Problem 69
Question
Find the direction angle of \(\mathbf{v}\). \(\mathbf{v}=-3 \sqrt{3} \mathbf{i}+3 \mathbf{j}\)
Step-by-Step Solution
Verified Answer
The direction angle of the vector \( \mathbf{v} \) is \( 150^\circ \).
1Step 1: Understand the Vector Components
Given vector \( \mathbf{v} = -3\sqrt{3} \mathbf{i} + 3 \mathbf{j} \), identify its components. Here, the \(x\)-component (\( \mathbf{i} \)) is \(-3\sqrt{3} \), and the \(y\)-component (\( \mathbf{j} \)) is \(3\).
2Step 2: Set Up the Direction Angle Formula
The direction angle \( \theta \) of a vector can be found using the formula \( \theta = \tan^{-1}\left( \frac{y}{x} \right) \). Here, \( x = -3\sqrt{3} \) and \( y = 3 \).
3Step 3: Calculate the Tangent Ratio
Substitute the values into the formula: \( \theta = \tan^{-1}\left( \frac{3}{-3\sqrt{3}} \right) \). Simplify the ratio to \( \theta = \tan^{-1}\left( -\frac{1}{\sqrt{3}} \right) \).
4Step 4: Solve the Tangent Inverse
Recognize that \( \tan^{-1}\left( -\frac{1}{\sqrt{3}} \right) = -30^\circ \). Since the vector is in the second quadrant, add \(180^\circ\) to find the correct direction: \( \theta = 180^\circ - 30^\circ = 150^\circ \).
Key Concepts
Vector ComponentsTangent InverseDirection Angle FormulaQuadrants
Vector Components
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.
Tangent Inverse
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 \(\textbackslash\textbackslashtan^{-1}\textbackslash\textbackslashleft( -\frac{1}{\textbackslash\textbackslashsqrt\textbraceleft3\textbraceright} \right)\).
\[\theta = \text{\textbackslash\textbackslashtan^{-1}\textbackslash\textbackslashleft( \frac{y}{x} \right)}\]
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 \(\textbackslash\textbackslashtan^{-1}\textbackslash\textbackslashleft( -\frac{1}{\textbackslash\textbackslashsqrt\textbraceleft3\textbraceright} \right)\).
Direction Angle Formula
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:
\(\theta = 180^\textbackslash\textbackslashcirc - 30^\textbackslash\textbackslashcirc = 150^\textbackslash\textbackslashcirc\).
\(\theta = \textbackslash\textbackslashtan^{-1}\textbackslasheft( \frac{y}{x} \right}\)
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:
\(\theta = 180^\textbackslash\textbackslashcirc - 30^\textbackslash\textbackslashcirc = 150^\textbackslash\textbackslashcirc\).
Quadrants
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.
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