The angle made by the vector (hat{i} + hat{j} | vedclass

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(C) Let the vector be $\vec{A} = \hat{i} + \hat{j}$.Here,the components are $A_x = 1$ and $A_y = 1$.The magnitude of the vector is $A = \sqrt{A_x^2 +

Vedclass · Accepted Answer

$45^\circ, 45^\circ$

Vedclass · Answer

(C) Let the vector be $\vec{A} = \hat{i} + \hat{j}$.Here,the components are $A_x = 1$ and $A_y = 1$.The magnitude of the vector is $A = \sqrt{A_x^2 + A_y^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.The angle $\alpha$ with the $x$-axis is given by $\cos \alpha = \frac{A_x}{A} = \frac{1}{\sqrt{2}}$.Therefore,$\alpha = 45^\circ$.The angle $\beta$ with the $y$-axis is given by $\cos \beta = \frac{A_y}{A} = \frac{1}{\sqrt{2}}$.Therefore,$\beta = 45^\circ$.

The angle made by the vector $(\hat{i} + \hat{j})$ with the $x$-axis and $y$-axis is:

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