Find |Delta overrightarrow{a}| and Delta a wh | vedclass

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(B) Let the initial vector be $\overrightarrow{OA}$ and the final vector after rotation be $\overrightarrow{OB}$.Given that the magnitude of the vecto

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$a\,d	heta, 0$

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(B) Let the initial vector be $\overrightarrow{OA}$ and the final vector after rotation be $\overrightarrow{OB}$.Given that the magnitude of the vector remains constant,$|\overrightarrow{OA}| = |\overrightarrow{OB}| = a$.The change in the vector is defined as $\Delta \overrightarrow{a} = \overrightarrow{OB} - \overrightarrow{OA}$.From the geometry of the triangle $OAB$,the magnitude of the change in the vector is the length of the chord $AB$.For a small angle $d	heta$,the arc length $AB$ is given by $AB = a \cdot d	heta$.Therefore,$|\Delta \overrightarrow{a}| = a \cdot d	heta$.Now,consider the change in the magnitude of the vector,$\Delta a = |\overrightarrow{OB}| - |\overrightarrow{OA}|$.Since the magnitude is constant,$\Delta a = a - a = 0$.Thus,$|\Delta \overrightarrow{a}| = a \cdot d	heta$ and $\Delta a = 0$.

Find $|\Delta \overrightarrow{a}|$ and $\Delta a$ when a vector $\overrightarrow{a}$ is rotated by an angle $d\theta$.

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