Problem 50 - Algebra and Trigonometry

Question

In Exercises \(47-52,\) write the vector \(\mathbf{v}\) in terms of i and \(\mathbf{j}\) whose magnitude livi and direction angle \(\theta\) are given. $$|\mathbf{v}|=10, \theta=330^{\circ}$$

Step-by-Step Solution

Answer

\( \mathbf{v} = 5\sqrt{3}i - 5j \)

Step 1: Convert Degree to Radian

First, the angle in degrees must be converted to radians because the trigonometric functions in the formulas to calculate the i and j components use radians. The formula to convert degrees to radians is \( \theta_{radians} = \theta_{degrees} \cdot \frac{\pi}{180} \). Thus, substituting \( \theta_{degrees} = 330 \) gives \( \theta_{radians} = 330 \cdot \frac{\pi}{180} = \frac{11\pi}{6} \).

Step 2: Calculate the Cartesian Components

Now calculate the i and j components of the vector. The i component (x-component) of the vector can be found by multiplying the magnitude of the vector by the cosine of the angle. The j component (y-component) of the vector can be found by multiplying the magnitude of the vector by the sine of the angle. Using the magnitute \(| \mathbf{v} | = 10 \) and \( \theta_{radians} = \frac{11\pi}{6} \), the i and j components are calculated as follows: \( \mathbf{v}_{i} = | \mathbf{v} | \cdot \cos(\theta) = 10 \cdot \cos(\frac{11\pi}{6}) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \) and \( \mathbf{v}_{j} = | \mathbf{v} | \cdot \sin(\theta) = 10 \cdot \sin(\frac{11\pi}{6}) = -10 \cdot \frac{1}{2} = -5 \).

Step 3: Express vector in terms of i and j

Finally, express the vector \( \mathbf{v} \) in terms of the i and j components: \( \mathbf{v} = \mathbf{v}_{i}i + \mathbf{v}_{j}j = 5\sqrt{3}i - 5j \).

Key Concepts

Converting Degrees to Radians

When working with vectors and trigonometry, it's essential to be comfortable with converting angles from degrees to radians. The formula to make this conversion is surprisingly straightforward:
\[ \theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180} \]
This is because a complete circle has 360 degrees or \(2\pi\) radians. For example, in the given exercise, converting \(330^\circ\) into radians involves multiplying \(330\) by \(\frac{\pi}{180}\), which simplifies to \(\frac{11\pi}{6}\). It's helpful to remember common conversions, like \(180^\circ = \pi\) radians, to quickly switch between units and ensure accuracy when performing calculations.

Calculating Vector Magnitude

The magnitude of a vector is a measure of its length and is denoted by vertical bars, akin to the absolute value sign. For a vector \( \mathbf{v} \) with components in terms of \( i \) and \( j \), the magnitude can be expressed as:
\[ | \mathbf{v} | = \sqrt{\mathbf{v}_{i}^2 + \mathbf{v}_{j}^2} \]
This formula emanates from the Pythagorean theorem, applying it to the right triangle formed by the vector components. In the case of the exercise, the magnitude \( |\mathbf{v}| \) is already provided as 10. This value is key in finding the component form of the vector and requires no further calculations in this particular instance. However, understanding how to compute it from component values is fundamental for tackling more complex problems.

Direction Angle of Vectors

The direction angle of a vector is the angle it makes with the positive x-axis, measured in a counterclockwise direction. It's an essential component in expressing vectors in their component form. The direction angle \( \theta \) is used in conjunction with trigonometric functions to determine the vector's components. Specifically:

• To find the x-component (also known as the i-component), multiply the magnitude of the vector by the cosine of the direction angle.
• To find the y-component (the j-component), multiply the magnitude of the vector by the sine of the direction angle.

In trigonometry, these functions relate the sides of a right triangle to its angles, which forms the connection between the direction angle and the vector components. Students may find it helpful to create a diagram and visually align their vector with a right triangle to better grasp these relationships.

Cartesian Components of Vectors

The Cartesian components of a vector break it down into its parts along the x and y axes, denoted as \( i \) and \( j \) respectively. These components tell us how far along each axis the vector extends. To find them, use the magnitude and direction of the vector, as shown in the exercise.

• The i-component (x-axis) is found by: \( \mathbf{v}_{i} = | \mathbf{v} | \cdot \cos(\theta) \)
• The j-component (y-axis) is found by: \( \mathbf{v}_{j} = | \mathbf{v} | \cdot \sin(\theta) \)

After computing these, the vector \( \mathbf{v} \) in terms of \( i \) and \( j \) will be \( \mathbf{v} = \mathbf{v}_{i}i + \mathbf{v}_{j}j \). This provides a clear and concise way to represent vector direction and magnitude in the Cartesian plane, paving the way for easier manipulation and visualization of vectors within this two-dimensional space.