What is the Direction of the Vector with Given Components?

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A vector with an x component of -25.0 units and a y component of 40.0 units has a calculated magnitude of 47.17 units. The direction was initially found to be 58 degrees, but this is incorrect as the vector lies in the second quadrant, where the correct angle is 122 degrees. The confusion arose from the textbook stating only the angle without specifying the quadrant, leading to frustration for the student. It was confirmed that if the angle were indeed 58 degrees, both components would have to be positive. The discussion highlights the importance of considering the quadrant when determining vector direction.
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Homework Statement



A vector has x component of (-25.0) units and y component of (40.0) units. Find the magnitude and direction of this vector.

Homework Equations





The Attempt at a Solution



Magnitude = \sqrt{}(-25.0)^2+(40.0)^2
= 47.17 units.

Direction = tan x = | 40.0 | / |-25.0|
= 1.6
x = tan^-1 (1.6)
= 58
since the vector is located in the IInd Quadrant theta = 180 -58 = 122 degrees.

But textbook answer is 58 degrees. I couldn't get it :confused:

Any suggestions ?? Thank you...
 
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hi thushanthan! :wink:
thushanthan said:
A vector has x component of (-25.0) units and y component of (40.0) units. Find the magnitude and direction of this vector.
…
But textbook answer is 58 degrees. I couldn't get it :confused:

Any suggestions ?? Thank you...

(which book is it?)

you're right, the text-book is wrong :smile: …

(as you know, if the direction was 58º, both x and y would be positive)
 
Thank you :smile:
 
What, exactly, did your textbook say? Anything like "58 degrees west of north" or "north 58 degrees west"?
 
No there is no direction mentioned. Only the value of theta is given, which is 58 degrees. :frown:
 
It's painful when the textbook is wrong and you've spent hours on one question you KNOW you did correctly... but there is some sense of victory when you find out they are wrong and you are right :)
 

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