Calculating Vector E Magnitude and Direction in Quad 4: Tips and Tricks

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Homework Help Overview

The discussion revolves around calculating the magnitude and direction of vector E, given as (125 i - 250 j), specifically focusing on its representation in the fourth quadrant.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the magnitude and direction of the vector, expressing confusion about the angle measurement and how to visualize the vector in a coordinate system. Participants discuss different methods for determining the angle, including using the tangent function and referencing the axes.

Discussion Status

Participants have provided supportive feedback on the original poster's approach, confirming the validity of the angle calculations and discussing alternative methods for determining the direction of the vector. Multiple interpretations of the angle's representation are being explored.

Contextual Notes

The discussion includes considerations of standard practices for measuring angles in a coordinate system, as well as the original poster's challenges with visualizing the vector's placement in the fourth quadrant.

klm
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vector E= (125 i - 250 j) what are the magnitude and direction?
ok so what i did was found out that vector E should be in Quad. 4. and that the magnitude will be 279.51 b/c you do the square root of 125^2 + 250^2 = 279.51. but i am having trouble understanding how to get the direction. i think my problem is trying to see how to draw the picture.
i know that you go +125 on the x-axis and then down -250 on the y axis, so the vector will be in quad 4. but what angle are you trying to find? what i did was tan^-1 (125/250) = 26 and then i added 270 and got 296 as my angle. is this the correct way to do this problem
 
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Nothing wrong with what you did. You found the angle with the -y-axis, then added 270. Sounds good to me! You are following standard practice of measuring the angle with respect to the +x axis using counterclockwise as +.

You can also, of course, start by finding the angle with respect to the x-axis using tan^-1(250/125) = 63.4 degrees below the x-axis. Subtract that from 360 and you get back to 296.6 degrees (same as you did).
 
Yes, that is a correct way to denote the direction of that vector. You could have also said, "26 degrees from the negative y axis" or "64 degrees from the positive x axis." That would have also been a correct representation of the direction.
 
thank you both very much!
 

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