Problem 52 - Trigonometry

Question

Vector \(\mathbf{U}\) is in standard position and makes an angle of \(120^{\circ}\) with the positive \(x\)-axis. Its magnitude is 24 . Write \(\mathbf{U}\) in component form \(\langle a, b\rangle\) and in vector component form \(a \mathbf{i}+b \mathbf{j}\).

Step-by-Step Solution

Answer

\( \mathbf{U} = \langle -12, 12\sqrt{3} \rangle = -12 \mathbf{i} + 12\sqrt{3} \mathbf{j} \).

Step 1: Use Trigonometry to Determine Components

In vector analysis, you can find the components of a vector using its angle with the positive x-axis and its magnitude. The formula we use here for a vector \( \mathbf{U} \) with magnitude \( |\mathbf{U}| = 24 \) and angle \( \theta = 120^{\circ} \) is as follows:- The x-component is calculated by \( a = |\mathbf{U}| \cos(\theta) \).- The y-component is calculated by \( b = |\mathbf{U}| \sin(\theta) \).By substituting the values, we get:\[ a = 24 \cos(120^{\circ}) \]\[ b = 24 \sin(120^{\circ}) \]

Step 2: Calculate Cosine and Sine Values

To determine the cosine and sine components, recall the properties of trigonometric functions at specific angles.For \( \theta = 120^{\circ} \):- \( \cos(120^{\circ}) = -\frac{1}{2} \)- \( \sin(120^{\circ}) = \frac{\sqrt{3}}{2} \)Substitute these values into the expression for \( a \) and \( b \).

Step 3: Calculate Components

Substitute the trigonometric values into the formulas for the components:\[ a = 24 \times \left(-\frac{1}{2}\right) = -12 \]\[ b = 24 \times \frac{\sqrt{3}}{2} = 12\sqrt{3} \]So, the components are \( a = -12 \) and \( b = 12\sqrt{3} \).

Step 4: Write in Vector Notation

Now that we have determined \( a \) and \( b \), write \( \mathbf{U} \) in both component forms:- **Component Notation**: \( \langle -12, 12\sqrt{3} \rangle \)- **Vector Component Form**: \( -12 \mathbf{i} + 12\sqrt{3} \mathbf{j} \)

Key Concepts

Trigonometry

Trigonometry connects angles and sides in triangles and is helpful in breaking down vectors into their components. When angles are involved, it helps us

• determine how long each side is, given an angle and one side length,
• analyze the position and direction of objects.

For vectors, trigonometry allows us to determine the precise horizontal (x-axis) and vertical (y-axis) movements of a vector. When given the angle a vector makes with the positive x-axis, along with its magnitude, we can use trigonometric principles to find its components.
When we talk about components of a vector like \[ \mathbf{U} = \langle a, b \rangle \]in trigonometry, we are essentially finding projections of the vector on the x-axis and y-axis. This involves using trigonometric functions of the given angle to split the vector into these parts.

Magnitude and Direction

When tackling vector problems, two main attributes define a vector: its magnitude and direction. Magnitude refers to the length or size of the vector, which is a scalar quantity and always positive.
Direction is about where the vector points in space, often described using angles from a reference direction—here, the positive x-axis.
To convert a vector's magnitude and direction into its x and y components, we rely upon the given angle (\( \theta \)) and the magnitude (\( |\mathbf{U}| \)). For example, the vector in our problem has a magnitude of 24 and a direction of \( 120^{\circ} \) from the positive x-axis. With these two pieces of information, you can ascertain how far, horizontally and vertically, the vector extends in the plane.

Sine and Cosine Functions

Sine and cosine functions are two of the primary trigonometric functions used in mathematical applications. These functions help determine the components of a vector based on its angle.
They are defined as:

• \( \cos(\theta) \): the horizontal component divided by the hypotenuse (adjacent/hypotenuse)
• \( \sin(\theta) \): the vertical component divided by the hypotenuse (opposite/hypotenuse)

In the vector scenario:

• The x-component comes from multiplying the magnitude by the cosine of the angle, \( a = |\mathbf{U}| \cos(\theta) \)
• The y-component is derived by multiplying the magnitude by the sine of the angle, \( b = |\mathbf{U}| \sin(\theta) \)

These calculations show how \( a = 24 \cos(120^{\circ}) = -12 \) and \( b = 24 \sin(120^{\circ}) = 12\sqrt{3} \), articulating the vector in component form as both \( \langle -12, 12\sqrt{3} \rangle \) and \( -12 \mathbf{i} + 12\sqrt{3} \mathbf{j} \).