Problem 15 - Precalculus

Question

Find the magnitude and direction angle of the given vector. $$\mathbf{u}=\langle\sqrt{3}, 3\rangle$$

Step-by-Step Solution

Answer

Magnitude is \( 2\sqrt{3} \) and direction angle is \( 60^\circ \).

Step 1: Magnitude Formula

To find the magnitude of a vector \( \mathbf{u} = \langle a, b \rangle \), use the formula: \[\| \mathbf{u} \| = \sqrt{a^2 + b^2}\]. In this case, the components are \( a = \sqrt{3} \) and \( b = 3 \).

Step 2: Apply Components to Magnitude Formula

Compute the magnitude by substituting the vector's components into the formula: \[\| \mathbf{u} \| = \sqrt{(\sqrt{3})^2 + (3)^2}\].

Step 3: Calculate Each Term in Magnitude

Calculate each term in the expression: \[ (\sqrt{3})^2 = 3 \quad \text{and} \quad (3)^2 = 9 \]. This gives:\[\| \mathbf{u} \| = \sqrt{3 + 9}\].

Step 4: Solve for Magnitude

Add the terms under the square root: \( \| \mathbf{u} \| = \sqrt{12} \). Simplify to obtain:\( \| \mathbf{u} \| = 2\sqrt{3} \).

Step 5: Direction Angle Formula

To find the direction angle \( \theta \) of the vector \( \mathbf{u} = \langle a, b \rangle \), use \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \). Here, \( a = \sqrt{3} \) and \( b = 3 \).

Step 6: Apply Components to Direction Angle Formula

Substitute the components into the direction angle formula: \[ \theta = \tan^{-1}\left(\frac{3}{\sqrt{3}}\right)\].

Step 7: Simplify the Fraction

Simplify the fraction inside the arctan: \[ \frac{3}{\sqrt{3}} = \sqrt{3} \]. Then, \( \theta = \tan^{-1}(\sqrt{3}) \).

Step 8: Calculate Direction Angle

Since \( \tan(60^\circ) = \sqrt{3} \), the direction angle \( \theta \) is:\( \theta = 60^\circ \).

Key Concepts

Direction Angle

When working with vectors, the direction angle gives us the direction in which the vector is pointing. For a vector \( \mathbf{u} = \langle a, b \rangle \), the direction angle \( \theta \) is calculated using the inverse tangent function:
\( \theta = \tan^{-1}\left(\frac{b}{a}\right) \). This formula helps us find the angle relative to the positive x-axis.

In this exercise, we used the components \( a = \sqrt{3} \) and \( b = 3 \) to find that \( \theta = \tan^{-1}(\sqrt{3}) \), which simplifies to the familiar angle of \( 60^\circ \). This angle is significant as it corresponds to one of the special angles in trigonometry, making calculations simpler.

Vector Components

Vectors are represented by their components, usually in the form \( \langle a, b \rangle \). The components tell us how much the vector moves along the x-axis (\( a \)) and the y-axis (\( b \)).

Understanding vector components is crucial because they help in determining both the magnitude and direction of the vector.

• The x-component (horizontal movement): is represented by \( a \).
• The y-component (vertical movement): is represented by \( b \).

In our example, \( a = \sqrt{3} \) and \( b = 3 \), indicating the vector \( \mathbf{u} \) moves \( \sqrt{3} \) units along the x-axis and 3 units along the y-axis.

Magnitude Formula

The magnitude, or length, of a vector \( \mathbf{u} = \langle a, b \rangle \), tells us how long the vector is, regardless of its direction. To find the magnitude, we use the formula:
\[ \| \mathbf{u} \| = \sqrt{a^2 + b^2} \]
This formula derives from the Pythagorean theorem, considering the vector as the hypotenuse of a right triangle with sides \( a \) and \( b \).

Applying this to our vector with \( a = \sqrt{3} \) and \( b = 3 \), we computed:

• \( (\sqrt{3})^2 = 3 \)
• \( (3)^2 = 9 \)
• The combined value is: \( 3 + 9 = 12 \)

Hence, the magnitude is \( \| \mathbf{u} \| = \sqrt{12} = 2\sqrt{3} \). This measurement illustrates the vector's total length or distance.

Trigonometric Functions

Trigonometric functions are mathematical functions related to the angles of triangles, especially useful in vector calculations. For vectors, the trigonometric function often used is the tangent (), specifically for finding the direction angle.

The tangent function is defined as the ratio of the opposite side to the adjacent side in a right triangle, which corresponds to \( \frac{b}{a} \) in vectors. To find the angle of the vector, we use the arctangent or inverse tangent function, \( \tan^{-1} \).

• \( \tan^{-1} \) helps us find the angle given a ratio of sides.

In essence, trigonometric functions bridge geometry and algebra, allowing us to solve complex problems involving angles and magnitudes, as seen in our vector exercise.