The angle between the lines bar{r}=(3 hat{i}+ | vedclass

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(D) The direction vectors of the two lines are $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.The angle $	h

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

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

The angle between the lines $\bar{r}=(3 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})$ and $\bar{r}=(5 \hat{i}-2 \hat{k})+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})$ is:

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