Find the angle between the vectors hat{i}-2 h | vedclass

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(A) Let the given vectors be $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$.The magnitude of $\vec{a}$ is $|\

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$\cos^{-1}\left(\frac{5}{7}\right)$

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(A) Let the given vectors be $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$.The magnitude of $\vec{a}$ is $|\vec{a}| = \sqrt{1^{2} + (-2)^{2} + 3^{2}} = \sqrt{1 + 4 + 9} = \sqrt{14}$.The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{3^{2} + (-2)^{2} + 1^{2}} = \sqrt{9 + 4 + 1} = \sqrt{14}$.The dot product is $\vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10$.We know that $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta$,where $	heta$ is the angle between the vectors.Substituting the values: $10 = \sqrt{14} \cdot \sqrt{14} \cos 	heta$.$10 = 14 \cos 	heta$.$\cos 	heta = \frac{10}{14} = \frac{5}{7}$.Therefore,$	heta = \cos^{-1}\left(\frac{5}{7}ight)$.

Find the angle between the vectors $\hat{i}-2 \hat{j}+3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$.

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