Problem 50 - Algebra and Trigonometry

Question

\(47-52\) . Find the magnitude and direction (in degrees) of the vector. $$ \mathbf{v}=\langle 40,9\rangle $$

Step-by-Step Solution

Answer

Magnitude: 41, Direction: \(12.68^\circ\).

Step 1: Understanding the vector components

The given vector \( \mathbf{v} = \langle 40, 9 \rangle \) consists of two components: 40 is the horizontal component (\( x \)-component) and 9 is the vertical component (\( y \)-component).

Step 2: Calculating the magnitude

The magnitude of a vector \( \mathbf{v} = \langle x, y \rangle \) is given by the formula \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \). For our vector, it's \( \| \langle 40, 9 \rangle \| = \sqrt{40^2 + 9^2} \).

Step 3: Computing squared values

Calculate the squared values: \( 40^2 = 1600 \) and \( 9^2 = 81 \).

Step 4: Adding squared values

Add the squared values: \( 1600 + 81 = 1681 \).

Step 5: Evaluating the square root

Find the square root of the sum to get the magnitude: \( \sqrt{1681} = 41 \).

Step 6: Calculating the direction

The direction \( \theta \) of the vector is given by \( \tan \theta = \frac{y}{x} \). Here, \( \theta = \tan^{-1} \left( \frac{9}{40} \right) \).

Step 7: Calculating arctangent value

Calculate \( \tan^{-1} \left( \frac{9}{40} \right) \) using a calculator to find \( \theta \approx 12.68^\circ \).

Step 8: Summary of results

The magnitude of the vector is 41, and the direction is approximately \( 12.68^\circ \).

Key Concepts

Vector Components

A vector is essentially a quantity that has both direction and magnitude. We can think of vectors as arrows pointing in a specific direction. The components of a vector are essential details that reveal how the vector behaves in a coordinate system. For the vector \( \mathbf{v} = \langle 40, 9 \rangle \), these components tell us how far the vector moves horizontally and vertically.

- The horizontal component, often referred to as the \( x \)-component, is 40. This means if you start at the origin, you would move 40 units to the right.- The vertical component, known as the \( y \)-component, is 9. This indicates an upward movement of 9 units.

Understanding vector components is a crucial step in dissecting the vector's behavior. It allows us to see exactly how the vector influences position in a 2D plane.

Magnitude Calculation

The magnitude of a vector is a measure of its length, or how far the tip of the vector is from the origin. It is calculated using the Pythagorean Theorem. For a vector \( \mathbf{v} = \langle x, y \rangle \), the magnitude \( \| \mathbf{v} \| \) is determined by the formula:\[ \| \mathbf{v} \| = \sqrt{x^2 + y^2} \]

Using our specific example vector \( \langle 40, 9 \rangle \):- Compute the squares: \( 40^2 = 1600 \) and \( 9^2 = 81 \).- Add them together: \( 1600 + 81 = 1681 \).- Finally, take the square root to find the magnitude: \( \sqrt{1681} = 41 \).

This calculation tells us that the vector's length or magnitude is 41 units.

Direction Angle

The direction of a vector is indicated by the angle it forms with the horizontal axis. This angle is typically measured in degrees and helps in understanding what direction the vector is pointing towards in the plane.

To find this direction angle \( \theta \), we rely on the function of tangent, which applies primarily to right triangles formed when vector components are considered. Given that \( \tan \theta = \frac{y}{x} \), for our vector \( \langle 40, 9 \rangle \):- We have \( \tan \theta = \frac{9}{40} \).- Solving with a calculator, we find \( \tan^{-1} \left( \frac{9}{40} \right) \approx 12.68^\circ \).

Thus, the direction the vector makes with the positive x-axis is approximately \( 12.68^\circ \), offering complete insight into its orientation.

Tangent Function

The tangent function is a fundamental concept in trigonometry, crucial for calculating angles in right triangles. We use it to determine the angle \( \theta \) that the vector makes with the positive x-axis. Defined as the ratio of the opposite side to the adjacent side in a right triangle, it is expressed as:\[ \tan \theta = \frac{y}{x} \]

Here, for our vector, the tangent function is:- The opposite side is the vertical component (\( y = 9 \)).- The adjacent side is the horizontal component (\( x = 40 \)).- Applying the tangent function gives: \( \tan \theta = \frac{9}{40} \).

When we use the arctangent (inverse tangent) function to find \( \theta \), we convert the result back into an angle measure. This entire process highlights how vital the tangent function is for dissecting the orientation of a vector in a coordinate system.