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Chapter 5: Problem 27

Find the exact magnitude and direction angle to the nearest tenth of a degree of each vector. $$ \langle\sqrt{3}, 1\rangle $$

Short Answer

Expert verified
Magnitude = 2, Direction angle = 30.0\^{\circ}

Step by step solution

01

Identify the Components of the Vector

The vector given is \(\textbf{v} = \langle \sqrt{3}, 1 \rangle\). The components are \(\textbf{v}_x = \sqrt{3}\) and \(\textbf{v}_y = 1\).
02

Calculate the Magnitude of the Vector

The magnitude of a vector \(\textbf{v} = \langle \textbf{v}_x, \textbf{v}_y \rangle\) is found using the formula: \[ \|\textbf{v}\| = \sqrt{\textbf{v}_x^2 + \textbf{v}_y^2} \] Substituting the given values: \[ \|\textbf{v}\| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]
03

Calculate the Direction Angle of the Vector

The direction angle \(\theta\) of a vector can be found using the tangent function: \[ \theta = \tan^{-1} \left( \frac{\textbf{v}_y}{\textbf{v}_x} \right) \] Substituting the given values: \[ \theta = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) \] We know \[ \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = 30\^{\circ} \]
04

Express the Angle to the Nearest Tenth of a Degree

The calculated angle is already in degrees. Therefore, the angle to the nearest tenth of a degree remains: \[ 30.0\^{\circ} \]

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Key Concepts

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

Vector Components
Vectors are often described in terms of their components. A vector has both a direction and a magnitude, and can be represented as \(\langle v_x, v_y \rangle\). Here, \(\sqrt{3}, 1 \) are the components, where \(\textbf{v}_x = \sqrt{3}\) is the horizontal component and \(\textbf{v}_y = 1\) is the vertical component.
Each component tells you how far the vector goes in that direction.
  • Horizontal component (\textbf{v}_x): the movement along the x-axis.
  • Vertical component (\textbf{v}_y): the movement along the y-axis.
Understanding the components is essential for further calculations of magnitude and direction.
Magnitude Calculation
The magnitude (or length) of a vector is a measure of how long the vector is. It can be calculated using the Pythagorean theorem.
For vector \(\textbf{v} = \langle \textbf{v}_x, \textbf{v}_y \rangle\), the formula is:
\[ \|\textbf{v}\| = \sqrt{ \textbf{v}_x^2 + \textbf{v}_y^2 } \]
For our example vector, \(\langle \sqrt{3}, 1 \rangle\), substituting the values gives us:
\[ \|\textbf{v}\| = \sqrt{ (\sqrt{3})^2 + 1^2 } = \sqrt{ 3 + 1 } = \sqrt{ 4 } = 2 \]
So, the magnitude of the vector is 2. This tells us how far the vector extends in space.
Direction Angle
The direction angle of a vector indicates its orientation. It can be found using trigonometric functions, specifically the tangent function.
The angle \(\theta\) with the positive x-axis can be calculated as:
\[ \theta = \tan^{-1} \left( \frac{\textbf{v}_y}{\textbf{v}_x} \right) \]
For our vector, substituting in the values gives:
\[ \theta = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) \]
We know that \(\tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = 30\^{\circ}\), so the direction angle is 30 degrees.
Always make sure to convert the result to the nearest tenth if needed!
Trigonometry
Trigonometry is key when dealing with vectors. It helps in connecting the components of a vector to its magnitude and direction.
Key trigonometric functions include:
  • \( \sin \theta \): opposite/hypotenuse
  • \( \cos \theta \): adjacent/hypotenuse
  • \( \tan \theta \): opposite/adjacent
For a given vector \(\textbf{v} = \langle \textbf{v}_x, \textbf{v}_y \rangle\), \(\tan \theta\) helps find the angle:
\[ \theta = \tan^{-1} \left( \frac{\textbf{v}_y}{\textbf{v}_x} \right) \]
Understanding trigonometry allows us to switch easily between vector components and polar form, involving magnitude and angle.

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Most popular questions from this chapter

Solve each problem. Surveying Triangular Property A surveyor locating the corners of a triangular piece of property started at one corner and walked \(480 \mathrm{ft}\) in the direction \(\mathrm{N} 36^{\circ} \mathrm{W}\) to reach the next corner. The surveyor turned and walked \(\mathrm{S} 21^{\circ} \mathrm{W}\) to get to the next corner of the property. Finally, the surveyor walked in the direction \(\mathrm{N} 82^{\circ} \mathrm{E}\) to get back to the starting point. What is the area of the property in square feet? Round to the nearest tenth.

Andrea and Carlos left the airport at the same time. Andrea flew at 180 mph on a course with bearing \(80^{\circ},\) and Carlos flew at 240 mph on a course with bearing \(210^{\circ} .\) How far apart were they after \(3 \mathrm{hr}\) ? Round to the nearest tenth of a mile.

Find the smallest positive angle between the vectors \langle-3,5\rangle and \langle 1,6\rangle .

A submarine sights a moving target at a distance of \(820 \mathrm{~m}\). A torpedo is fired \(9^{\circ}\) ahead of the target as shown in the drawing and travels \(924 \mathrm{~m}\) in a straight line to hit the target. How far has the target moved from the time the torpedo is fired to the time of the hit? Round to the nearest tenth of a meter.

Given that \(\mathbf{A}=\langle 3,1\rangle\) and \(\mathbf{B}=\langle-2,3\rangle,\) find the magnitude and direction angle for each of the following vectors. Give exact answers using radicals when possible. Otherwise round to the nearest tenth. $$ \text { 5B } $$
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