Problem 11 - College Trigonometry

Question

Find the component form of the vector \(\vec{v}\) using the information given about its magnitude and direction. Give exact values. \(\|\vec{v}\|=6 ;\) when drawn in standard position \(\vec{v}\) lies in Quadrant I and makes a \(60^{\circ}\) angle with the positive \(x\) -axis

Step-by-Step Solution

Answer

The component form is \( (3, 3\sqrt{3}) \).

Step 1: Understand the problem

We need to find the component form of vector \( \vec{v} \) given its magnitude \( \|\vec{v}\| = 6 \), and that it forms a \( 60^{\circ} \) angle with the positive x-axis in Quadrant I.

Step 2: Use the formula for vector components

A vector \( \vec{v} \) with a magnitude of \( \|\vec{v}\| \) and an angle \( \theta \) with the positive x-axis has components \((\|\vec{v}\| \cos \theta, \|\vec{v}\| \sin \theta)\).

Step 3: Calculate the x-component

The x-component of \( \vec{v} \) is calculated as \( x = \|\vec{v}\| \cos \theta = 6 \cos 60^{\circ} \). We know that \( \cos 60^{\circ} = \frac{1}{2} \), so \( x = 6 \times \frac{1}{2} = 3 \).

Step 4: Calculate the y-component

The y-component of \( \vec{v} \) is calculated as \( y = \|\vec{v}\| \sin \theta = 6 \sin 60^{\circ} \). We know that \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \), so \( y = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \).

Step 5: Write the component form of the vector

Thus, the component form of the vector \( \vec{v} \) is \( \vec{v} = (3, 3\sqrt{3}) \).

Key Concepts

Magnitude of a Vector

Vectors are essential in physics and mathematics, helping us represent quantities that have both magnitude and direction. The magnitude of a vector, often described as its length, reveals how "long" the vector is. It is a non-negative value and provides insight into the vector's strength. For a vector \( \vec{v} \) given by components \( (x, y) \), the magnitude is calculated using the Pythagorean theorem, expressed as:

• \( \|\vec{v}\| = \sqrt{x^2 + y^2} \)

In this exercise, the magnitude of vector \( \vec{v} \) is given as 6. Hence, any component form we find should ideally be consistent with this magnitude. The magnitude connects to other key vector features, such as its orientation in space and direction.Non-zero magnitudes confirm the presence and influence of vectors, highlighting their role in many fields, from engineering to computer graphics.

Angles in Coordinate Plane

Angles are crucial to understanding vectors. In the coordinate plane, they determine the vector's direction relative to the axes. An angle is measured starting from the positive x-axis, moving counterclockwise. The specific angle can tell us the vector's orientation and, when combined with its magnitude, provide a complete description. For instance, a vector in this exercise makes a \( 60^{\circ} \) angle with the positive x-axis, placing it in a specific orientation. The angle helps in calculating vector components, key in expressing vectors in component form. By defining the angle \( \theta \), we align the vector's representation with its physical meaning, making computations and visualizations intuitive and predictable. The angle lets us access the trigonometric ratios needed for translating the vector into its components, which blend into practical uses like navigation, graphic distortions, and mechanical rotations.

Trigonometric Functions in Vector Analysis

Trigonometric functions are core to breaking down vectors into manageable pieces—the "x" and "y" components. These components make using vectors in calculations straightforward. When a vector forms an angle \( \theta \) with the positive x-axis, the trigonometric functions \( \cos \theta \) and \( \sin \theta \) become essential tools. They relate the vector's components to its magnitude. Specifically:

• The x-component: \( x = \|\vec{v}\| \cos \theta \)
• The y-component: \( y = \|\vec{v}\| \sin \theta \)

These functions translate the vector's direction into concrete values. In our scenario, with an angle of \( 60^{\circ} \), the cosine and sine values are readily computed (\( \frac{1}{2} \) and \( \frac{\sqrt{3}}{2} \), respectively). This relation underlines how trigonometric functions bridge the gap between a vector's positional angle and its real-world significance in tasks like force distribution and signal path delineation.

Quadrant of a Vector

The quadrant in which a vector lies provides immediate insight into the sign and general direction of its components. The Cartesian plane is divided into four quadrants:

• Quadrant I: Both x and y are positive.
• Quadrant II: x is negative, y is positive.
• Quadrant III: Both x and y are negative.
• Quadrant IV: x is positive, y is negative.

In the given problem, vector \( \vec{v} \) is in Quadrant I. Knowing the quadrant not only helps ascertain the signs of the vector's components (positive in both cases for Quadrant I) but also immediately provides an intuitive check when calculating components from the magnitude and angle. In the context of vector analysis, understanding the quadrant simplifies decision-making on component validity, helping ensure accurate interpretations and calculations in real-world applications like navigation systems and graphical representations.