Problem 47 - Algebra and Trigonometry

Question

\(47-52\) . Find the magnitude and direction (in degrees) of the vector. $$ \mathbf{v}=\langle 3,4\rangle $$

Step-by-Step Solution

Answer

Magnitude is 5, direction is approximately 53.13 degrees.

Step 1: Understanding the Vector Components

First, notice that the vector \( \mathbf{v} = \langle 3, 4 \rangle \) has components \(3\) and \(4\), where \(3\) is the horizontal component (along the x-axis) and \(4\) is the vertical component (along the y-axis).

Step 2: Calculating the Magnitude

To find the magnitude (length) of the vector \( \mathbf{v} \), we use the formula for the Euclidean norm: \( \| \mathbf{v} \| = \sqrt{3^2 + 4^2}.\) Compute this to find \( \| \mathbf{v} \| = \sqrt{9 + 16} = \sqrt{25} = 5 \). Thus, the magnitude of the vector is \(5\).

Step 3: Determining the Direction

The direction of a vector is typically calculated using the angle \( \theta \) between the vector and the positive x-axis. Use the tangent function for this, \( \tan \theta = \frac{{4}}{{3}} \). Use the arctangent function to find \( \theta = \tan^{-1} \left( \frac{4}{3} \right) \), which is approximately \(53.13^\circ\).

Step 4: Final Step: Verify Direction and Description

Since both components \(3\) and \(4\) are positive, the vector is in the first quadrant. Therefore, the angle \(53.13^\circ\) with the positive x-axis, and upwards from the horizontal, correctly represents the direction of the vector.

Key Concepts

Vector Components

Vectors are often represented as a pair of numbers in a form like \( \mathbf{v} = \langle 3, 4 \rangle \). Here, the components \(3\) and \(4\) signify the direction and magnitude along the x-axis and y-axis, respectively. These are the horizontal and vertical parts that make up the vector.

• The horizontal component \(3\) shows how far along the x-axis the vector extends.
• The vertical component \(4\) tells us how far along the y-axis it reaches.

Understanding these components is crucial for vector operations like addition, where you combine corresponding components, or for breaking down the vector into its basic parts. For our specific vector \( \langle 3, 4 \rangle \), it means that you would move 3 units to the right and 4 units up. This forms a right triangle with these components acting as the legs.

Euclidean Norm

The Euclidean norm, or magnitude, of a vector is akin to measuring the length of a vector from its tail to its head. For a vector \( \mathbf{v} = \langle x, y \rangle \), the Euclidean norm can be found using the formula \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \).
For our vector \( \langle 3, 4 \rangle \), the computation is straightforward:

• Square the horizontal component: \(3^2 = 9\).
• Square the vertical component: \(4^2 = 16\).
• Add these squares: \(9 + 16 = 25\).
• Take the square root: \(\sqrt{25} = 5\).

Thus, the magnitude of \( \langle 3, 4 \rangle \) is 5, representing the length of the vector in a graphical space. It's important because it informs us about the vector's reach when considered apart from its direction.

Tangent Function

The tangent function is crucial when dealing with vectors and their directions. It relates the opposite and adjacent sides of a right triangle formed by the vector components.
For a vector \( \langle x, y \rangle \), the tangent of the angle \( \theta \) between the vector and the positive x-axis is given by \( \tan \theta = \frac{y}{x} \). In our scenario, the vector \( \langle 3, 4 \rangle \) forms a right triangle:

• Opposite side (y-component): 4 units.
• Adjacent side (x-component): 3 units.

This gives us \( \tan \theta = \frac{4}{3} \), indicating how steep or shallow the vector's slope is relative to the x-axis. The tangent function helps determine this angle's size, which will be used to find the vector's direction.

Arctangent

The arctangent, sometimes marked as \( \tan^{-1} \), is the inverse function of tangent. It is used to find the angle whose tangent is a given number.
To find the angle \(\theta\) for the vector \( \langle 3, 4 \rangle \), where \( \tan \theta = \frac{4}{3} \), we calculate \( \theta = \tan^{-1} \left( \frac{4}{3} \right) \). This process yields an angle of approximately \(53.13^\circ\).

• Step 1: Calculate \( \frac{4}{3} \).
• Step 2: Apply the arctangent function.
• Result: \( \theta = 53.13^\circ \).

The arctangent gives us the direction of the vector relative to the x-axis in degrees. A positive angle indicates the vector rises above the x-axis as expected in the first quadrant, where both x and y components are positive.