Easy Tips for Describing Vectors | Learn How to Describe Vectors Accurately

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SUMMARY

This discussion focuses on effective methods for accurately describing vectors, particularly in terms of their direction and magnitude. Key techniques include establishing a Cartesian coordinate system to visualize vectors, using the Pythagorean theorem to calculate angles, and employing the clock method for intuitive understanding. Recommended resources for further learning include Khan Academy and Math is Fun, which provide interactive exercises and explanations. Consistent practice is emphasized as essential for mastering vector description skills.

PREREQUISITES
  • Understanding of basic geometry and trigonometry
  • Familiarity with Cartesian coordinate systems
  • Knowledge of the Pythagorean theorem
  • Ability to use inverse tangent functions for angle calculation
NEXT STEPS
  • Explore Khan Academy's vector description tutorials
  • Practice using the Pythagorean theorem for vector calculations
  • Learn about vector components and their applications in physics
  • Investigate interactive activities on Math is Fun related to vectors
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Students, educators, and anyone interested in mastering vector descriptions in mathematics and physics.

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I am having problems with describing vectors. I understand the part that vectors have magnitude and direction, but that's not the part I'm having issues with. I can't look at a figure and find out whether or not the vector is, for example, 30 degrees south of east or 30 degrees east of south. Does anybody know a good way to find out how to describe vectors? Or maybe a website I could visit to help me?

Thanks.
 
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I'm sure there are plenty of websites out there, but none come to mind (try a google search).
Descriptions of vectors such as "30 degrees south of east" only give the direction of the vector and tell us nothing about it's magnitude. This can be seen by setting up a cartesian coordinate system with two axis (the typical x-y grid). Label each end of the axis North, South, East, and West (for example the +y direction could be North, +x could be East, the -y south, and -x west). To obtain a vector pointing 30 degrees south of east, start with an arbitrary non zero position vector pointing east ( in the direction of the +x axis ) and rotate it 30 degrees south (that is toward the -y axis). The resulting vector should have an angle of -30 degrees from the +x axis, and based on this coordinate system we call that 30 degrees south of east.
 


Describing vectors accurately can be challenging, but with some practice and helpful tips, you can improve your skills. One useful tip is to always start by identifying the direction of the vector. This can be done by looking at the figure and determining which way the vector is pointing. Once you have identified the direction, you can then use basic geometry to determine the angle of the vector.

To find the angle, imagine a right triangle with the vector as the hypotenuse. Use the Pythagorean theorem to find the length of the adjacent and opposite sides. The angle can then be found using the inverse tangent function (tan^-1). This will give you the angle in degrees or radians, depending on the calculator you are using.

Another helpful tip is to use the clock method. Imagine the vector as the hour hand of a clock. The direction it is pointing towards will correspond to the hour on the clock. For example, if the vector is pointing towards 3 o'clock, it is 90 degrees east of north.

You can also use online resources such as Khan Academy or Math is Fun to practice and improve your vector description skills. These websites offer explanations, examples, and interactive activities to help you understand and describe vectors accurately.

Remember, practice makes perfect. Keep practicing and using these tips, and you will become more confident in describing vectors accurately. Don't get discouraged if it takes time to master this skill. With patience and persistence, you will get there. Good luck!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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