Two vectors vec A and vec B ha | vedclass

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Question

$\overrightarrow{\mathrm{A}}=\cos 60^{\circ} \hat{\mathrm{i}}+\sin 60^{\circ} \hat{\mathrm{j}} \quad$ and $\overrightarrow{\mathrm{B}}=\hat{\mathrm{i}

Vedclass · Accepted Answer

Vedclass · Answer

$\overrightarrow{\mathrm{A}}=\cos 60^{\circ} \hat{\mathrm{i}}+\sin 60^{\circ} \hat{\mathrm{j}} \quad$ and $\overrightarrow{\mathrm{B}}=\hat{\mathrm{i}}$

$\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}=-\frac{1}{2} \hat{\mathrm{i}}+\frac{\sqrt{3}}{2} \hat{\mathrm{j}} \Rightarrow$ $II$ quadrant

Two vectors $\vec A$ and $\vec B$ have magnitudes $2$ and $1$ respectively. If the angle between $\vec A$ and $\vec B$ is $60^o$, then which of the following vectors may be equal to $\frac{{\vec A}}{2} - \vec B$

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