Problem 28 - Precalculus

Question

Find the vector, given its magnitude and direction angle. $$|\mathbf{u}|=8, \theta=200^{\circ}$$

Step-by-Step Solution

Answer

The vector is \( \langle -7.5176, -2.736 \rangle \).

Step 1: Understanding the Components of the Vector

To find a vector given its magnitude and direction, we need to decompose it into its components. The vector in question is denoted as \( \mathbf{u} \) with a magnitude of 8 and a direction angle of \( 200^{\circ} \). The general formula to convert polar to Cartesian coordinates (components) is:\[\mathbf{u} = \langle |\mathbf{u}|\cos\theta, |\mathbf{u}|\sin\theta \rangle\]where \( \theta \) is the direction angle of the vector.

Step 2: Calculate the X-component of the Vector

The X-component of the vector, \( u_x \), is calculated by multiplying the magnitude of the vector by the cosine of the direction angle.\[u_x = |\mathbf{u}| \cdot \cos(\theta) = 8 \cdot \cos(200^{\circ})\]Using a calculator, compute \( \cos(200^{\circ}) = -0.9397 \) (rounded to 4 decimal places), thus:\[u_x = 8 \cdot (-0.9397) = -7.5176\]

Step 3: Calculate the Y-component of the Vector

The Y-component, \( u_y \), is calculated by multiplying the magnitude of the vector by the sine of the direction angle.\[u_y = |\mathbf{u}| \cdot \sin(\theta) = 8 \cdot \sin(200^{\circ})\]Using a calculator, compute \( \sin(200^{\circ}) = -0.3420 \) (rounded to 4 decimal places), thus:\[u_y = 8 \cdot (-0.3420) = -2.736\]

Step 4: Combine Components into the Vector

Now, combine the calculated X and Y components to express the vector \( \mathbf{u} \) in component form:\[\mathbf{u} = \langle -7.5176, -2.736 \rangle\]Thus, the vector is represented in its component form with the calculated X and Y components.

Key Concepts

Magnitude of a Vector

The magnitude of a vector is one of the fundamental concepts in vector mathematics. Think of the magnitude as the "length" or "size" of the vector. It tells us how far away the endpoint of the vector is from its starting point.

• To find the magnitude, we usually use the formula for a vector \( \mathbf{v} = \langle x, y \rangle \), which is \( |\mathbf{v}| = \sqrt{x^2 + y^2} \). However, here it's directly provided as 8.
• The magnitude helps determine the scale and is crucial when converting between coordinate systems or working with vector components.

In our exercise, you're given that the vector \( \mathbf{u} \) has a magnitude of 8. This means the length from the origin to the point on a graph where this vector ends, is precisely 8 units.
Understanding this helps in the decomposition of the vector, where this magnitude gets split into parts along the X and Y axes.

Direction Angle of a Vector

The direction angle of a vector helps to determine its orientation on a plane. It's measured from the positive x-axis towards the vector's line, usually counterclockwise.

• In the exercise, the vector has a direction angle \( \theta = 200^{\circ} \).
• An angle of 200 degrees is more than 180 degrees, meaning the vector points into the third quadrant of the Cartesian plane.

When you understand the direction angle, you're better equipped to decompose the vector into its horizontal (X) and vertical (Y) components.
The angle is vital when using trigonometric functions like cosine and sine to find these components. Here it’s used to explain to which extent the vector leans towards a new direction, affecting the signs of the vector components.

Polar to Cartesian Coordinates

Converting polar to Cartesian coordinates involves changing a vector from one form of notation to another. Polar coordinates describe a vector in terms of magnitude and angle.

• Formula: To convert, use \( \mathbf{u} = \langle |\mathbf{u}|\cos\theta, |\mathbf{u}|\sin\theta \rangle \).
• The cosine function helps find how far the vector extends horizontally (X-axis) and the sine function for vertical extension (Y-axis).

In this exercise, we calculated:- **X-component**: \( u_x = 8 \cdot \cos(200^{\circ}) = -7.5176 \)- **Y-component**: \( u_y = 8 \cdot \sin(200^{\circ}) = -2.736 \)
The combination of these components gives us the vector in Cartesian form, allowing us to understand its exact placement on a graph. This process is pivotal when vectors need to be used in further calculations involving different coordinate systems or when working in real-world applications.