The angle between the pair of lines vec{r} = | vedclass

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(C) The angle $	heta$ between two lines with direction vectors $\vec{b_1} = 3\hat{i} + 5\hat{j} + 4\hat{k}$ and $\vec{b_2} = \hat{i} + \hat{j} + 2\ha

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$\cos^{-1}(\frac{8\sqrt{3}}{15})$

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(C) The angle $	heta$ between two lines with direction vectors $\vec{b_1} = 3\hat{i} + 5\hat{j} + 4\hat{k}$ and $\vec{b_2} = \hat{i} + \hat{j} + 2\hat{k}$ is given by $\cos 	heta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.$\vec{b_1} \cdot \vec{b_2} = (3)(1) + (5)(1) + (4)(2) = 3 + 5 + 8 = 16$.$|\vec{b_1}| = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$.$|\vec{b_2}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.$\cos 	heta = \frac{16}{5\sqrt{2} \cdot \sqrt{6}} = \frac{16}{5\sqrt{12}} = \frac{16}{5(2\sqrt{3})} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}$.Rationalizing the denominator: $\frac{8}{5\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{5 \cdot 3} = \frac{8\sqrt{3}}{15}$.Therefore, $	heta = \cos^{-1}(\frac{8\sqrt{3}}{15})$.

The angle between the pair of lines $\vec{r} = -3\hat{i} + \hat{j} + 3\hat{k} + \lambda(3\hat{i} + 5\hat{j} + 4\hat{k})$ and $\vec{r} = -\hat{i} + 4\hat{j} + 5\hat{k} + \mu(\hat{i} + \hat{j} + 2\hat{k})$ is . . . . . . .

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