91Ó°ÊÓ

A vector is defined by its components, which can be thought of as its coordinates in a Cartesian plane. For a 2-dimensional vector, these components are shown as \(\textbf{v} = \langle x, y \rangle\).
The 'x-component' represents how far the vector moves along the x-axis, while the 'y-component' shows the movement along the y-axis. In the given problem, the vector \(\textbf{v} = \langle 8, -8 \sqrt{3} \rangle\) has:

• x-component: 8
• y-component: -8 \sqrt{3}

Recognizing these components is the first step in understanding the full properties of a vector, such as its magnitude and direction.

The magnitude of a vector gives us an idea of the vector's length and is always a positive quantity. You can calculate the magnitude of a vector \(\textbf{v} = \langle x, y \rangle\) using the Pythagorean theorem formula: \| \textbf{v} \| = \sqrt{x^2 + y^2}
For the vector \(\textbf{v} = \langle 8, -8 \sqrt{3} \rangle\), substitute the components into the formula: \ \sqrt{8^2 + (-8 \sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16
So, the magnitude of this vector is 16. This result tells us that the 'length' of the vector in the plane is 16 units.

The direction angle of a vector describes the angle it makes relative to the positive x-axis. For a vector \(\textbf{v} = \langle a, b \rangle\), this angle \(\theta\) can be found using the tangent function: \tan( \theta ) = \frac{b}{a}
Plug in the components of the vector to find the angle: \ \tan( \theta ) = \frac{-8 \sqrt{3}}{8} = - \sqrt{3}
Using the inverse trigonometric function, we have: \ \theta = \tan^{-1}(- \sqrt{3}) \approx -60^{\text{\degree}}
Since our vector is located in the fourth quadrant due to its positive x-component and negative y-component, we need to adjust the angle to find the correct direction:
\360^{\text{\degree}} - 60^{\text{\degree}} = 300^{\text{\degree}}
Thus, the direction angle is 300 degrees.

Inverse trigonometric functions are crucial when finding angles from given trigonometric ratios. The three main inverse trigonometric functions are \(\tan^{-1}\), \(\text{sin}^{-1}\), and \(\text{cos}^{-1}\). In this problem, we used the inverse tangent function to find the direction angle.
\(\tan( \theta ) = \frac{b}{a}\) allows us to find \(\theta\) when we know the lengths of the opposite side (y-component) and adjacent side (x-component).
So for \(\tan^{-1}( \frac{b}{a} )\), we solved: \ \theta = \tan^{-1}( - \sqrt{3} )
This function gave us an angle in the fourth quadrant (\ \approx -60^{\text{\degree}}), and we adjusted it based on vector direction to get a positive angle (300 degrees). These functions are fundamental tools for transitioning between trigonometric ratios and angles.