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. \(\|\mathbf{v}\|=3, \quad \alpha=240^{\circ}\)

Short answer

The vector is \(\backslash mathbf{v} = -1.5\backslash mathbf{i} - 1.5\backslash sqrt{3}\backslash mathbf{j}\).

Step-by-step solution

Understand the Problem

We need to express a vector \(\backslash mathbf{v}\) in the form \(\backslash mathbf{ai}+ \backslash mathbf{bj}\), given its magnitude (length) \(\backslash |\backslash mathbf{v}\backslash |\) and the angle \(\backslash alpha\) it makes with the positive \(\backslash x\)-axis.

Use Trigonometry to Find Components

Using trigonometry, the components of the vector \(\backslash mathbf{v}\) can be found as follows: \(\backslash mathbf{a} = \backslash |\backslash mathbf{v}\backslash | \backslash cos \backslash alpha\) and \(\backslash mathbf{b} = \backslash |\backslash mathbf{v}\backslash | \backslash sin \backslash alpha\).

Plug in the Given Values

Substitute \(\backslash |\backslash mathbf{v}\backslash | = 3\) and \(\backslash alpha = 240^\backslash ci \) into the equations: \(\backslash mathbf{a} = 3\backslash cos(240^\backslash ci)\) and \(\backslash mathbf{b} = 3\backslash sin(240^\backslash ci)\).

Evaluate Trigonometric Functions

Evaluate \(\backslash cos(240^\backslash ci)\) and \(\backslash sin(240^\backslash ci)\). From the unit circle, \(\backslash cos(240^\backslash ci) = -0.5\) and \(\backslash sin(240^\backslash ci) = -\backslash sqrt{3}/2\).

Calculate the Components

Calculate \(\backslash mathbf{a} = 3 \backslash cdot (-0.5) = -1.5\) and \(\backslash mathbf{b} = 3 \backslash cdot (-\backslash sqrt{3}/2) = -1.5\backslash sqrt{3}\).

Write the Vector in Component Form

Combine the components to write the vector in the required form: \(\backslash mathbf{v} = -1.5\backslash mathbf{i} - 1.5\backslash sqrt{3}\backslash mathbf{j}\).

Key concepts

trigonometry

To understand how we find the components of a vector, we need to delve into trigonometry. Trigonometry is the study of the relationships between the sides and angles of triangles. It helps us to relate angles to the lengths of sides, which is essential when working with vectors.

For instance, when we have a vector \(\mathbf{v}\) with a given magnitude and angle \((\alpha)\) relative to the positive x-axis, we can use trigonometric functions cosine (cos) and sine (sin). These functions arise from the unit circle, which we’ll explore in another section.

To find the x-component (a) of the vector, we use the formula \(\mathbf{a} = \|\mathbf{v}\| \cos\alpha\). Similarly, to find the y-component (b), we use \(\mathbf{b} = \|\mathbf{v}\| \sin\alpha\). These components come directly from the definitions of the trigonometric functions, where:

• \(\cos\alpha = \) adjacent side (x-component) divided by the hypotenuse (magnitude of the vector)
• \(\sin\alpha = \) opposite side (y-component) divided by the hypotenuse

Trigonometry thus provides a precise way to convert polar coordinates (angle and magnitude) into Cartesian coordinates (x and y).”

magnitude

The magnitude of a vector is a measure of its length. It tells us how long the vector is without giving us any information about its direction. Mathematically, the magnitude of a vector \(\mathbf{v}\), denoted \(|\mathbf{v}|\), can be found using the Pythagorean theorem if we know its components.

For a vector \((a \mathbf{i} + b \mathbf{j})\), the magnitude is calculated as:
\[\|\mathbf{v}\| = \sqrt{a^2 + b^2}\]
In our example, we are already provided with the magnitude, which is 3. Therefore, we did not need to calculate it but rather used it directly in our formulas for vector components.

Understanding the concept of magnitude is crucial because it gives the scale of the vector, and when combined with the direction (angle), it can completely describe the vector in 2D space.

unit circle

The unit circle is a fundamental concept in trigonometry and is crucial for understanding how to evaluate trigonometric functions. A unit circle is a circle with a radius of exactly 1, centered at the origin of a coordinate system. It serves as a reference for defining trigonometric functions for all angles.

Each point on the unit circle corresponds to an angle measured from the positive x-axis, and it has coordinates \( (\cos\alpha, \sin\alpha) \). This means that the cosine of an angle \((\alpha)\) is the x-coordinate, and the sine is the y-coordinate.

In the given exercise, we needed to evaluate \(\cos(240^\circ)\) and \(\sin(240^\circ)\). On the unit circle:

• \(\cos(240^\circ) = -0.5\)
• \(\sin(240^\circ) = -\sqrt{3}/2\)

These values are derived from the unit circle and are necessary to find the vector components. The unit circle simplifies these evaluations because it provides the exact trigonometric values for standard angles, helping us quickly solve for the components of any vector based on its angle and magnitude.

By combining these trigonometric values with the vector’s magnitude, we derived the final components to properly write the vector in component form.