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Chapter 8: Problem 13

Find the magnitude and direction of the vector. $$ \langle 6,5\rangle $$

Short Answer

Expert verified
Magnitude: \( \sqrt{61} \); Direction: \( 39.81^\circ \).

Step by step solution

01

Understand the Vector

You are given the vector \( \langle 6, 5 \rangle \), which has an x-component of 6 and a y-component of 5. The task is to find its magnitude and direction.
02

Calculate the Magnitude

The magnitude of a vector \( \langle a, b \rangle \) is calculated using the formula \( \sqrt{a^2 + b^2} \). Substitute the given components into the formula: \[magnitude = \sqrt{6^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61}\]
03

Calculate the Direction (Angle)

The direction of a vector is given by the angle \( \theta \) it makes with the positive x-axis. This can be calculated using the tangent function: \[ \tan \theta = \frac{b}{a} = \frac{5}{6} \] To find \( \theta \), compute the inverse tangent: \[ \theta = \tan^{-1} \left( \frac{5}{6} \right) \approx 39.81^\circ \]
04

Summarize the Results

You have found the magnitude and direction of the vector \( \langle 6, 5 \rangle \). The magnitude is \( \sqrt{61} \), and the direction is approximately \( 39.81^\circ \) from the positive x-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Components
Vectors, like our given vector \( \langle 6, 5 \rangle \), are mathematical objects defined by both a direction and a magnitude. They are often described using their components, which tell us how much the vector moves along each axis. For our vector, the components are:
  • x-component: 6, which means the vector moves 6 units in the positive x-direction.
  • y-component: 5, indicating the vector moves 5 units in the positive y-direction.
Think of vector components as the building blocks that define the vector's movement and direction in space. They allow us to easily perform calculations involving vectors by breaking down complex movements into more straightforward linear parts. Imagine breaking down diagonal movement into horizontal and vertical steps. Indeed, this simplification makes analyzing vectors straightforward and intuitive.
Magnitude Calculation
The magnitude of a vector is a measure of its length or size. To find the magnitude of a vector \( \langle a, b \rangle \), we use the Pythagorean theorem. This theorem emerges because a vector's x and y components create a right triangle where the vector is the hypotenuse. Here's how to calculate it:
  • We start with the formula: \( \text{Magnitude} = \sqrt{a^2 + b^2} \).
  • In our example, using \( a = 6 \) and \( b = 5 \), we substitute these values to get: \( \sqrt{6^2 + 5^2} \).
  • Calculating further, we find \( \sqrt{36 + 25} = \sqrt{61} \).
  • Thus, the magnitude of vector \( \langle 6, 5 \rangle \) is \( \sqrt{61} \), reflecting its length in space.
The magnitude gives you a single positive number that represents how long the vector is, without any reference to direction. This can be particularly useful in physics, where understanding the force, speed, or distance represented by a vector is vital.
Angle with x-axis
Determining the angle a vector makes with the x-axis provides insights about the vector's direction. This angle, often denoted by \( \theta \), shows precisely how a vector is oriented relative to the horizontal axis.
  • The formula used to find \( \theta \) is \( \tan \theta = \frac{b}{a} \), where \( b \) is the y-component and \( a \) is the x-component.
  • For our vector \( \langle 6, 5 \rangle \), substitute \( b = 5 \) and \( a = 6 \) into the formula: \( \tan \theta = \frac{5}{6} \).
  • To find \( \theta \), apply the inverse tangent function: \( \theta = \tan^{-1} \left( \frac{5}{6} \right) \).
  • This calculation results in an angle of approximately \( 39.81^\circ \).
Understanding the angle helps in visualizing and applying the vector in physical scenarios such as navigation, where direction matters. Hence, calculating the angle with the x-axis is a core step in comprehending how vectors project in different orientations.

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Most popular questions from this chapter

Eliminate the parameter \(t\) to rewrite the parametric equation as a Cartesian equation. $$ \left\\{\begin{array}{l} x(t)=2 e^{t} \\ y(t)=1-5 t \end{array}\right. $$

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