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(A) Let $Q$ be the angle between the given lines.The angle between the given lines is given by the formula,$\cos Q = \left| \frac{\vec{b}_{1} \cdot \v

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$Q=\cos ^{-1}\left(\frac{19}{21}\right)$

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(A) Let $Q$ be the angle between the given lines.The angle between the given lines is given by the formula,$\cos Q = \left| \frac{\vec{b}_{1} \cdot \vec{b}_{2}}{|\vec{b}_{1}| |\vec{b}_{2}|} ight|$.The given lines are parallel to the vectors $\vec{b}_{1} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ and $\vec{b}_{2} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ respectively.Calculate the magnitudes of the vectors:$|\vec{b}_{1}| = \sqrt{3^{2} + 2^{2} + 6^{2}} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.$|\vec{b}_{2}| = \sqrt{1^{2} + 2^{2} + 2^{2}} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.Calculate the dot product of the vectors:$\vec{b}_{1} \cdot \vec{b}_{2} = (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k})$$= (3 	imes 1) + (2 	imes 2) + (6 	imes 2) = 3 + 4 + 12 = 19$.Substitute the values into the formula:$\cos Q = \frac{19}{7 	imes 3} = \frac{19}{21}$.Therefore,$Q = \cos ^{-1}\left(\frac{19}{21}ight)$.

Find the angle between the following pairs of lines:
$\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})$ and
$\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$

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Find the angle between the following pairs of lines:$\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})$ and$\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$