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Chapter 8: Problem 19

Find the magnitude and direction of the vector. $$ \langle-4,-6\rangle $$

Short Answer

Expert verified
Magnitude: \(2\sqrt{13}\), Direction: 3rd quadrant angle.

Step by step solution

01

Identify the components

The vector given is \( \langle -4, -6 \rangle \). We identify the components as \( x = -4 \) and \( y = -6 \).
02

Calculate the magnitude

The magnitude of a vector \( \langle x, y \rangle \) is calculated using the formula \( \sqrt{x^2 + y^2} \). Substitute \( x \) and \( y \) values to get \( \sqrt{(-4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \).
03

Find the direction angle

The direction angle \( \theta \) is found using the formula \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \). Substitute \( x = -4 \) and \( y = -6 \) to get \( \theta = \tan^{-1} \left( \frac{-6}{-4} \right) = \tan^{-1} \left( \frac{3}{2} \right) \).
04

Adjust the angle for correct direction

Since both components are negative, the vector is in the third quadrant. The reference angle is given by \( \tan^{-1} \left( \frac{3}{2} \right) \), but the actual angle must be \( 180^\circ + \theta \). Thus, find the final direction angle which depends on the quadrant.

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Key Concepts

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

Vector Components
The concept of vector components is fundamental when analyzing vectors. A vector is expressed in terms of its horizontal and vertical parts, often noted as \( \langle x, y \rangle \). In this exercise, our vector is \( \langle -4, -6 \rangle \), where the individual components help us understand the direction and positioning of the vector. Here:
  • \( x = -4 \) represents the component along the x-axis.
  • \( y = -6 \) represents the component along the y-axis.
These components give a comprehensive picture of how the vector behaves in a coordinate system.
Magnitude Formula
Calculating the magnitude of a vector provides insight into its length or size. The magnitude formula is derived from the Pythagorean theorem, reflecting the distance from the origin to the point \( \langle x, y \rangle \). The formula is given by:\[\text{Magnitude} = \sqrt{x^2 + y^2}\]Inserting the given components, we compute:\[\sqrt{(-4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\]This tells us how much ground the vector covers, irrespective of its direction.
Direction Angle
The direction angle of a vector tells us the orientation or direction of the vector in a plane. It's calculated using the arctangent function and is expressed in degrees or radians. The formula for the direction angle \( \theta \) is:\[\theta = \tan^{-1} \left( \frac{y}{x} \right)\]For our vector \( \langle -4, -6 \rangle \), we substitute the components:\[\theta = \tan^{-1} \left( \frac{-6}{-4} \right) = \tan^{-1} \left( \frac{3}{2} \right)\]This gives us the reference angle, needed to guide us in finding the precise direction.
Trigonometry
Trigonometry plays a crucial role in understanding vector angles and their applications. Utilizing trigonometric functions allows us to relate the sides and angles of a right triangle. For vectors, the tangent function specifically helps connect the vertical and horizontal components, aiding in direction angle calculations.
  • \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
In our case, \( \frac{-6}{-4} = \frac{3}{2} \), demonstrates this relation.
Third Quadrant Adjustment
When dealing with vectors that fall into different quadrants, angle adjustments are necessary. A vector in the third quadrant, like \( \langle -4, -6 \rangle \), has both components negative.
  • The initial angle, found using \( \tan^{-1} \), is a reference.
  • For the third quadrant, where angles range from \(180^\circ\) to \(270^\circ\), adjustments are crucial.
To get the correct direction, add \( 180^\circ \) to the reference angle:\[\text{Direction angle} = 180^\circ + \theta\]This ensures the angle accurately reflects its true positioning in the coordinate system.

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