A vector of length ell is turned by an angle | vedclass

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(B) Let the initial vector be $\vec{\ell}_1$ and the final vector be $\vec{\ell}_2$. Both have magnitude $\ell$,so $|\vec{\ell}_1| = |\vec{\ell}_2| =

Vedclass · Accepted Answer

$2\ell \sin \frac{	heta}{2}$

Vedclass · Answer

(B) Let the initial vector be $\vec{\ell}_1$ and the final vector be $\vec{\ell}_2$. Both have magnitude $\ell$,so $|\vec{\ell}_1| = |\vec{\ell}_2| = \ell$.The change in the position vector of the tip is given by $\Delta \vec{\ell} = \vec{\ell}_2 - \vec{\ell}_1$.The magnitude of the change is $|\Delta \vec{\ell}| = \sqrt{|\vec{\ell}_2|^2 + |\vec{\ell}_1|^2 - 2|\vec{\ell}_2||\vec{\ell}_1| \cos 	heta}$.Substituting the magnitudes,we get $|\Delta \vec{\ell}| = \sqrt{\ell^2 + \ell^2 - 2\ell^2 \cos 	heta}$.$|\Delta \vec{\ell}| = \sqrt{2\ell^2(1 - \cos 	heta)}$.Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$,we get:$|\Delta \vec{\ell}| = \sqrt{2\ell^2 \cdot 2 \sin^2 \frac{	heta}{2}} = \sqrt{4\ell^2 \sin^2 \frac{	heta}{2}}$.$|\Delta \vec{\ell}| = 2\ell \sin \frac{	heta}{2}$.

$A$ vector of length $\ell$ is turned by an angle $\theta$. Find the change in the position vector of the tip.

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