Problem 20 - College Trigonometry

Question

Find the component form of the vector \(\vec{v}\) using the information given about its magnitude and direction. Give exact values. \(\|\vec{v}\|=4 \sqrt{3} ;\) when drawn in standard position \(\vec{v}\) lies in Quadrant IV and makes a \(30^{\circ}\) angle with the positive \(x\) -axis

Step-by-Step Solution

Answer

The component form of the vector \(\vec{v}\) is \(\langle 6, -2\sqrt{3} \rangle\).

Step 1: Identify the magnitude and direction

The magnitude of the vector \(\vec{v}\) is given as \(4 \sqrt{3}\), and it makes a \(30^{\circ}\) angle with the positive \(x\)-axis in Quadrant IV.

Step 2: Determine the components using trigonometric functions

In Quadrant IV, the cosine of \(30^{\circ}\) determines the \(x\)-component and the sine of \(30^{\circ}\) determines the \(y\)-component. However, the sine function will yield a negative component because the vector is in Quadrant IV.

Step 3: Calculate the x-component

The \(x\)-component is given by \(\|\vec{v}\| \cdot \cos(30^{\circ})\). Thus, the \(x\)-component is \(4 \sqrt{3} \cdot \frac{\sqrt{3}}{2} = 6\).

Step 4: Calculate the y-component

The \(y\)-component is given by \(\|\vec{v}\| \cdot \sin(30^{\circ})\). Thus, the \(y\)-component is \(4 \sqrt{3} \cdot -\frac{1}{2} = -2\sqrt{3}\).

Step 5: Construct the vector in component form

Combine the components to write the vector \(\vec{v}\) in component form: \(\vec{v} = \langle 6, -2\sqrt{3} \rangle\).

Key Concepts

Magnitude and Direction

The magnitude of a vector is a measure of its length, often represented as \( \|\vec{v}\| \). In our example, the magnitude is given as \( 4\sqrt{3} \), indicating the length of the vector. Knowing the magnitude is essential to find vector components.
Direction is described by the angle the vector makes with the positive \( x \)-axis. This angle helps determine how the vector is split into components along the \( x \) and \( y \) axes. Here, the vector makes a \( 30^{\circ} \) angle.

• Magnitude: \( 4\sqrt{3} \)
• Direction (Angle): \( 30^{\circ} \) with the \( x \)-axis

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are key tools for finding vector components. For a vector lying at an angle \( \theta \) from the \( x \)-axis:

• Cosine: determines the \( x \)-component
• Sine: determines the \( y \)-component

For our vector, the cosine of \( 30^{\circ} \), equivalent to \( \frac{\sqrt{3}}{2} \), rivals the \( x \)-component, while sine, \( \frac{1}{2} \), rivals the \( y \)-component.
Thus:

• \( \text{x-component} = \|\vec{v}\| \cdot \cos(30^{\circ}) = 4\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 6 \)
• \( \text{y-component} = \|\vec{v}\| \cdot \sin(30^{\circ}) = 4\sqrt{3} \cdot \frac{1}{2} = 2\sqrt{3} \)

Angle with the x-axis

The angle that a vector makes with the positive \( x \)-axis is crucial to determining its orientation. This angle is used in conjunction with trigonometric functions to find the vector components.
In our example, with a \( 30^{\circ} \) angle, the vector’s placement in Quadrant IV affects its component signs, with the \( y \)-component being negative.
Remember:

• Positive \( x \)-axis: typically aligned with the cosine of the angle.
• Quadrant adjustments: adjust sign of components based on the quadrant location.

Quadrant Analysis

Each quadrant in the coordinate system influences the signs and values of a vector's components. For vectors in standard position:

• **Quadrant I:** both components are positive.
• **Quadrant II:** negative \( x \)-component, positive \( y \)-component.
• **Quadrant III:** both components are negative.
• **Quadrant IV:** positive \( x \)-component, negative \( y \)-component.

In our example, the vector lies in Quadrant IV, meaning its \( x \)-component is positive, while its \( y \)-component is negative. This is confirmed by the calculations:

• \( \text{x-component} = 6 \)
• \( \text{y-component} = -2\sqrt{3} \)