Problem 26 - APEX Calculus

Question

Find the unit vector in the first quadrant of \(\mathbb{R}^{2}\) that makes a \(50^{\circ}\) angle with the \(x\) -axis.

Step-by-Step Solution

Answer

The unit vector is \((0.6428, 0.7660)\).

Step 1: Understand the Concept of a Unit Vector

A unit vector is a vector with a magnitude of 1. It points in the same direction as the original vector but has been scaled to have a length of 1. In this problem, we want to find a unit vector that makes a 50° angle with the x-axis.

Step 2: Express the Vector in Component Form

A vector in two-dimensional space that makes an angle θ with the x-axis can be expressed as \((\cos(θ), \sin(θ))\). For our problem, the angle θ is 50°. Therefore, the vector can be expressed as \((\cos(50^\circ), \sin(50^\circ))\).

Step 3: Calculate the Components

Using trigonometric functions, calculate the components of the vector. - \(\cos(50^\circ) ≈ 0.6428\)- \(\sin(50^\circ) ≈ 0.7660\)Thus, the vector is approximately \((0.6428, 0.7660)\).

Step 4: Verify the Unit Vector Property

To ensure that this vector is a unit vector, check its magnitude:\[\sqrt{(\cos(50^\circ))^2 + (\sin(50^\circ))^2} = \sqrt{0.6428^2 + 0.7660^2} = \sqrt{1} = 1\]This confirms that the vector is indeed a unit vector.

Step 5: Final Result

The unit vector in the first quadrant of \(\mathbb{R}^{2}\) making a 50° angle with the x-axis is \((0.6428, 0.7660)\).

Key Concepts

Magnitude of a Vector

First, let's dive into understanding what magnitude means in the context of vectors. The magnitude of a vector is essentially its length, much like measuring a line segment in geometry. If you consider a vector as an arrow, its magnitude indicates how long that arrow is. For a vector expressed as \((x, y)\) in two-dimensional space, the magnitude is calculated using the formula:
\[ \text{Magnitude} = \sqrt{x^2 + y^2} \]
This formula comes from the Pythagorean theorem since the vector components \(x\) and \(y\) form the legs of a right triangle with the vector as the hypotenuse.
For a unit vector, however, we specifically want this magnitude to be exactly 1. This is because a unit vector's length is always normalized to one, ensuring it only reflects direction and not magnitude.

• This aspect makes unit vectors particularly useful in many applications, such as simplifying vector calculations and representing directional quantities.

Angle Calculation for Vectors

Next, we have a look at how angles play a crucial role in defining a vector's direction. In our exercise, we're focusing on a vector that forms a \(50^{\circ}\) angle with the x-axis. Angles help determine the orientation of a vector.
For any vector in a two-dimensional space that makes an angle, say \(θ\), with the x-axis, we can use this angle to describe the vector in terms of its components:
\[ ( \cos(θ), \sin(θ) ) \]
This way of expressing the vector components can be extremely helpful when you need the direction fixed by the angle.

• Knowing the angle allows you to easily calculate the components of the unit vector using trigonometric functions, which we'll expand upon further.

Understanding how to calculate and use angles in vectors frees you to convert angular information into practical vector components, crucial for problems you might face later in both physics and engineering.

Trigonometric Functions in Vector Calculation

Finally, let's explore how trigonometric functions come into play in vector calculations, especially when working with angles and magnitudes. Trigonometric functions are the backbone of expressing vector components based on angles.
When you have an angle \(θ\) and you want to find a unit vector in that direction, you use the cosine and sine functions to determine the vector's x and y components, respectively:
- \(\cos(θ)\) gives you the x-coordinate of the unit vector.
- \(\sin(θ)\) provides the y-coordinate of the unit vector.
These functions are determined by projecting the vector onto the x and y axes, reflecting the proportions that belong to each direction.

• In our exercise, using \(\cos(50^\circ) \approx 0.6428\) and \(\sin(50^\circ) \approx 0.7660\), we find the components of the unit vector are approximately \((0.6428, 0.7660)\).

Trigonometric functions allow us to break down an angle into concrete, usable components that specify direction while maintaining the unity of magnitude for the unit vector.