Problem 38 - Algebra and Trigonometry

Question

Find the magnitude and direction (in degrees) of the vector. $$ \mathbf{v}=\mathbf{i}+\mathbf{j} $$

Step-by-Step Solution

Answer

Magnitude is \( \sqrt{2} \), direction is 45° from positive x-axis.

Step 1: Identify Components of the Vector

The vector \( \mathbf{v} = \mathbf{i} + \mathbf{j} \) can be written in component form as \( \mathbf{v} = (1, 1) \). This means the vector has an x-component of 1 (coefficient of \( \mathbf{i} \)) and a y-component of 1 (coefficient of \( \mathbf{j} \)).

Step 2: Magnitude Calculation

To find the magnitude of \( \mathbf{v} \), use the formula: \[ \| \mathbf{v} \| = \sqrt{x^2 + y^2} \].Substitute the components (1, 1):\[ \| \mathbf{v} \| = \sqrt{1^2 + 1^2} = \sqrt{2} \].

Step 3: Determine the Angle of the Vector

The direction \( \theta \) of a vector can be found using the formula: \[ \theta = \tan^{-1} \left( \frac{y}{x} \right) \].Substituting for \( y = 1 \) and \( x = 1 \), we get:\[ \theta = \tan^{-1} (1) = 45^\circ \].

Step 4: Check the Quadrant for the Angle

Since both components of the vector are positive, the vector lies in the first quadrant where the angle 45° is correct relative to the standard position from the positive x-axis.

Key Concepts

Vector Components

When dealing with vectors, imagining them as arrows in a coordinate plane can be helpful. A vector described using unit vectors, such as \( \mathbf{i} \) and \( \mathbf{j} \), is broken down into its horizontal and vertical components. For the vector \( \mathbf{v} = \mathbf{i} + \mathbf{j} \), simply identify the coefficients of these unit vectors to obtain the vector components. Here, the x-component (horizontal) is 1, and the y-component (vertical) is 1. This representation in component form is \( \mathbf{v} = (1, 1) \). Understanding how to identify these components is crucial when calculating both the vector's magnitude and direction.

Magnitude Calculation

The magnitude of a vector refers to its length, which is calculated using the Pythagorean theorem. For a vector with components \( (x, y) \), the magnitude is given by the formula \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \). This formula is derived from the distance formula in the coordinate plane. For our vector \( \mathbf{v} = (1, 1) \), plug in the component values:

• \( \sqrt{1^2 + 1^2} \)
• This simplifies to \( \sqrt{1 + 1} \)
• Resulting in a magnitude of \( \sqrt{2} \)

This value represents the vector's distance from the origin to its endpoint in the coordinate plane.

Direction Angle

The direction of a vector is defined by the angle it makes with the positive x-axis. This angle is important for understanding the vector's orientation. For a vector in component form, the direction angle \( \theta \) should be calculated using the arc tangent function. Particularly, the angle is given by \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \). For our vector \( \mathbf{v} = (1, 1) \), the calculation would be:

• \( \tan^{-1} \left( \frac{1}{1} \right) \)
• This simplifies to \( \tan^{-1} (1) \)
• Resulting in \( 45^\circ \)

This angle indicates that the vector is directed at a 45-degree angle from the positive x-axis.

Inverse Tangent Function

The inverse tangent, or arctan function, is essential for finding the angle of a vector. It allows us to determine the angle given a ratio of the vector's y-component to its x-component. The function \( \tan^{-1} \left( \frac{y}{x} \right) \) outputs the angle \( \theta \) in either radians or degrees. Using \( (x, y) = (1, 1) \) from our example, \( \tan^{-1} (1) \) yields \( 45^\circ \), highlighting how an equal y and x produce a 45-degree angle. This operation leverages the symmetrical properties of the tangent graph, making it a go-to calculation for finding vector angles.

Quadrant Analysis

Understanding in which quadrant a vector resides is crucial for accurate angle interpretation. The Cartesian plane is divided into four quadrants that help specify the vector's direction. Generally, the sign of the vector's components guides us:

• First quadrant: Both components positive.
• Second quadrant: Negative x-component, positive y-component.
• Third quadrant: Both components negative.
• Fourth quadrant: Positive x-component, negative y-component.

For the vector \( \mathbf{v} = (1, 1) \), both components are positive, confirming its location in the first quadrant. Here, the calculated 45° angle accurately describes the direction since the first quadrant is where angles range from 0° to 90° in standard position.