Chapter 9: Problem 3
Find the magnitude and direction angle for each vector. a) \(\mathbf{u}=(-4,7)\) b) \(\mathbf{v}=(2,5)\) c) \(3 u\) d) \(-2 \mathbf{v}\) e) \(3 u+2 v\) \(f) u-v\)
Short Answer
Expert verified
Magnitudes: \(\sqrt{65}, \sqrt{29}, 3\sqrt{65}, 2\sqrt{29}, \sqrt{1081}, \sqrt{40}\). Angles adjusted for quadrants.
Step by step solution
01
Finding Magnitude of Vector u
The magnitude of a vector \(\mathbf{u} = (a, b)\) is found using the formula \( \|\mathbf{u}\| = \sqrt{a^2 + b^2} \). For \(\mathbf{u} = (-4, 7)\), calculate \(\|\mathbf{u}\| = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}\).
02
Finding Direction Angle of Vector u
The direction angle \(\theta\) for a vector \(\mathbf{u} = (a, b)\) is found using \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\). For \(\mathbf{u} = (-4, 7)\), \(\theta = \tan^{-1}\left(\frac{7}{-4}\right)\). Since this vector is in the second quadrant, adjust the angle accordingly.
03
Finding Magnitude of Vector v
The magnitude of vector \(\mathbf{v} = (2, 5)\) is \(\|\mathbf{v}\| = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}\).
04
Finding Direction Angle of Vector v
The direction angle \(\theta\) for \(\mathbf{v} = (2, 5)\) is \(\theta = \tan^{-1}\left(\frac{5}{2}\right)\). This vector is in the first quadrant, so no further adjustment is needed for the angle.
05
Magnitude and Direction of Vector 3u
Multiply each component of vector \(\mathbf{u} = (-4, 7)\) by 3 to get \(3u = (-12, 21)\). The magnitude \(\|3u\| = 3\|\mathbf{u}\| = 3\sqrt{65}\). The direction angle is the same as \(\mathbf{u}\)'s direction angle.
06
Magnitude and Direction of Vector -2v
Multiply each component of vector \(\mathbf{v} = (2, 5)\) by -2 to get \(-2v = (-4, -10)\). The magnitude \(\|-2v\| = 2\|\mathbf{v}\| = 2\sqrt{29}\). Calculate the direction angle \(\theta = \tan^{-1}\left(\frac{-10}{-4}\right)\), and adjust as it's in the third quadrant.
07
Vector Sum 3u + 2v
Multiply and add vectors: \(3u = (-12, 21)\) and \(2v = (4, 10)\), resulting in \(3u + 2v = (-12 + 4, 21 + 10) = (-8, 31)\). Calculate the magnitude \(\sqrt{(-8)^2 + 31^2}\) and direction angle \(\tan^{-1}\left(\frac{31}{-8}\right)\), adjust since it's in the second quadrant.
08
Vector Difference u - v
Calculate \(u - v = (-4 - 2, 7 - 5) = (-6, 2)\). The magnitude \(\sqrt{(-6)^2 + 2^2}\). The direction angle \(\tan^{-1}\left(\frac{2}{-6}\right)\), adjust for the second quadrant.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnitude of a Vector
Calculating the magnitude of a vector is like finding its length in space. Imagine a vector as a directed line segment that starts from the origin (0,0) and extends to a point (a,b) on a coordinate plane. To find how "long" this vector is, you use the magnitude formula:
- For a vector \( \mathbf{u} = (a, b) \), the magnitude \( \| \mathbf{u} \| = \sqrt{a^2 + b^2} \).
Direction Angle
The direction angle of a vector tells you the angle that the vector makes with the positive x-axis in a 2D plane. To find this angle, use the inverse tangent function, which helps determine angles in right triangles. For a vector \( \mathbf{u} = (a, b) \), the direction angle \( \theta \) is calculated as:
- \( \theta = \tan^{-1}\left( \frac{b}{a} \right) \)
- First Quadrant: Both a and b are positive; no adjustment needed.
- Second Quadrant: a is negative, b is positive; add 180° to \( \theta \) if needed.
- Third Quadrant: Both a and b are negative; add 180° to \( \theta \) to correct the direction.
- Fourth Quadrant: a is positive, b is negative; add 360° or subtract from 0 if required.
Vector Addition and Subtraction
Adding or subtracting vectors is crucial in understanding their combined effect or net movement. With addition, you're essentially combining the inputs of two vectors to form a result vector. For vectors \( \mathbf{u} = (a_1, b_1) \) and \( \mathbf{v} = (a_2, b_2) \), the processes work as follows:
- Addition: \( \mathbf{u} + \mathbf{v} = (a_1 + a_2, b_1 + b_2) \)
- Subtraction: \( \mathbf{u} - \mathbf{v} = (a_1 - a_2, b_1 - b_2) \)
Quadrants in Coordinate Geometry
Understanding quadrants in a coordinate plane is key to comprehending vector direction and location. The plane is divided into four sections called quadrants, separated by the x and y axes:
- First Quadrant (I): Both x (a) and y (b) are positive.
- Second Quadrant (II): x (a) is negative, y (b) is positive.
- Third Quadrant (III): Both x (a) and y (b) are negative.
- Fourth Quadrant (IV): x (a) is positive, y (b) is negative.