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Vector A⃗ has a magnitude of 3.00 and is directed parallel to the negative y-axis and vector B⃗ has a magnitude of 3.00 and is directed parallel to the positive y-axis. Determine the magnitude and direction angle (as measured counterclockwise from the positive x-axis) of vector C⃗ , if C⃗ =A⃗ −B⃗ .
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|dw:1443490284799:dw|
Now, \(\sf |\vec{C}|=|\vec{A}- \vec{B}| = \sqrt{A^2+B^2-2ABCos180}\)
Angle will be 45 degrees from the +x axis in clockwise sense.
|dw:1443491207007:dw|
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