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. \(|\mathbf{v}|=5, \quad \alpha=60^{\circ}\)

Short answer

\( \mathbf{v} = 2.5 \mathbf{i} + 2.5 \sqrt{3} \mathbf{j} \)

Step-by-step solution

Understand the Problem

We are given the magnitude \(\|\bf{v}\| = 5\) and the angle \(\alpha = 60^{\circ}\) that the vector \(\bf{v}\) makes with the positive \(\bf{x}\)-axis. We need to express the vector in the form \(\bf{ai} + \bf{bj}\).

Determine the Components

The vector can be represented in component form using trigonometric functions. The components are: \[ a = \|\bf{v}\| \cos\alpha\] \[ b = \|\bf{v}\| \sin\alpha \] \Thus, \(\bf{ai} + \bf{bj}\)

Calculate the Components

Plugging in the given values: \[ a = 5 \cos 60^{\circ} \ = 5 \cdot \frac{1}{2} = 2.5\] \[ b = 5 \sin 60^{\circ} \ = 5 \cdot \frac{\sqrt{3}}{2} = 2.5 \sqrt{3}\]

Write the Vector

The vector \(\bf{v}\) in the form \(\bf{ai} + \bf{bj}\) is \[ \bf{v} = 2.5 \bf{i} + 2.5 \sqrt{3} \bf{j}\]

Key concepts

Vector Components

Vectors can be split into two parts: components along the x-axis and components along the y-axis. When we talk about vector \(\textbf{v}\), we often represent it in the form \(\textbf{ai} + \textbf{bj}\). Here, \(\textbf{i}\) and \(\textbf{j}\) are unit vectors along the x-axis and y-axis, respectively. This means \(\textbf{i}\) points in the direction of the x-axis and has a magnitude of 1, while \(\textbf{j}\) points in the direction of the y-axis and also has a magnitude of 1.

To find the components \(\textbf{a}\) and \(\textbf{b}\), we use trigonometric functions. The component along the x-axis, \(\textbf{a}\), can be found using cosine, and the component along the y-axis, \(\textbf{b}\), can be found using sine. These calculations ensure that we know exactly how much of the vector's length lies along each axis.

Trigonometric Functions

Trigonometric functions like sine and cosine help us relate the angles and sides of a triangle. When dealing with vectors, these functions are essential for breaking down the vector into its components.

To find the x-component (\(\textbf{a}\)) of a vector with magnitude \(\|\textbf{v}\|\) and angle \(\alpha\):
We use the equation \[a = \|\textbf{v}\| \cos\alpha\]
Similarly, to find the y-component (\(\textbf{b}\)):
We use the equation \[b = \|\textbf{v}\| \sin\alpha\]
In our exercise, we had \(\|\textbf{v}\| = 5\) and \(\alpha = 60^{\circ}\). Plugging these values in:
\[a = 5 \cos 60^{\circ} = 5 \cdot \frac{1}{2} = 2.5\]
\[b = 5 \sin 60^{\circ} = 5 \cdot \frac{\sqrt{3}}{2} = 2.5\sqrt{3}\]

These trigonometric functions allow us to convert the magnitude and direction of the vector into its horizontal and vertical components.

Magnitude and Direction

The magnitude of a vector is its length. It tells us how strong or long the vector is. The direction is given by the angle it makes with the positive x-axis.

In our example, the vector had a magnitude of \(5\) and an angle \(\alpha = 60^{\circ}\). Magnitude and direction are crucial to understanding a vector's impact in space. Using magnitude and direction with trigonometric functions, we could determine that our vector, \(\textbf{v}\), is \(2.5 \textbf{i} + 2.5 \sqrt{3} \textbf{j}\).This final form \(\textbf{ai} + \textbf{bj}\) tells us the exact positioning: \(2.5\) units along the x-axis and \(2.5 \sqrt{3}\) units along the y-axis. This breakdown is what makes vectors so practical in fields ranging from physics to engineering, as you can predict and manipulate the direction and magnitude components easily.