Problem 16 - College Trigonometry

Question

Find the component form of the vector \(\vec{v}\) using the information given about its magnitude and direction. Give exact values. \(\|\vec{v}\|=2 \sqrt{3} ;\) when drawn in standard position \(\vec{v}\) lies in Quadrant II and makes a \(30^{\circ}\) angle with the positive \(y\) -axis

Step-by-Step Solution

Answer

The vector \(\vec{v}\) in component form is \((-\sqrt{3}, 3)\).

Step 1: Understanding Vector Components in Quadrant II

In Quadrant II, vectors have a negative x-component and a positive y-component. Since the vector makes a \(30^\circ\) angle with the positive y-axis, we need to translate this angle to the relevant x-axis angle. The angle from the positive x-axis in standard position, counterclockwise, would be \(90^\circ + 30^\circ = 120^\circ\).

Step 2: Apply Trigonometric Functions for Components

For a vector \(\vec{v}\) making an angle \(\theta\) with the positive x-axis, and with magnitude \(\|\vec{v}\|\), the component form is given by \(\vec{v} = (\|\vec{v}\| \cos(\theta), \|\vec{v}\| \sin(\theta))\). In this case, \(\theta = 120^\circ\) and \(\|\vec{v}\| = 2\sqrt{3}\).

Step 3: Calculate X-Component Using Cosine

The x-component is found using \(\vec{v}_x = \|\vec{v}\| \cos(\theta)\). Calculate \(\cos(120^\circ)\), which is \(-\frac{1}{2}\). Thus, \(\vec{v}_x = 2\sqrt{3} \times (-\frac{1}{2}) = -\sqrt{3}\).

Step 4: Calculate Y-Component Using Sine

The y-component is found using \(\vec{v}_y = \|\vec{v}\| \sin(\theta)\). Calculate \(\sin(120^\circ)\), which is \(\frac{\sqrt{3}}{2}\). Thus, \(\vec{v}_y = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3\).

Step 5: Combine Components into Vector Form

Combining the calculated components, the vector in component form is \(\vec{v} = (-\sqrt{3}, 3)\).

Key Concepts

Quadrant System

When working with vectors in coordinate geometry, understanding the quadrant system is essential. The coordinate plane is divided into four quadrants:

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

Identifying the correct quadrant helps in determining the signs of the vector components. For instance, in Quadrant II, where our vector is located, the x-component will always be negative, while the y-component remains positive. This directional understanding is crucial when translating an angle relative to an axis to find vector components.

Magnitude and Direction of Vectors

The magnitude is essentially the length of the vector and can be thought of as the distance it covers. It is always a non-negative value. In this exercise, the vector magnitude \( \|\vec{v}\| \) is given as \( 2\sqrt{3} \). The direction is indicated by the angle that the vector makes with a reference axis. When a vector is said to lie in Quadrant II and makes a 30-degree angle with the positive y-axis, it's important to convert this angle to its reference from the x-axis for component calculations. As a result, we see that the angle with the positive x-axis is 120 degrees. This angle conversion plays a pivotal role in utilizing trigonometric functions to find the exact vector components.

Trigonometric Functions

Trigonometric functions link angles with ratios of right triangle sides and help determine vector components using magnitude and direction. When you know the magnitude \( \|\vec{v}\| \) and an angle \( \theta \), the components on the x- and y-axes can be computed.For our vector, trigonometric functions are employed as follows:

• The cosine function provides the x-component: \( \vec{v}_x = \|\vec{v}\| \cos(\theta) \).
• The sine function provides the y-component: \( \vec{v}_y = \|\vec{v}\| \sin(\theta) \).

In the given problem, \( \theta = 120^\circ \). You compute \( \cos(120^\circ) = -\frac{1}{2} \) and \( \sin(120^\circ) = \frac{\sqrt{3}}{2} \).Hence, substituting these values: \( \vec{v}_x = 2\sqrt{3} \times (-\frac{1}{2}) = -\sqrt{3} \) and \( \vec{v}_y = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3 \). These calculations give you the vector in component form: \( \vec{v} = (-\sqrt{3}, 3) \).