The angle between the pair of lines given by | vedclass

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(B) The direction vectors of the two lines are $\vec{b_1} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b_2} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.First, cal

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$\cos^{-1}(\frac{19}{21})$

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(B) The direction vectors of the two lines are $\vec{b_1} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b_2} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.First, calculate the magnitudes of the direction vectors:$|\vec{b_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.$|\vec{b_2}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.Next, calculate the dot product of the direction vectors:$\vec{b_1} \cdot \vec{b_2} = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$.The angle $	heta$ between the lines is given by the formula $\cos 	heta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.Substituting the values, we get $\cos 	heta = \frac{19}{3 \cdot 7} = \frac{19}{21}$.Therefore, $	heta = \cos^{-1}(\frac{19}{21})$.

The angle between the pair of lines given by $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{i} - 2\hat{k} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$ is . . . . . . .

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