Problem 35 - College Trigonometry

Question

For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta<360^{\circ}\) so that \(\vec{v}=\|\vec{v}\|\langle\cos (\theta), \sin (\theta)\rangle\) (See Definition 11.8.) Round approximations to two decimal places. $$ \vec{v}=\langle-\sqrt{2}, \sqrt{2}\rangle $$

Step-by-Step Solution

Answer

The magnitude is 2, and the angle \( \theta \) is 135 degrees.

Step 1: Understand the problem

We are given a vector \( \vec{v} = \langle -\sqrt{2}, \sqrt{2} \rangle \). The task is to find its magnitude and an angle \( \theta \) such that the vector can be expressed in the form \( \|\vec{v}\|\langle\cos(\theta), \sin(\theta)\rangle \), with the angle in the range from 0 to 360 degrees.

Step 2: Calculate the magnitude of the vector

The magnitude of a vector \( \vec{v} = \langle x, y \rangle \) is calculated using the formula \( \|\vec{v}\| = \sqrt{x^2 + y^2} \). Here, \( x = -\sqrt{2} \) and \( y = \sqrt{2} \). So, \( \|\vec{v}\| = \sqrt{(-\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \).

Step 3: Use inverse trigonometry to find the angle

Since the vector has components related to \( \cos(\theta) \) and \( \sin(\theta) \), we use atan2 to find the angle: \( \theta = \text{atan2}(y, x) = \text{atan2}(\sqrt{2}, -\sqrt{2}) \). In most programming languages or calculators, \( \text{atan2}(\sqrt{2}, -\sqrt{2}) \) gives an angle of \(rac{3\pi}{4}\) radians or 135 degrees. Thus, \( \theta = 135^\circ \).

Step 4: Verify the angle lies within the specified range

We found \( \theta = 135^\circ \), which is within the range \( 0 \leq \theta < 360^\circ \). Therefore, no adjustments are needed for \( \theta \).

Key Concepts

Understanding Inverse Trigonometry

When dealing with vectors in any dimension, it's crucial to determine the direction these vectors are pointing. This is where inverse trigonometry comes into play. Using trigonometric functions such as sine, cosine, and tangent allows us to find angles from the known ratios of the sides of right-angled triangles. However, when angles need to be recalculated from known sine or cosine values, we apply inverse trigonometric functions.

• Inverse Sine (rcsin): This function helps find an angle whose sine value we already know.
• Inverse Cosine (rccos): This is used to find an angle when its cosine value is known.
• Inverse Tangent (rctan): Ideal for calculating an angle by using the quotient of the opposite side to the adjacent side of a triangle.

For vectors, we rely more on the atan2 function, especially when dealing with two-dimensional space. The tan2(y, x) function provides the correct positive or negative angle based on the quadrant of the angle, avoiding ambiguities that might arise from simple arc functions.

Angle Calculation in Vector Spaces

Calculating the angle of a vector relative to the horizontal axis is a fundamental aspect of vector analysis. In our specific case, we use the atan2 function, which is more robust than the basic arctan function as it considers the signs of both components of the vector.
To find theta:

• Input the y and x components of the vector into atan2. This automatically recognizes which quadrant the angle is in, providing an accurate angle measurement.
• In our example with {v} = lang -{{2}} lang, {{2},_, atan2( takes the input as atan2({2},-{2}) and directly outputs _expr _expr{{{2pi/4}} radians,_ or 135 degrees.

This method ensures that the angle you calculate fits the typical interval of [0, 360] degrees, giving a perfect directional sense to the vector. For anyone handling vectors in physics or engineering, understanding and using this function is crucial.

Finding and Using Unit Vectors

Unit vectors play a vital role in vector mathematics. A unit vector is essentially a vector that has a magnitude of 1 but maintains its direction. They're handy because they simplify calculations by focusing only on the direction without worrying about the length of the vector.
To convert a vector to its unit vector form:

• First, calculate the magnitude of the original vector, as done in the provided solution with  v = mag 2.__
• Then, divide each component of the vector by that magnitude. For the vector _expr = 2,{{", we divide each part to get  {{{ the unit vector x 0}, 0<00,`.

By representing vectors this way, we always have a directional indicator at hand, irrespective of the original size of the vector. This concept is crucial in physics, computer graphics, and any field where directional data is critical.