The angle between the vectors 2 hat{k} - 3 ha | vedclass

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(B) Let $\vec{a} = -3 \hat{j} + 2 \hat{k}$ and $\vec{b} = \hat{i} - 2 \hat{k}$.The dot product formula is $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}|

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$\cos^{-1}\left(\frac{-4}{\sqrt{65}}\right)$

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(B) Let $\vec{a} = -3 \hat{j} + 2 \hat{k}$ and $\vec{b} = \hat{i} - 2 \hat{k}$.The dot product formula is $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$,where $	heta$ is the angle between the vectors.First,calculate the dot product: $\vec{a} \cdot \vec{b} = (0)(1) + (-3)(0) + (2)(-2) = -4$.Next,calculate the magnitudes:$|\vec{a}| = \sqrt{0^2 + (-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$.$|\vec{b}| = \sqrt{1^2 + 0^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.Thus,$|\vec{a}| |\vec{b}| = \sqrt{13} 	imes \sqrt{5} = \sqrt{65}$.Substituting these into the formula: $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-4}{\sqrt{65}}$.Therefore,$	heta = \cos^{-1}\left(\frac{-4}{\sqrt{65}}ight)$.

The angle between the vectors $2 \hat{k} - 3 \hat{j}$ and $\hat{i} - 2 \hat{k}$ is

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