The angle between the vectors vec{a} = hat{i} | vedclass

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(D) Let $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.The dot product of two vectors is given by $\vec{a} \cdot

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$\frac{\pi}{2}$

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(D) Let $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.The dot product of two vectors is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$,where $	heta$ is the angle between them.First,calculate the dot product: $\vec{a} \cdot \vec{b} = (1)(1) + (-1)(2) + (1)(1) = 1 - 2 + 1 = 0$.Next,calculate the magnitudes: $|\vec{a}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$ and $|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$.Substitute these values into the formula: $0 = \sqrt{3} \cdot \sqrt{6} \cdot \cos 	heta$.Since $\sqrt{3} \cdot \sqrt{6} 
eq 0$,we must have $\cos 	heta = 0$.Therefore,$	heta = \cos^{-1}(0) = \frac{\pi}{2}$.

The angle between the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ is

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