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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 is \( \sqrt{61} \) and direction is approximately \( 39.81^\circ \).

Step by step solution

01

Determine the Magnitude of the Vector

The magnitude of a vector \( \langle a, b \rangle \) is calculated using the formula \( \sqrt{a^2 + b^2} \). For the given vector \( \langle 6, 5 \rangle \), this becomes \( \sqrt{6^2 + 5^2} \).
02

Simplify the Magnitude Calculation

First, calculate \( 6^2 = 36 \) and \( 5^2 = 25 \). Add these values to get \( 36 + 25 = 61 \). Therefore, the magnitude is \( \sqrt{61} \).
03

Calculate the Direction Angle

The direction of the vector is given by \( \theta = \tan^{-1}\left( \frac{b}{a} \right) \). Substituting in the given values, \( \theta = \tan^{-1}\left( \frac{5}{6} \right) \).
04

Evaluate the Direction

Use a calculator to find the angle: \( \theta = \tan^{-1}(0.8333) \approx 39.81^\circ \).

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Key Concepts

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

Magnitude of a Vector
The magnitude of a vector is like finding its length or size. Think of a vector like an arrow. The magnitude is just how long that arrow is.
To find this for a vector \( \langle a, b \rangle \), we use the formula:
  • \( \sqrt{a^2 + b^2} \)
For our vector \( \langle 6, 5 \rangle \), you calculate as follows:
  • Square each component: \( 6^2 = 36 \), \( 5^2 = 25 \).
  • Add them: \( 36 + 25 = 61 \).
  • Get the square root: \( \sqrt{61} \).
This gives us the magnitude, which tells us just how long the vector is in terms of units.
Direction of a Vector
Knowing a vector鈥檚 direction is like knowing where an arrow is pointing. We find this direction using an angle, typically measured from the positive x-axis in the counterclockwise direction.
For the vector \( \langle a, b \rangle \), the direction angle \( \theta \) can be determined using trigonometry.
Remember, the key is to use the tangent of the angle, which connects both components \( a \) and \( b \) of the vector.
Trigonometry
Trigonometry helps us understand the relationships between angles and sides of triangles.
In vectors, it's essential when finding directions. Think of the vector as forming a right triangle with the x-axis and itself as the hypotenuse.
The roles of opposite, adjacent, and hypotenuse sides are crucial. When finding direction, the tangent function compares the opposite side (b) over the adjacent side (a). This gives a ratio that helps find angles and thus the direction of the vector.
By understanding trigonometry and these relationships, calculating the direction becomes a straightforward task using right triangle concepts.
Inverse Tangent Function
The inverse tangent function, denoted as \( \tan^{-1} \), is crucial for finding a vector's direction based on its components.
This function reverses the normal tangent process: instead of finding a ratio from an angle, it finds an angle from a ratio.
For the vector \( \langle 6, 5 \rangle \), the tangent of the direction angle \( \theta \) is given as the ratio:
  • \( \frac{5}{6} \)
Using \( \tan^{-1} \), we find \( \theta \):
  • \( \theta = \tan^{-1} \left( \frac{5}{6} \right) \approx 39.81^\circ \)
This numerical angle tells us precisely where the vector points.

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Most popular questions from this chapter

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