The angle between force vec{F}=3 hat{i}+4 hat | vedclass

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(B) The angle $	heta$ between two vectors $\vec{F}$ and $\vec{d}$ is given by the formula: $\cos 	heta = \frac{\vec{F} \cdot \vec{d}}{|\vec{F}| |\ve

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$\cos ^{-1}(0.32)$

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(B) The angle $	heta$ between two vectors $\vec{F}$ and $\vec{d}$ is given by the formula: $\cos 	heta = \frac{\vec{F} \cdot \vec{d}}{|\vec{F}| |\vec{d}|}$First,calculate the dot product: $\vec{F} \cdot \vec{d} = (3)(5) + (4)(4) + (-5)(3) = 15 + 16 - 15 = 16$Next,calculate the magnitudes: $|\vec{F}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$$|\vec{d}| = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50}$Substitute these values into the formula: $\cos 	heta = \frac{16}{\sqrt{50} 	imes \sqrt{50}} = \frac{16}{50}$Therefore,$\cos 	heta = 0.32$,which implies $	heta = \cos^{-1}(0.32)$.

The angle between force $\vec{F}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ and displacement $\vec{d}=5 \hat{i}+4 \hat{j}+3 \hat{k}$ is

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