Problem 24 - College Trigonometry

Question

Find the component form of the vector \(\vec{v}\) using the information given about its magnitude and direction. Give exact values. \(\|\vec{v}\|=5 ;\) when drawn in standard position \(\vec{v}\) lies in Quadrant III and makes an angle measuring \(\arctan \left(\frac{4}{3}\right)\) with the negative \(x\) -axis

Step-by-Step Solution

Answer

The component form of the vector \( \vec{v} \) is \((-3, -4)\).

Step 1: Determine the reference angle

The given vector makes an angle of \( \arctan \left(\frac{4}{3}\right) \) with the negative \(x\)-axis. First, we calculate this angle, which is the reference angle. This angle in radian terms is \( \theta = \arctan \left(\frac{4}{3}\right) \approx 0.93 \) radians.

Step 2: Identify the angle in standard position

Since \( \vec{v} \) is in Quadrant III, the angle measured in standard position from the positive \(x\)-axis is \( \pi + \theta \). Using the reference angle from Step 1, this becomes \( \pi + 0.93 = 3.07 \) radians roughly.

Step 3: Use the magnitude and direction to find vector components

Use trigonometric functions to find the components of \( \vec{v} \) based on the angle determined. The component form of the vector is given by: \( \vec{v} = \|\vec{v}\| (\cos \theta, \sin \theta) \). Here, \( \theta = 3.07 \) and \( \|\vec{v}\| = 5 \). Thus, \( \vec{v} = 5 (\cos 3.07, \sin 3.07) \).

Step 4: Calculate the cosine and sine values

Since our standard angle is in Quadrant III, cosine and sine will both be negative: \( \cos(3.07) \approx -\frac{3}{5} \) and \( \sin(3.07) \approx -\frac{4}{5} \). This gives the components of \( \vec{v} \) as: \( \vec{v} = (5 \times -\frac{3}{5}, 5 \times -\frac{4}{5}) = (-3, -4) \).

Step 5: Verify the vector form

Check if both magnitude and direction match the given conditions. The calculated magnitude is \( \sqrt{(-3)^2 + (-4)^2} = 5 \), and the direction matches the angle specified for Quadrant III.

Key Concepts

Magnitude

The magnitude of a vector, often denoted as \( \|\vec{v}\| \), is essentially the length of the vector. Think of it like measuring the distance from the vector’s starting point (usually the origin) to its ending point in a plane. Magnitude is always a non-negative value. In our exercise, the vector \( \vec{v} \) has a given magnitude of 5.

To calculate the magnitude of a vector \( \vec{v} = (x, y) \), we use the Pythagorean theorem:

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

In our step-by-step process, we verified the magnitude by confirming it equals 5. This shows that the vector indeed corresponds to the given conditions.

Direction

Direction indicates the angle at which a vector is oriented with respect to the positive x-axis. In a two-dimensional plane, this is often expressed as an angle measured counter-clockwise from the positive x-axis. Knowing the direction is vital because it affects the vector's components in the x and y directions.

In the given problem, the vector’s direction is determined by the angle it makes with the negative x-axis in Quadrant III. This direction is calculated using the angle given by \( \arctan \left(\frac{4}{3}\right) \), which measures approximately 0.93 radians. By adjusting this angle to fit the standard position measured from the positive x-axis, we determine it as \( \pi + 0.93 \) radians for Quadrant III.

When working with direction:

• Account for the relevant quadrant to adjust the angle properly.
• Convert your angle to a measure from the positive x-axis when finding components.

Doing so ensures the vector accurately points in the intended direction.

Trigonometry

Trigonometry helps us break down a vector into its components along the x and y axes. These components are crucial because they allow us to understand and use the vector in calculations and visualizations. For a vector \( \vec{v} \) with a magnitude \( \|\vec{v}\| \) and an angle \( \theta \), the components can be found as follows:

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

In our exercise, trigonometric functions were used to find the values of \( \cos \) and \( \sin \) for an angle of 3.07 radians. Since the vector is in Quadrant III, both components end up being negative (\( -3 \) and \( -4 \)). This quadrant-specific behavior is important in vector analysis. Remember, the use of cosine and sine helps us efficiently locate the vector in the coordinate system.

Quadrant III

In a Cartesian plane, Quadrant III is the section where both x and y values are negative. It stretches from \( \pi \) to \( \frac{3\pi}{2} \) radians, which translates to angles between 180° and 270°. This is critical to our problem because:

• The vector component values in this quadrant are negative.
• It affects the calculation of the vector's angle in standard position.

When solving vector problems like this, acknowledge the quadrant to decide the sign of components and ensure the vector is plotted correctly. This directly impacts the final solution, ensuring the direction and magnitude fit the given criteria.