Problem 9 - Precalculus

Question

Find the magnitude and direction angle of the given vector. $$u=(5,-1)$$

Step-by-Step Solution

Answer

Magnitude is \( \sqrt{26} \) and direction angle is approximately \( 348.69^\circ \).

Step 1: Find the Magnitude of the Vector

The magnitude of a vector \( u = (a, b) \) is calculated using the formula: \( \|u\| = \sqrt{a^2 + b^2} \). For vector \( u = (5, -1) \), substitute \( a = 5 \) and \( b = -1 \) into the formula.\[ \|u\| = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \]This gives us a magnitude of \( \sqrt{26} \).

Step 2: Determine the Direction Angle

The direction angle \( \theta \) of a vector \( u = (a, b) \) is found using the formula: \( \theta = \tan^{-1}(\frac{b}{a}) \). Here \( a = 5 \) and \( b = -1 \). Calculate:\[ \theta = \tan^{-1}\left(\frac{-1}{5}\right) = \tan^{-1}(-0.2) \]This angle \( \theta \) is the direction angle measured counter-clockwise from the positive x-axis. Use a calculator to find \( \theta \approx -11.31^\circ \). Since the angle is negative, to find a positive angle, add \( 360^irc \): \( 360 - 11.31 = 348.69^\circ \).

Key Concepts

Magnitude of a Vector

The magnitude of a vector is essentially its length. It tells you how long the vector is when measured from its start point to its endpoint. This can be particularly important in various applications, such as physics, engineering, and computer graphics. To find the magnitude, we use the formula:

• \( \|u\| = \sqrt{a^2 + b^2} \)

Where \( u \) is the vector \((a, b)\). Simply substitute the given values into the formula. For the vector \( u = (5, -1) \), you calculate as follows:

• \( \|u\| = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \)

This means the magnitude of the vector \( u \) is \( \sqrt{26} \), which is a little over 5, as \( \sqrt{25}=5 \). This understanding of magnitude helps visualize the vector's size on a graph.

Direction Angle

The direction angle of a vector gives us the orientation of the vector relative to the positive x-axis. It's crucial because it allows us to see in which direction the vector is pointing. The direction angle \( \theta \) can be found using the formula:

• \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \)

This approach uses the inverse tangent function, which helps you find an angle whose tangent is \( \frac{b}{a} \). For our vector \( u = (5, -1) \):

• \( \theta = \tan^{-1}\left(\frac{-1}{5}\right) = \tan^{-1}(-0.2) \)

Upon finding \( \theta \), you get approximately \( -11.31^\circ \). This value is negative because the vector points in the "below the x-axis" direction. To convert it to a positive angle often used in navigation and graphics, add \( 360^\circ \):

• \( 360^\circ - 11.31^\circ = 348.69^\circ \)

Trigonometric Functions

To understand vectors better, trigonometric functions are very useful. These functions relate the angles and sides of triangles and are not only pivotal in geometry, but also in analyzing vectors.

• Tangent (\( \tan \)): Used to relate the opposite side and adjacent side of a right triangle. In vectors, \( \tan θ = \frac{b}{a} \). This relationship is key in finding the direction angle of a vector.
• Inverse Tangent (\( \tan^{-1} \)): Also known as \( \arctan \), is used to find the angle \( θ \) when you know the ratio of the sides.

When we apply \( \tan^{-1} \) in the context of a vector, we are solving for the angle that would give us the slope or inclination of the vector. This is why knowing trigonometric functions enhances our capability to interpret and manipulate vectors in various fields of study.