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

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(D) The angle $	heta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec

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None of these

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(D) The angle $	heta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.First,calculate the dot product: $\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1$.Next,calculate the magnitudes: $|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ and $|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.Now,substitute these values into the formula: $\cos 	heta = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3}$.Therefore,$	heta = \cos^{-1}\left(\frac{1}{3}ight)$.Comparing this with the given options,the correct answer is not explicitly listed as $\cos^{-1}(1/3)$,so the correct choice is $D$.

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

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