Problem 43 - Algebra and Trigonometry

Question

\(41-46\) . Find the horizontal and vertical components of the vector with given length and direction, and write the vector in terms of the vectors \(\mathbf{i}\) and \(\mathbf{j}\) . $$ |\mathbf{v}|=1, \quad \theta=225^{\circ} $$

Step-by-Step Solution

Answer

The vector is \(-\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j}\).

Step 1: Understanding the Vector Components

The horizontal and vertical components of a vector can be found using trigonometric functions of the angle, given the magnitude of the vector. These components for vector \( \mathbf{v} \) can be represented as \( v_x \mathbf{i} + v_y \mathbf{j} \), where \( v_x \) and \( v_y \) are the horizontal and vertical components, respectively.

Step 2: Calculate the Horizontal Component

The horizontal component \( v_x \) is calculated using the cosine of the angle \( \theta \). The formula is given by \( v_x = |\mathbf{v}| \cos \theta \). Substitute the given values: \( v_x = 1 \cos 225^{\circ} \). Since \( \cos 225^{\circ} = -\frac{\sqrt{2}}{2} \), thus \( v_x = -\frac{\sqrt{2}}{2} \mathbf{i} \).

Step 3: Calculate the Vertical Component

The vertical component \( v_y \) is found using the sine of the angle \( \theta \). The formula is \( v_y = |\mathbf{v}| \sin \theta \). Substitute the values: \( v_y = 1 \sin 225^{\circ} \). Since \( \sin 225^{\circ} = -\frac{\sqrt{2}}{2} \), thus \( v_y = -\frac{\sqrt{2}}{2} \mathbf{j} \).

Step 4: Write the Vector in Terms of \( \mathbf{i} \) and \( \mathbf{j} \)

Combine the components to express the vector \( \mathbf{v} \) as \(-\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \).

Key Concepts

Horizontal Component

The horizontal component of a vector is a crucial aspect when it comes to breaking down vectors into more manageable parts. Think of it like a shadow that a vector casts on the x-axis. This helps in simplifying calculations in physics and engineering. To find the horizontal component, we make use of trigonometric functions.
Typically, for any vector with a certain angle, the horizontal component can be determined using the cosine function. Why cosine? Because it relates directly to how much of the vector lies along the horizontal axis.

• Formula: The horizontal component, often denoted as \( v_x \), is calculated with the formula: \( v_x = |\mathbf{v}| \cos \theta \), where \( |\mathbf{v}| \) is the magnitude of the vector, and \( \theta \) is the angle it makes with the positive x-axis.
• Example: For a vector with magnitude 1 and angle \( 225^{\circ} \), since \( \cos 225^{\circ} = -\frac{\sqrt{2}}{2} \), the horizontal component is \( -\frac{\sqrt{2}}{2} \mathbf{i} \).

Understanding how to calculate the horizontal component is fundamental for tasks that require decomposing vectors into vector sums.

Vertical Component

The vertical component of a vector is comparable to how high or low the vector moves along the vertical, or y-axis. This component helps when trying to determine how a vector affects things moving up or down or when projecting forces in vertical directions.
To pinpoint the vertical component, we employ the sine function. It is employed here because it measures how far upwards or downwards the vector stretches, based on its angle.

• Formula: If you want to find the vertical part, labelled as \( v_y \), you would use the formula \( v_y = |\mathbf{v}| \sin \theta \), where \( |\mathbf{v}| \) is the vector's size, and \( \theta \) gives the vector's orientation.
• Example: As seen with a magnitude of 1 at \( 225^{\circ} \), since \( \sin 225^{\circ} = -\frac{\sqrt{2}}{2} \), the vertical component turns out to be \( -\frac{\sqrt{2}}{2} \mathbf{j} \).

The ability to determine the vertical component is essential for analyzing vector-induced movements in vertical directions.

Trigonometric Functions

Trigonometric functions are pivotal when working with vectors since they allow us to express vectors in more tangible components. When dealing with vector problems, understanding these functions can simplify calculations significantly.
Vectors are often described in terms of their magnitude and direction, and trigonometric functions let us translate these into clear horizontal and vertical components.

• Cosine Function: This is used to find the horizontal component of a vector. It determines how much of a vector's direction is devoted to moving horizontally.
• Sine Function: This is utilized for the vertical component. It tells us how much of the vector does in moving vertically.
• Common Angles: Understanding values of sine and cosine for typical angles, like \( 0^{\circ}, 90^{\circ}, 180^{\circ}, 270^{\circ} \), greatly aids in quick calculations.

Mastering trigonometric functions not only makes finding vector components easier but also equips you with the skills for more advanced geometry and physics-related problems.