crucial information about where the vector "points" relative to the origin.

Inverse Tangent Function

The inverse tangent function, denoted as \(\tan^{-1} \) or \(\arctan \), is pivotal in determining the angle of a vector relative to the axes. It calculates the angle whose tangent is a given number. When dealing with vectors, the ratio \(\frac{b}{a}\) of the components provides input to this function. For example,

• Given \(\tan^{-1}\left(\frac{24}{-7}\right)\), the inverse tangent yields an initial angle of \(-73.7^{\circ}\).

It’s important to note that the result often depends on the quadrant. Adjustments may be necessary, as shown by adding \(180^{\circ}\) when the vector lies in the second quadrant. The inverse tangent function thus simplifies finding angles in trigonometry, aiding in understanding vector direction and orientation.