Problem 47 - 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. $$ \vec{v}=\langle-2,-6\rangle $$

Step-by-Step Solution

Answer

The magnitude is approximately 6.32 and the angle \(\theta \approx 251.57^{\circ}\).

Step 1: Calculate the Magnitude of Vector \(\vec{v}\)

To find the magnitude of \(\vec{v}\), use the formula for the magnitude of a vector: \[ \|\vec{v}\| = \sqrt{(-2)^2 + (-6)^2} \]. Simplify this to \( \|\vec{v}\| = \sqrt{4 + 36} = \sqrt{40} \). Further simplify to get \( \|\vec{v}\| = 2\sqrt{10} \). As an approximate decimal, \( \|\vec{v}\| \approx 6.32 \).

Step 2: Determine the Components of the Unit Vector

We need to express \(\vec{v}\) in the form \(\|\vec{v}\|\langle\cos(\theta), \sin(\theta)\rangle\). We have \( \|\vec{v}\| \approx 6.32 \), so the unit vector \(\langle \cos(\theta), \sin(\theta) \rangle\) can be found by dividing the components of \(\vec{v}\) by \( \|\vec{v}\| \). Thus, \(\langle \cos(\theta), \sin(\theta) \rangle = \langle -\frac{2}{6.32}, -\frac{6}{6.32} \rangle \approx \langle -0.32, -0.95 \rangle\).

Step 3: Calculate the Angle \(\theta\)

The angle \(\theta\) can be found using the arctangent function: \( \theta = \arctan\left(\frac{-b}{-a}\right) \), where \(a = -2\) and \(b = -6\). So, \( \theta = \arctan\left(\frac{6}{2}\right) = \arctan(3) \). Calculate \( \arctan(3) \approx 71.57^{\circ} \). Since the vector is in the third quadrant, the actual angle is \(\theta = 180^{\circ} + 71.57^{\circ} = 251.57^{\circ}\).

Step 4: Round the Angle \(\theta\)

Since we need to round to two decimal places, the angle \(\theta\) is approximately \(251.57^{\circ}\) when rounded properly.

Key Concepts

Unit Vector

A unit vector is a vector that has a magnitude of exactly 1. It maintains the direction of the original vector but scales it down to a standard size. Finding a unit vector is very useful because it allows us to focus on direction rather than size.

To obtain the unit vector of a given vector \( \vec{v} \), we divide each component of \( \vec{v} \) by the magnitude of the vector \( \|\vec{v}\| \). Using the vector \( \vec{v} = \langle -2, -6 \rangle \) as an example, the magnitude is calculated as \( \sqrt{4 + 36} = \sqrt{40} \approx 6.32 \). Therefore, the unit vector is given by:

• Divide the first component: \( \frac{-2}{6.32} \approx -0.32 \)
• Divide the second component: \( \frac{-6}{6.32} \approx -0.95 \)

Thus, the unit vector is approximately \( \langle -0.32, -0.95 \rangle \). This unit vector indicates the direction of \( \vec{v} \) on a unit circle.

Trigonometric Angles

Trigonometric angles play a significant role in breaking down vectors into their components. They are the angles used when working with trigonometric functions like sine and cosine, which help in expressing vectors in polar coordinates.

Using trigonometric angles, any vector can be expressed as a combination of its magnitude and direction. For example, a vector \( \vec{v} \) can be expressed in terms of its magnitude \( \|\vec{v}\| \) and angle \( \theta \) such that:

• The horizontal component is \( \|\vec{v}\| \cos(\theta) \)
• The vertical component is \( \|\vec{v}\| \sin(\theta) \)

For the vector \( \vec{v} = \langle -2, -6 \rangle \), \( \theta \) has been calculated as approximately \( 251.57^\circ \). The negative components indicate that the angles use sine and cosine corresponding to the orientation of the third quadrant of the coordinate plane. This quadrant is identified by negative cosine and sine values.

Arctangent Function

The arctangent function, often represented as \( \arctan \), is an inverse trigonometric function used to find the angle whose tangent is a given number. It is extremely useful when calculating angles from vector components, especially when dealing with vectors that are not aligned along the axes.

To determine the angle \( \theta \) for the vector \( \vec{v} = \langle -2, -6 \rangle \), we use \( \arctan \) to find the angle whose tangent equals the ratio of the opposite side to the adjacent side of our vector:

• \( \theta = \arctan\left(\frac{6}{2}\right) = \arctan(3) \approx 71.57^\circ \)

Since the vector is in the third quadrant, where both components are negative, \( \theta \) must be adjusted by adding \( 180^\circ \) to reflect its true direction. Therefore, \( \theta \approx 251.57^\circ \). This angle takes into account the standard position of angles in the coordinate system, ensuring the vector is accurately represented in its proper quadrant.