If O is the origin and the position vector of | vedclass

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(C) The position vector of point $A$ is given by $\overrightarrow{OA} = 4\,i + 5\,j$.To find the unit vector parallel to $\overrightarrow{OA}$,we use

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$\frac{1}{\sqrt{41}}(4\,i + 5\,j)$

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(C) The position vector of point $A$ is given by $\overrightarrow{OA} = 4\,i + 5\,j$.To find the unit vector parallel to $\overrightarrow{OA}$,we use the formula $\hat{a} = \frac{\overrightarrow{a}}{|\overrightarrow{a}|}$.First,calculate the magnitude of $\overrightarrow{OA}$:$|\overrightarrow{OA}| = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$.Now,the unit vector is:$\hat{OA} = \frac{4\,i + 5\,j}{\sqrt{41}} = \frac{1}{\sqrt{41}}(4\,i + 5\,j)$.Thus,the correct option is $C$.

If $O$ is the origin and the position vector of $A$ is $4\,i + 5\,j$,then a unit vector parallel to $\overrightarrow{OA}$ is:

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