Problem 51 - Trigonometry

Question

Vector \(\mathbf{V}\) is in standard position and makes an angle of \(30^{\circ}\) with the positive \(x\)-axis. Its magnitude is 18 . Write \(\mathbf{V}\) in component form \(\langle a, b\rangle\) and in vector component form \(a \mathbf{i}+b \mathbf{j}\).

Step-by-Step Solution

Answer

The vector \( \mathbf{V} \) in component form is \( \langle 9\sqrt{3}, 9 \rangle \) and in vector component form is \( 9\sqrt{3} \mathbf{i} + 9 \mathbf{j} \).

Step 1: Understand The Problem

We need to write vector \( \mathbf{V} \) in component form \( \langle a, b \rangle \) and as a vector using \( \mathbf{i} \) and \( \mathbf{j} \), given that \( \mathbf{V} \) is 18 units long and makes an angle of \( 30^{\circ} \) with the positive x-axis.

Step 2: Expression for the Vector Components

To find the components \( a \) and \( b \), use the formulas for a vector given its magnitude and direction: \( a = \| \mathbf{V} \| \cos \theta \) and \( b = \| \mathbf{V} \| \sin \theta \), where \( \theta \) is the angle with the positive x-axis and \( \| \mathbf{V} \| \) is the vector's magnitude.

Step 3: Calculate the x-component

Substitute the given values into the formula for \( a \): \( a = 18 \cos(30^{\circ}) \). Recall that \( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \). Thus, \( a = 18 \times \frac{\sqrt{3}}{2} = 9\sqrt{3} \).

Step 4: Calculate the y-component

Substitute the given values into the formula for \( b \): \( b = 18 \sin(30^{\circ}) \). Recall that \( \sin(30^{\circ}) = \frac{1}{2} \). Thus, \( b = 18 \times \frac{1}{2} = 9 \).

Step 5: Present the Vector in Component Form

The vector \( \mathbf{V} \) in component form is \( \langle 9\sqrt{3}, 9 \rangle \).

Step 6: Present the Vector in Vector Component Form

The vector \( \mathbf{V} \) in vector component form using \( \mathbf{i} \) and \( \mathbf{j} \) is \( 9\sqrt{3} \mathbf{i} + 9 \mathbf{j} \).

Key Concepts

Trigonometric Functions

Trigonometric functions are a cornerstone in understanding vector calculations and their components. Often, vectors are given in terms of their direction and magnitude rather than their exact position on a coordinate plane. To find the components of such vectors, the functions cosine and sine are immensely helpful.

• **Cosine (\( \cos \theta \))**: This function helps determine the adjacent side (x-component) of a right triangle when the hypotenuse and an angle are known. In the context of vectors, if we know the magnitude of a vector and the angle it makes with the x-axis, the cosine of this angle will allow us to find its projection along the x-axis.
• **Sine (\( \sin \theta \))**: Similarly, sine allows for finding the opposite side (y-component) of the triangle. With vectors, it calculates the component parallel to the y-axis when you have the vector’s magnitude and the angle.

In our exercise, we used the angle \(30^{\circ}\) and vector's magnitude to find its components using these functions. Remember, both cosine and sine can be calculated using trigonometric tables or a calculator to provide exact component values for given angles.

Magnitude of a Vector

The magnitude of a vector is essentially a measure of its length. Imagine it as the distance traversed from the vector's initial to its terminal point. Understanding magnitude is crucial for determining how 'strong' or 'large' a vector reflects a quantity in physics or defines points in space.

• To compute the magnitude, we generally take the square root of the sum of the squares of its components. But when you're given the magnitude directly, like in this problem (which is 18), it simplifies finding direction-based components.
• It is often represented as \( \| \mathbf{V} \| \), and it’s a scalar. This means it doesn’t have direction, only size.

In our exercise, this known value allowed us to partition the vector into its horizontal and vertical parts efficiently by multiplying it by relevant trigonometric functions.

Standard Position Angle

Vectors are often described by their angle in the standard position, meaning their starting point is at the origin (0,0), and it makes a definite angle with the positive x-axis. This concept is critical because it provides a consistent, universal way to describe vector orientation relative to standard coordinate axes.

• By defining an angle in this manner, calculations involving vector direction become less complex and far more systematic.
• This angle serves as a reference, simplifying the prediction of where a vector points just by knowing its angle and magnitude.

In the given exercise, the vector is described as making an angle of \(30^{\circ}\) with the x-axis in standard position. Such angles are common in trigonometry, and they give us direct pathways to apply trigonometric functions when calculating vector components.