6(-i + \sqrt{3} j)
6(-\sqrt{3} i - j)
6(i - \sqrt{3} j)
6(i + \sqrt{3} j)
Correct answer is B
F = F \cos \theta i + F \sin \theta j
r = (12, 210°) = 12 \cos 210 i + 12 \sin 210 j
= - 6\sqrt{3} i - 6j
= 6(-\sqrt{3} i - j)
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