Find the angle the vector sqrt{3} hat{i} - ha | vedclass

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(B) The vector is given by $\vec{A} = \sqrt{3} \hat{i} - \hat{j}$.Comparing this with $\vec{A} = A_x \hat{i} + A_y \hat{j}$,we get $A_x = \sqrt{3}$ an

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(B) The vector is given by $\vec{A} = \sqrt{3} \hat{i} - \hat{j}$.Comparing this with $\vec{A} = A_x \hat{i} + A_y \hat{j}$,we get $A_x = \sqrt{3}$ and $A_y = -1$.The angle $	heta$ that the vector makes with the positive $x$-axis is given by $	an 	heta = \frac{A_y}{A_x}$.Substituting the values,we have $	an 	heta = \frac{-1}{\sqrt{3}}$.Since $	an(30^{\circ}) = \frac{1}{\sqrt{3}}$,$	an(-30^{\circ}) = -\frac{1}{\sqrt{3}}$.Therefore,$	heta = -30^{\circ}$.

Find the angle the vector $\sqrt{3} \hat{i} - \hat{j}$ makes with the positive $x$-axis. (in $^{\circ}$)

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