Problem 51 - Algebra and Trigonometry

Question

\(47-52\) . Find the magnitude and direction (in degrees) of the vector. $$ \mathbf{v}=\mathbf{i}+\sqrt{3} \mathbf{j} $$

Step-by-Step Solution

Answer

Magnitude is 2, direction is 60 degrees.

Step 1: Identify the Vector Components

Start by identifying the components of the vector \( \mathbf{v} \). The vector \( \mathbf{v} = \mathbf{i} + \sqrt{3} \mathbf{j} \) consists of:- An \( \mathbf{i} \) component of 1, which is the horizontal component.- A \( \sqrt{3} \mathbf{j} \) component, which is the vertical component.

Step 2: Calculate the Magnitude of the Vector

The magnitude \( |\mathbf{v}| \) of a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \) is given by the formula \( |\mathbf{v}| = \sqrt{a^2 + b^2} \). Substituting the values, we get:\[ |\mathbf{v}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \].So, the magnitude of the vector is 2.

Step 3: Determine the Direction of the Vector

The direction \( \theta \) of a vector is given by \( \theta = \tan^{-1}\left( \frac{b}{a} \right) \), where \( a = 1 \) and \( b = \sqrt{3} \). Thus,\[ \theta = \tan^{-1}\left( \frac{\sqrt{3}}{1} \right) = \tan^{-1}(\sqrt{3}) \].Knowing \( \tan(60^\circ) = \sqrt{3} \), we find \( \theta = 60^\circ \).

Key Concepts

Understanding Vector Components

Vectors are important mathematical entities used to describe both magnitude and direction. In any two-dimensional vector expressed in the form \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \), the components are crucial for various calculations.
- The horizontal component, represented by \( a\mathbf{i} \), tells us how far the vector travels horizontally. In the original exercise, this component is \( 1 \).
- The vertical component, given by \( b\mathbf{j} \), indicates how much the vector moves vertically. Here it is \( \sqrt{3} \).
These components allow us to understand the vector structure better and perform calculations like the magnitude and direction. Extracting these values is the first step to working with vectors effectively.

Applying the Magnitude Formula

The magnitude of a vector is a measure of its total length regardless of direction. It provides a scalar quantity describing how long or large the vector is. To find the magnitude of a vector, we utilize the magnitude formula:\[ |\mathbf{v}| = \sqrt{a^2 + b^2} \]- For our vector \( \mathbf{v} = 1\mathbf{i} + \sqrt{3}\mathbf{j} \), substitute the components where \( a = 1 \) and \( b = \sqrt{3} \).
- The calculation becomes:
\[ |\mathbf{v}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]
Thus, the magnitude is 2. This calculation shows that the vector stretches across a distance of 2 units. Knowing the magnitude helps in visualizing the vector's true scale on a coordinate system.

Calculating the Direction Angle

The direction angle of a vector allows us to understand where the vector is pointing relative to the positive x-axis. For vectors, the direction is given by:\[ \theta = \tan^{-1}\left( \frac{b}{a} \right) \]- In our case, with \( a = 1 \) and \( b = \sqrt{3} \), plug in these values:
\[ \theta = \tan^{-1}\left( \frac{\sqrt{3}}{1} \right) = \tan^{-1}(\sqrt{3}) \]
- Recognizing that \( \tan(60^\circ) = \sqrt{3} \), it follows that \( \theta = 60^\circ \).
This angle indicates that the vector is pointing at a 60-degree angle above the positive x-axis. Knowing the direction angle is crucial for applications across physics and engineering, helping to precisely determine vector orientation.