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 $

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Answer:

Correct Answer: B

Solution:

$ \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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