If |vec{a}|=4, |vec{b}|=5, |vec{a}-vec{b}|=3 | vedclass

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(D) Given: $|\vec{a}|=4, |\vec{b}|=5, |\vec{a}-\vec{b}|=3$. Squaring the magnitude equation: $|\vec{a}-\vec{b}|^2 = 3^2 = 9$. Using the property $|\ve

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$\frac{16}{9}$

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(D) Given: $|\vec{a}|=4, |\vec{b}|=5, |\vec{a}-\vec{b}|=3$. Squaring the magnitude equation: $|\vec{a}-\vec{b}|^2 = 3^2 = 9$. Using the property $|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}||\vec{b}| \cos 	heta = 9$. Substituting the values: $16 + 25 - 2(4)(5) \cos 	heta = 9$. $41 - 40 \cos 	heta = 9$. $40 \cos 	heta = 32$. $\cos 	heta = \frac{32}{40} = \frac{4}{5}$. Since $\cos 	heta = \frac{4}{5}$,we have $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$. Therefore,$\cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{4/5}{3/5} = \frac{4}{3}$. Thus,$\cot^2 	heta = (\frac{4}{3})^2 = \frac{16}{9}$.

If $|\vec{a}|=4, |\vec{b}|=5, |\vec{a}-\vec{b}|=3$ and $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$,then $\cot^2 \theta=$

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