Problem 52 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

The magnitude is \( \sqrt{2} \) and the direction is 45 degrees.

Step 1: Understand the Components

The given vector \( \mathbf{v} = \mathbf{i} + \mathbf{j} \) means that the vector has components 1 along the x-axis and 1 along the y-axis. This can be represented as \( \mathbf{v} = (1, 1) \).

Step 2: Calculate the Magnitude

The magnitude of a vector \( (a, b) \) is calculated using the formula \( \sqrt{a^2 + b^2} \). Here, \( a = 1 \) and \( b = 1 \). Substitute these into the formula to get the magnitude:\[ |\mathbf{v}| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \].Therefore, the magnitude of the vector is \( \sqrt{2} \).

Step 3: Determine the Direction Using the Inverse Tangent

The direction of a vector is found by calculating the angle \( \theta \) it makes with the positive x-axis. This angle can be found using the inverse tangent function:\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{1}{1}\right) \].Thus, \( \theta = \tan^{-1}(1) \).

Step 4: Calculate the Angle (Direction)

The inverse tangent of 1 is an angle of 45 degrees, as \( \tan(45^\circ) = 1 \). Therefore, the direction \( \theta \) of the vector is 45 degrees.

Key Concepts

Vector Components

Vectors are mathematical objects that have both magnitude and direction. In this section, we'll focus on understanding vector components. A vector component shows the influence of the vector in a particular direction.For a two-dimensional vector like the one given in the exercise, represented as \( \mathbf{v} = \mathbf{i} + \mathbf{j} \), it means the vector has components along the \( x \)-axis and \( y \)-axis. The notation \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors pointing in the direction of the \( x \)-axis and \( y \)-axis respectively. Therefore, for our vector \( \mathbf{v} = (1, 1) \), each of these components is 1:- The \( x \)-component is 1.- The \( y \)-component is also 1.This means the vector moves 1 unit in the horizontal direction and 1 unit in the vertical direction. Understanding that a vector can be broken down into components helps in visualizing how it interacts within the coordinate system.

Magnitude Calculation

The magnitude of a vector tells us how long the vector is. It gives you the length of the vector from the origin to the endpoint. To calculate the magnitude of a vector \((a, b)\), we apply the formula \[ \sqrt{a^2 + b^2} \].Let's apply this to our vector \( \mathbf{v} = (1, 1) \):- We substitute \( a = 1 \) and \( b = 1 \) into the formula: - \[ \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \].So, the magnitude of our vector is \( \sqrt{2} \). This tells us the distance from the point \( (0,0) \) to \( (1, 1) \) is \( \sqrt{2} \) units. The magnitude reflects the overall size or extent of the vector, independent of its direction.

Inverse Tangent Function

The inverse tangent function is used to calculate the angle that the vector makes with the positive \( x \)-axis. This is essential for determining the vector's direction.The inverse tangent, denoted as \( \tan^{-1} \) or \( \arctan \), helps find this angle \( \theta \). For a vector with components \((a, b)\), the formula to find the direction is:- \[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \].In our exercise:- We substitute \( a = 1 \) and \( b = 1 \) into the formula: - \[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1) \].The inverse tangent of 1 is known to be 45 degrees. Therefore, our vector's direction is 45 degrees in relation to the positive \( x \)-axis.Understanding how to calculate this angle is crucial when working with vectors, as it helps in visualizing and understanding the vector’s orientation within a plane.