A vector overrightarrow{a} is turned without | vedclass

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(B) From the figure,let the initial vector be $\overrightarrow{OA}$ and the final vector be $\overrightarrow{OB}$. Since the length does not change,$|

Vedclass · Accepted Answer

$a\,d	heta, 0$

Vedclass · Answer

(B) From the figure,let the initial vector be $\overrightarrow{OA}$ and the final vector be $\overrightarrow{OB}$. Since the length does not change,$|\overrightarrow{OA}| = |\overrightarrow{OB}| = a$.By the triangle law of vector addition,$\overrightarrow{OB} = \overrightarrow{OA} + \Delta \overrightarrow{a}$,which implies $\Delta \overrightarrow{a} = \overrightarrow{OB} - \overrightarrow{OA}$.The magnitude $|\Delta \overrightarrow{a}|$ represents the length of the chord $AB$ in the circular arc of radius $a$ subtending an angle $d	heta$ at the center $O$. For a small angle $d	heta$,the chord length is approximately equal to the arc length.Using the relation $	ext{angle} = \frac{	ext{arc}}{	ext{radius}}$,we have $d	heta = \frac{AB}{a}$,which gives $AB = a\,d	heta$.Therefore,$|\Delta \overrightarrow{a}| = a\,d	heta$.$\Delta a$ represents the change in the magnitude of the vector,which is $\Delta a = |\overrightarrow{OB}| - |\overrightarrow{OA}| = a - a = 0$.Thus,the values are $a\,d	heta$ and $0$ respectively.

$A$ vector $\overrightarrow{a}$ is turned without a change in its length through a small angle $d\theta$. The values of $|\Delta \overrightarrow{a}|$ and $\Delta a$ are respectively:

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