If $\overrightarrow{A}$ is a vector with magnitude $(3, 4)$,show that the magnitude of its unit vector is $1$.
(N/A) The vector is given as $\overrightarrow{A} = 3\hat{i} + 4\hat{j}$.
The magnitude of $\overrightarrow{A}$ is $|\overrightarrow{A}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The unit vector $\hat{A}$ is defined as $\hat{A} = \frac{\overrightarrow{A}}{|\overrightarrow{A}|} = \frac{3\hat{i} + 4\hat{j}}{5} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}$.
The magnitude of the unit vector is $|\hat{A}| = \sqrt{(\frac{3}{5})^2 + (\frac{4}{5})^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1$.
Thus,the magnitude of the unit vector is $1$.
The magnitude of $\overrightarrow{A}$ is $|\overrightarrow{A}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The unit vector $\hat{A}$ is defined as $\hat{A} = \frac{\overrightarrow{A}}{|\overrightarrow{A}|} = \frac{3\hat{i} + 4\hat{j}}{5} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}$.
The magnitude of the unit vector is $|\hat{A}| = \sqrt{(\frac{3}{5})^2 + (\frac{4}{5})^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1$.
Thus,the magnitude of the unit vector is $1$.
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