Problem 52 - Algebra and Trigonometry

Question

\(47-52\) . Find the magnitude and direction (in degrees) of the vector. $$ \mathbf{v}=\mathbf{i}+\mathbf{j} $$

Step-by-Step Solution

Answer

Magnitude: \( \sqrt{2} \), Direction: \( 45^\circ \).

Step 1: Understand the Vector Components

The vector \( \mathbf{v} = \mathbf{i} + \mathbf{j} \) has components \( (1, 1) \) since \( \mathbf{i} \) represents \( x = 1 \) and \( \mathbf{j} \) represents \( y = 1 \).

Step 2: Calculate the Magnitude of the Vector

The magnitude of a vector \( \mathbf{v} = (x, y) \) is calculated by the formula: \[ |\mathbf{v}| = \sqrt{x^2 + y^2} \]Substitute the components:\[ |\mathbf{v}| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

Step 3: Determine the Direction of the Vector

The direction \( \theta \) of a vector can be found using the arctangent of the ratio of its components:\[ \theta = \text{atan2}(y, x) = \text{atan2}(1, 1) \]This simplifies to \( \theta = 45^\circ \), which is the angle above the positive \(x\)-axis.

Key Concepts

Vector Components

When analyzing vectors, an essential step is identifying their components. Vector components distinguish each vector by breaking it into its horizontal and vertical influences. Think of a vector as an arrow originating from the origin, its position expressed in terms of its components.

• In the equation \( \mathbf{v} = \mathbf{i} + \mathbf{j} \), each term represents a component of the vector.
• \( \mathbf{i} \) is the unit vector along the x-axis, providing a value of 1 in the horizontal direction.


• \( \mathbf{j} \) is the unit vector along the y-axis, contributing a value of 1 vertically.

Consequently, the vector \( \mathbf{v} \) in its component form is \((1, 1)\), where 1 is the displacement along both the x-axis and y-axis. Recognizing these components is crucial for further calculations, such as finding magnitude and direction.

Magnitude Formula

Magnitude measures the size or length of a vector. To determine how "long" a vector is, we use the magnitude formula derived from the Pythagorean Theorem:\[ |\mathbf{v}| = \sqrt{x^2 + y^2} \]This formula helps calculate the diagonal length of the vector triangle.

• Given \( \mathbf{v} = (1, 1) \), each component squared is added: \( 1^2 + 1^2 \).
• Sum the squares to get 2, then take the square root to find the magnitude \( \sqrt{2} \).

The magnitude \( \sqrt{2} \) gives us a sense of how far the vector extends in its current direction. It’s slightly longer than 1, showing both x and y influences.

Direction Angle

The direction of a vector is vital in understanding where it points relative to the positive x-axis. Often expressed in degrees, the direction angle tells us the orientation of a vector in the plane:

• For the vector \( \mathbf{v} = (1, 1) \), the angle is between the vector and the positive x-axis.
• This angle, known as \( \theta \), is the direction angle.

To find this, compute the arctan of the y-component over the x-component. In our case, this ratio is \( 1/1 = 1 \). The angle associated with an arctan of 1 is 45 degrees, indicating the vector points northeast at a 45-degree angle relative to the x-axis.

Arctangent Function

The arctangent function is a trigonometric function used to find angles from ratios. Specifically, \( \tan^{-1} \) or \( \text{atan2} \) is used to find an angle \( \theta \) given a ratio of opposite to adjacent sides in a right triangle.In vector analysis:

• The arctan function converts a vector's slope into its directional angle.
• The ratio for \( \mathbf{v} = (1, 1) \) is \( 1 \), signifying a slope of 1 upward for each unit across, matching the 45-degree angle.

It’s a handy tool to translate from rectangular coordinates (x, y) into a more intuitive angle \( \theta \). The arctangent is especially useful when the vector lays neither horizontally nor vertically, translating slopes efficiently into comprehensible angles for navigation and representation.