Problem 38 - 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. $$ \vec{v}=\langle 6,0\rangle $$

Step-by-Step Solution

Answer

The magnitude of \( \vec{v} \) is 6, and the angle \( \theta \) is \( 0^{\circ} \).

Step 1: Find the Magnitude of the Vector

The magnitude \( \| \vec{v} \| \) of a vector \( \vec{v} = \langle x, y \rangle \) is found using the formula \( \| \vec{v} \| = \sqrt{x^2 + y^2} \). For the vector \( \vec{v} = \langle 6, 0 \rangle \), the magnitude is:\[\| \vec{v} \| = \sqrt{6^2 + 0^2} = \sqrt{36} = 6.\]

Step 2: Understand the Angle in the Vector Notation

We need to express the vector \( \vec{v} = \langle 6, 0 \rangle \) in the form \( \| \vec{v} \| \langle \cos(\theta), \sin(\theta) \rangle \). Given that \( \vec{v} = \langle x, y \rangle \), we have \( x = 6 \) and \( y = 0 \).

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

Since \( y = 0 \), the vector points entirely along the x-axis. When a vector lies on the x-axis and points in the positive direction, \( \theta = 0^{\circ} \). Since the vector is exactly \( 6 \) units along the positive x-axis, our angle \( \theta \) is:\[\theta = 0^{\circ}.\]

Step 4: Verify the Vector Expression

Verify that: \[ \vec{v} = \| \vec{v} \| \langle \cos(\theta), \sin(\theta) \rangle = 6 \langle \cos(0^{\circ}), \sin(0^{\circ}) \rangle. \]Calculate:\[ \cos(0^{\circ}) = 1, \quad \sin(0^{\circ}) = 0. \]Thus:\[ \vec{v} = 6 \langle 1, 0 \rangle = \langle 6, 0 \rangle, \] which is correct.

Key Concepts

Vector Notation

To properly understand vectors, it's essential to grasp their notation. A vector is usually represented by a pair of components in the form \(\vec{v} = \langle x, y \rangle \). These components correspond to the vector's direction along the x-axis and y-axis, respectively.

Think of a vector as an arrow in a plane; the components \(x\) and \(y\) describe precisely where this arrow points. In our exercise, the vector \(\vec{v} = \langle 6, 0 \rangle\) has a length (or magnitude) in the x-direction but none in the y-direction.

• The first component (6 in our case) shows the vector's length along the x-axis.
• The second component indicates the vector's length along the y-axis (which is 0 here).

This notation helps simplify complex geometry problems by breaking down vectors into their core, manageable elements.

Trigonometric Form of Vectors

Vectors can also be represented in a trigonometric form which uses angles and trigonometric functions instead of direct component values. This form is expressed as \( \| \vec{v} \| \langle \cos(\theta), \sin(\theta) \rangle \), where:

• \( \| \vec{v} \| \) is the magnitude of the vector.
• \( \theta \) is the angle the vector makes with the positive x-axis.
• \( \langle \cos(\theta), \sin(\theta) \rangle \) gives the direction of the vector relative to this angle.

In the exercise, our vector \(\vec{v} = \langle 6, 0 \rangle\) converts to the trigonometric form as \(6 \langle \cos(0^{\circ}), \sin(0^{\circ}) \rangle\). Since \(\theta = 0^{\circ}\), it indicates our vector lies along the x-axis, heading directly to the right.

The trigonometric form is valuable for operations like rotations and understanding vector orientations, as it emphasizes the angle and length rather than just position.

Angle Calculation

Calculating the angle \(\theta\) that a vector makes with the positive x-axis is an important skill in vector mathematics. This angle tells you which direction the vector is pointing.

In our example, with \(\vec{v} = \langle 6, 0 \rangle\), the vector points entirely in the positive x-direction. Thus, the angle is \(\theta = 0^{\circ}\).

• If the vector were in the y-direction, \(\theta\) would be \(90^{\circ}\) for positive y-direction or \(270^{\circ}\) for negative y-direction.
• For vectors in the negative x-direction, \(\theta\) would be \(180^{\circ}\).

Knowing the angle might be straightforward in simple cases like ours, but for other vectors, you can use a little trigonometry:

• \(\tan(\theta) = \frac{y}{x}\) helps in finding the angle.
• Using inverse trigonometric functions like \(\arctan\) can assist in calculating \(\theta\).

This skill ensures you can accurately describe and manipulate vectors in various applications, from physics to computer graphics.