Problem 43 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Vector Components

A vector in plane coordinate can be represented with its magnitude and direction. If a vector has a magnitude of \(|\mathbf{v}|\) and a direction angle \(\theta\), it can be split into horizontal and vertical components using trigonometric functions. The horizontal component is \(\mathbf{v}_x = |\mathbf{v}| \cdot \cos(\theta)\) and the vertical component is \(\mathbf{v}_y = |\mathbf{v}| \cdot \sin(\theta)\).

Step 2: Calculate the Horizontal Component

Calculate the horizontal component using the cosine function: \[ \mathbf{v}_x = |\mathbf{v}| \cdot \cos(225^{\circ}) \] Given \(|\mathbf{v}| = 1\) and \(\theta = 225^{\circ}\), we have \[ \cos(225^{\circ}) = -\frac{\sqrt{2}}{2} \] So, \[ \mathbf{v}_x = 1 \cdot -\frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} \]

Step 3: Calculate the Vertical Component

Calculate the vertical component using the sine function: \[ \mathbf{v}_y = |\mathbf{v}| \cdot \sin(225^{\circ}) \] Given \(|\mathbf{v}| = 1\) and \(\theta = 225^{\circ}\), we have \[ \sin(225^{\circ}) = -\frac{\sqrt{2}}{2} \] So, \[ \mathbf{v}_y = 1 \cdot -\frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} \]

Step 4: Express the Vector in Terms of i and j

Combine the horizontal and vertical components to express the vector in terms of \(\mathbf{i}\) and \(\mathbf{j}\). The vector is given by: \[ \mathbf{v} = \mathbf{v}_x \mathbf{i} + \mathbf{v}_y \mathbf{j} \] Substitute \(\mathbf{v}_x\) and \(\mathbf{v}_y\): \[ \mathbf{v} = -\frac{\sqrt{2}}{2}\mathbf{i} - \frac{\sqrt{2}}{2}\mathbf{j} \]

Key Concepts

Magnitude and Direction

Vectors have two key attributes: magnitude and direction.
The magnitude is the length or size of the vector, representing how strong or significant the vector is. In the problem at hand, the given vector has a magnitude of 1.

Direction refers to where the vector is pointing in a coordinate plane. It's commonly expressed using an angle from a standard reference direction, like the positive x-axis. Here, the direction is given as \(225^{\circ}\). This implies the vector is directed into the third quadrant of the coordinate plane, where both x and y components are negative.

Understanding these two aspects is crucial, as they determine the vector's position and influence in a 2D space.

Trigonometric Functions

Trigonometric functions, such as cosine and sine, are essential for breaking a vector's magnitude into its horizontal and vertical components.
Using the angle \(\theta\) provided, these functions allow us to determine how much of the vector's magnitude lies along the x-axis (horizontal) and y-axis (vertical).

The horizontal component, \(\mathbf{v}_x\), is found using cosine:

• \(\mathbf{v}_x = |\mathbf{v}| \cdot \cos(\theta)\)
• For the given problem: \(\cos(225^{\circ}) = -\frac{\sqrt{2}}{2}\), so \(\mathbf{v}_x = -\frac{\sqrt{2}}{2}\)

The vertical component, \(\mathbf{v}_y\), is determined using sine:

• \(\mathbf{v}_y = |\mathbf{v}| \cdot \sin(\theta)\)
• In this exercise: \(\sin(225^{\circ}) = -\frac{\sqrt{2}}{2}\), which gives us \(\mathbf{v}_y = -\frac{\sqrt{2}}{2}\)

These calculations demonstrate how trigonometric functions are invaluable for converting a vector's overall magnitude and direction into specific coordinate impacts.

Coordinate Representation

Once the horizontal and vertical components are found, vectors can be clearly described in a coordinate format.
This format incorporates unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) which represent the standard basis vectors along the x and y axes, respectively.

The general way to express a vector \(\mathbf{v}\) in these terms is by combining its components:

• \(\mathbf{v} = \mathbf{v}_x \mathbf{i} + \mathbf{v}_y \mathbf{j}\)
• From our calculation, it's \(\mathbf{v} = -\frac{\sqrt{2}}{2}\mathbf{i} - \frac{\sqrt{2}}{2}\mathbf{j}\)

This expression clearly delineates the vector's influence along each axis and is vital for solving problems involving vector additions, transformations, or projections.