Problem 65
Question
Write the vector \(\mathbf{v}\) in the form \(\mathbf{ai}+ \mathbf{bj}\), given its magnitude \(\|\mathbf{v}\|\) and the angle \(\alpha\) it makes with the positive \(x\) -axis. \(\|\mathbf{v}\|=25, \quad \alpha=330^{\circ}\)
Step-by-Step Solution
Verified Answer
The vector \( \textbf{v} \) in the form \(\textbf{ai} + \textbf{bj} \) is \( \textbf{v} = 12.5 \sqrt{3} \textbf{i} - 12.5 \textbf{j} \).
1Step 1 - Understand the components of the vector
Given the vector \(\textbf{v}\) with magnitude \(\textbar \textbf{v} \textbar = 25\) and angle \( \alpha = 330^{\circ} \) with the positive x-axis, we need to determine its components along the x and y axes.
2Step 2 - Calculate the x-component
The x-component of the vector can be found using \(\textbf{a} = \textbar \textbf{v} \textbar \cos(330^{\circ}) \). So, \(\textbf{a} = 25 \cos(330^{\circ})\). Since \( \cos(330^{\circ}) = \cos(360^{\circ} - 30^{\circ}) = \cos(-30^{\circ}) = \cos(30^{\circ}) = \dfrac{\textbf{\sqrt{3}}}{2} \), we have: \(\textbf{a} = 25 \dfrac{\sqrt{3}}{2} = 12.5 \sqrt{3} \).
3Step 3 - Calculate the y-component
The y-component of the vector can be found using \(\textbf{b} = \textbar \textbf{v} \textbar \sin(330^{\circ}) \). So, \(\textbf{b} = 25 \sin(330^{\circ})\). Since \( \sin(330^{\circ}) = \sin(360^{\circ} - 30^{\circ}) = \sin(-30^{\circ}) = - \sin(30^{\circ}) = -\textbf{\dfrac{1}{2}} \), we have: \(\textbf{b} = 25 \dfrac{-1}{2} = -12.5 \).
4Step 4 - Write the vector in the form \(\textbf{ai} + \textbf{bj}\)
Substitute the values of \( \textbf{a} \) and \( \textbf{b} \) into the form \(\textbf{ai} + \textbf{bj}\): \( \textbf{v} = 12.5 \sqrt{3} \textbf{i} - 12.5 \textbf{j} \).
Key Concepts
magnitude of a vectortrigonometric functionsvector representation
magnitude of a vector
The magnitude of a vector represents its length or size. It is a scalar quantity, meaning it only has magnitude and no direction. The magnitude of a vector \(\textbf{v}\) is denoted as \(\textbar \textbf{v} \textbar \). For example, if the magnitude is given as 25, it means the length of the vector is 25 units. To compute the magnitude of a vector \( \textbf{v} = \textbf{ai} + \textbf{bj} \), we use the formula: \[ \textbar \textbf{v} \textbar = \sqrt{a^2 + b^2} \]. This allows us to understand the strength or impact of the vector in physical terms. For vectors in physics and engineering, magnitude provides insight into quantities like force, velocity, or displacement.
trigonometric functions
Trigonometric functions like cosine and sine are essential for working with vectors, particularly for finding their components along different axes. Cosine (\(\text{cos}\)) and sine (\(\text{sin}\)) functions relate the angles of a right triangle to the lengths of its sides. Given an angle \(\theta\), these functions are defined as:
\[ \text{a} = 25 \text{cos} (330^{\text{\tiny ∘}}) = 25 \frac{\text{\tiny √3}}{2} = 12.5 \text{\tiny √3} \] \[ \text{b} = 25 \text{sin} (330^{\text{\tiny ∘}}) = 25 (-\text{\tiny 1/2}) = -12.5 \]. These trigonometric identities help break down vectors into manageable parts.
- \textrm{cosine:} \text{cos} ({{\theta}}) = \frac{{\text{adjacent side}}}{{\text{hypotenuse}}}
- \textrm{sine:} \text{sin} ({{\theta}}) = \frac{{\text{opposite side}}}{{\text{hypotenuse}}}
\[ \text{a} = 25 \text{cos} (330^{\text{\tiny ∘}}) = 25 \frac{\text{\tiny √3}}{2} = 12.5 \text{\tiny √3} \] \[ \text{b} = 25 \text{sin} (330^{\text{\tiny ∘}}) = 25 (-\text{\tiny 1/2}) = -12.5 \]. These trigonometric identities help break down vectors into manageable parts.
vector representation
A vector can be represented in different forms, but one of the most common notations is in terms of its components along the axes. The vector \( \textbf{v} \) is represented as \( \textbf{ai} + \textbf{bj} \), where \( \textbf{a} \) is the x-component and \( \textbf{b} \) is the y-component. This representation helps to clearly depict the direction and magnitude of the vector in a 2D plane. For our example, we have:
\( \textbf{v} = 12.5 \text{\tiny √3} \textbf{i} - 12.5 \textbf{j} \), where:
\( \textbf{v} = 12.5 \text{\tiny √3} \textbf{i} - 12.5 \textbf{j} \), where:
- \textbf{ai}\text{i component:} 12.5 \text{\tiny √3} units along the x-axis
- \textbf{bj}\text{j component:} -12.5 units along the y-axis
Other exercises in this chapter
Problem 64
Find all the complex roots. Write your answers in exponential form. The complex fifth roots of \(-i\)
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The rectangular coordinates of a point are given. Find polar coordinates for each point. $$ (5,5 \sqrt{3}) $$
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Find the four complex fourth roots of unity, \(1,\) and plot them.
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The rectangular coordinates of a point are given. Find polar coordinates for each point. $$ \left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right) $$
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