Question

Find the direction angle of \(\mathbf{v}\). \(\mathbf{v}=-\mathbf{i}-5 \mathbf{j}\)

Short answer

The direction angle is approximately \(258.69^{\circ}\).

Step-by-step solution

Identify the Components of the Vector

The given vector \(\textbf{v} = -\textbf{i} - 5\textbf{j}\) has components \(v_x = -1\) and \(v_y = -5\).

Write the Formula for the Direction Angle

The direction angle \(\theta\) of a vector \(\textbf{v} = (v_x, v_y)\) can be found using the formula \[\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right).\]

Substitute the Components into the Formula

Substitute \(v_x = -1\) and \(v_y = -5\) into the formula so it's \[\theta = \tan^{-1}\left(\frac{-5}{-1}\right) = \tan^{-1}(5).\]

Calculate the Arctangent

Calculate \(\theta\) using arctangent: \[\theta = \tan^{-1}(5).\]

Adjust for the Correct Quadrant

Since both components \(v_x\) and \(v_y\) are negative, the vector \(\textbf{v}\) lies in the third quadrant. Therefore, we add 180 degrees to find the actual direction angle: \[\theta = \tan^{-1}(5) + 180^{\circ}.\]

Final Answer

The direction angle \(\theta = \tan^{-1}(5) + 180^{\circ}\). Use a calculator to find \(\tan^{-1}(5)\), which is approximately \(78.69^{\circ}\). Therefore, the direction angle is \[\theta \approx 78.69^{\circ} + 180^{\circ} = 258.69^{\circ}.\]

Key concepts

vector components

To solve vector problems, we first need to identify its components. A vector in the plane is often expressed in the form of \(\textbf{v} = v_x \textbf{i} + v_y \textbf{j}\). Here, \(\textbf{i}\) and \(\textbf{j}\) are basic unit vectors in the horizontal and vertical directions respectively, while \(v_x\) and \(v_y\) are the vector's components along these axes.

In our problem, the vector \(\textbf{v} = -\textbf{i} - 5\textbf{j}\) has components \(v_x = -1\) and \(v_y = -5\). These components tell us that the vector moves one unit left and five units down.

arctangent

The arctangent function, denoted \(\tan^{-1}\), is an inverse trigonometric function used to find an angle whose tangent is a given number. For our vector, the direction angle \(\theta\) is found using the formula:
\[ \theta = \tan^{-1}\bigg(\frac{v_y}{v_x}\bigg) \]
By substituting the components, we get: \[ \theta = \tan^{-1}\bigg(\frac{-5}{-1}\bigg) = \tan^{-1}(5) \]
This calculation helps us find an angle based on the ratio of the vector's vertical to horizontal components.

quadrants in coordinate plane

Understanding the coordinate plane is crucial for determining the correct angle direction. The plane is divided into four quadrants:

• Quadrant I: \(v_x > 0, v_y > 0\)
• Quadrant II: \(v_x < 0, v_y > 0\)
• Quadrant III: \(v_x < 0, v_y < 0\)
• Quadrant IV: \(v_x > 0, v_y < 0\)

In our case, both components \(v_x = -1\) and \(v_y = -5\) are negative, which places the vector in the third quadrant. This is important for accurately determining the direction angle.

vector direction

The direction of a vector is given by its angle relative to a reference direction, usually the positive x-axis. To account for the correct quadrant when finding the direction:

• If the vector is in Quadrant I or IV, the angle found from \(\tan^{-1}\) is the actual direction angle.
• If the vector is in Quadrant II or III, we must adjust by adding 180°.

As our vector is in the third quadrant, we add 180°:
\[ \theta = \tan^{-1}(5) + 180^{\text{°}} \]

inverse trigonometric functions

Inverse trigonometric functions, like \(\tan^{-1}\), \(\frac{(\text{arcsin})}{(\text{sin}^{-1})}\), and \( \text{arccos} \)are used to find angles when we have the trigonometric ratios. For instance, \(\tan^{-1}(x)\) returns the angle whose tangent is \(x\). These functions are essential tools in vector analysis.
Using a calculator, we find:
\[ \theta \text{≈} 78.69^{\text{°}} \]
Finally, adjusting for the third quadrant, we have:
\[ \theta \text{≈} 78.69^{\text{°}} + 180^{\text{°}} \text{≈} 258.69^{\text{°}} \]