Can Vectors with an angle 180(degrees+) have a negative magnitude?

AI Thread Summary
Vectors with angles greater than 180 degrees can have a negative magnitude depending on the context and the chosen coordinate system. In this case, a vector of 260 km at 48 degrees south of east is analyzed, leading to confusion about whether the magnitude should be considered positive or negative. The convention used states that counterclockwise angles are positive while clockwise angles are negative. The sine function is negative in the fourth quadrant, where this vector lies, resulting in a negative y-component and a positive x-component. Understanding these conventions clarifies the vector's components and their signs.
SaltyBriefs
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Homework Statement


Can vectors with 180+(degrees) have a negative magnitude? I'm trying to find components of a vector that is going 260 km, 48 (degrees) south of east. So I'm confused whether the 260 is positive or not because of the -48 degrees.

Homework Equations


(\vec{V}) (sin\theta)
(\vec{V}) (cos\theta)

The Attempt at a Solution


(-260km)(sin(48)) or (260km)(sin(48)) =\vec{V}_{y}

(-260km)(cos(48)) or (260km)(cos(48)) =\vec{V}_{x}

Which one??
 
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sin(x) = -sin(-x)

cos(x) = cos(-x)

Hope that helps.

Just use the convention that counterclockwise is postive and clockwise is negative in terms of measuring an angle.
 
Last edited:
WatermelonPig said:
sin(x) = -sin(-x)

cos(x) = cos(-x)

Hope that helps.

Just use the convention that counterclockwise is postive and clockwise is negative in terms of measuring an angle.
Thank you this helped so much! But um just a quick question, why is sin negative? O.o
 
SaltyBriefs said:
Thank you this helped so much! But um just a quick question, why is sin negative? O.o

Oh wait is this from cos, sin
and since it is in the 4th quadrant, y is negative (sin) and x is positive (cos)
 
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