If the magnitudes of vectors vec P,vec Q,and | vedclass

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(C) Given that the magnitudes are $P = 5$,$Q = 12$,and $R = 13$.Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the vectors form a right-angled triangle wh

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$\cos^{-1}(\frac{12}{13})$

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(C) Given that the magnitudes are $P = 5$,$Q = 12$,and $R = 13$.Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the vectors form a right-angled triangle where $\vec R$ is the hypotenuse.In the triangle formed by vectors $\vec P$,$\vec Q$,and $\vec R$,the angle $	heta$ is between $\vec Q$ and $\vec R$.Using trigonometry in the right-angled triangle,$\cos 	heta = \frac{	ext{Adjacent side}}{	ext{Hypotenuse}} = \frac{Q}{R}$.Substituting the values,$\cos 	heta = \frac{12}{13}$.Therefore,$	heta = \cos^{-1}(\frac{12}{13})$.

If the magnitudes of vectors $\vec P$,$\vec Q$,and $\vec R$ are $5$,$12$,and $13$ units respectively,and if $\vec P + \vec Q = \vec R$,then the angle between $\vec Q$ and $\vec R$ is:

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