Kinematics Question 139
Question: The angle made by the vector $ A=\hat{i}+\hat{j} $ with x- axis it’s [EAMCET (Engg.) 1999]
Options:
A) $ 90{}^\circ $
B) $ 45{}^\circ $
C) $ 22.5{}^\circ $
D) $ 30{}^\circ $
Correct Answer: B $ \vec{A}=\hat{i}+\hat{j} $
therefore $ |A|=\sqrt{1^{2}+1^{2}}=\sqrt{2} $ $ \cos \alpha =\frac{A _{x}}{|A|}=\frac{1}{\sqrt{2}}=\cos 45{}^\circ $ \ $ \alpha =45{}^\circ $
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