The magnitudes of displacements overrightarrow (a) and overrightarrow (b) are 3 m and 4 m, respectively and overrightarrow (c)=overrightarrow (a)+overrightarrow (b) . Considering various orientations of overrightarrow (a) and overrightarrow (b) what are (a) the maximum possible magnitude for overrightarrow (c) and (b) the minimum possible magnitude?

To find the maximum and minimum possible magnitudes of the vector $\overrightarrow{c} = \overrightarrow{a} + \overrightarrow{b}$, we need to consider the different orientations of $\overrightarrow{a}$ and $\overrightarrow{b}$.<br /><br />(a) Maximum possible magnitude:<br />The maximum possible magnitude of $\overrightarrow{c}$ occurs when the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are aligned in the same direction. In this case, the magnitudes of $\overrightarrow{a}$ and $\overrightarrow{b}$ can be added directly.<br /><br />Maximum magnitude of $\overrightarrow{c}$ = magnitude of $\overrightarrow{a}$ + magnitude of $\overrightarrow{b}$ = 3 m + 4 m = 7 m.<br /><br />(b) Minimum possible magnitude:<br />The minimum possible magnitude of $\overrightarrow{c}$ occurs when the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular to each other. In this case, the magnitudes of $\overrightarrow{a}$ and $\overrightarrow{b}$ can be subtracted using the Pythagorean theorem.<br /><br />Minimum magnitude of $\overrightarrow{c}$ = $\sqrt{(\text{magnitude of } \overrightarrow{a})^2 + (\text{magnitude of } \overrightarrow{b})^2}$ = $\sqrt{(3 \, \text{m})^2 + (4 \, \text{m})^2}$ = $\sqrt{9 \, \text{m}^2 + 16 \, \text{m}^2}$ = $\sqrt{25 \, \text{m}^2}$ = 5 m.<br /><br />Therefore, the maximum possible magnitude of $\overrightarrow{c}$ is 7 m, and the minimum possible magnitude is 5 m.