91Ó°ÊÓ

Vectors can be broken down into their components to make it easier to handle and understand their direction and magnitude. A vector has two main parts: the x-component and the y-component. These components are represented as coefficients in front of the unit vectors \(\textbf{i}\) and \(\textbf{j}\) respectively. For example, in the given vector \(\textbf{v} = -3\textbackslash\textbackslashsqrt\textbraceleft3\textbraceright \textbackslash\textbackslashmathbf\textbracelefti\textbraceright + 3 \textbackslash\textbackslashmathbf\textbraceleft j\textbraceright\), the x-component is -3\(\textbackslash\textbackslashsqrt\textbraceleft3\textbraceright\) and the y-component is 3. Understanding these components will help in calculating the direction angle of the vector.

The tangent inverse function, denoted as \(\textbackslash\textbackslashtan^{-1}\) or arctan, is crucial in finding the direction angle of a vector. Given a right triangle where the opposite side is the \(\text{y-component}\) and the adjacent side is the \(\text{x-component}\), you can find the angle \(\textbackslash\textbackslashtheta\) using the formula:

\[\theta = \text{\textbackslash\textbackslashtan^{-1}\textbackslash\textbackslashleft( \frac{y}{x} \right)}\]

For our vector \(\textbf{v}\), we substitute the components to get:

\(\textbackslash\textbackslashtheta = \textbackslash\textbackslashtan^{-1}\textbackslash\textbackslashleft( \frac{3}{-3\textbackslash\textbackslashsqrt\textbraceleft3\textbraceright} \right}\)

This simplifies to finding \(\textbackslash\textbackslashtan^{-1}\textbackslash\textbackslashleft( -\frac{1}{\textbackslash\textbackslashsqrt\textbraceleft3\textbraceright} \right)\).

The direction angle of a vector, known as the angle it makes with the positive x-axis, can be determined using the direction angle formula. This formula is:

\(\theta = \textbackslash\textbackslashtan^{-1}\textbackslasheft( \frac{y}{x} \right}\)

After substituting the values, if the angle is negative or doesn't match the expected quadrant, you might need to adjust it. For our vector's components, substituting gives:

\(\theta = \textbackslash\textbackslashtan^{-1}\textleft( -\frac{1}{\textbackslash\textbackslashsqrt\textleft{3\textbraceright} \right)}\)

This calculation leads to\(\theta = -30^\textbackslash\textbackslashcirc \). Since the vector is in the second quadrant, 180° is added:

\(\theta = 180^\textbackslash\textbackslashcirc - 30^\textbackslash\textbackslashcirc = 150^\textbackslash\textbackslashcirc\).

The coordinate plane is divided into four sections called quadrants. These help in determining the correct direction angle of a vector. The quadrants are:

• First Quadrant: both x and y-components are positive.
• Second Quadrant: x-component is negative, and y-component is positive.
• Third Quadrant: both x and y-components are negative.
• Fourth Quadrant: x-component is positive, and y-component is negative.

For our case, since the vector's x-component is \(-3\textbackslash\textbackslashsqrt\textbraceleft3\textbraceright\) (negative) and the y-component is 3 (positive), the vector lies in the Second Quadrant. Therefore, the direction angle is adjusted by adding 180° to the calculated angle, leading to the final angle being 150°.