Discussion
The angle between the lines \(\vec{r}=(3\hat{i}+2\hat{j}-4\hat{k})+\lambda(\hat{i}+2\hat{j}+2\hat{k})\) and \(\vec{r}=(5\hat{i}-2\hat{j})+\mu(3\hat{i}+2\hat{j}+6\hat{k})\) is
Answer: C
The angle is \(\cos^{-1}\left(\frac{19}{21}\right)\).
The angle between two lines is the angle between their direction vectors, here \(\vec{b}_{1} = \hat{i} + 2\hat{j} + 2\hat{k}\) and \(\vec{b}_{2} = 3\hat{i} + 2\hat{j} + 6\hat{k}\). The position vectors play no part.
\[ \begin{aligned} \vec{b}_{1} \cdot \vec{b}_{2} &= 3 + 4 + 12 = 19 \\ |\vec{b}_{1}| &= \sqrt{1 + 4 + 4} = 3 \\ |\vec{b}_{2}| &= \sqrt{9 + 4 + 36} = 7 \\ \cos\theta &= \frac{19}{3 \times 7} = \frac{19}{21} \end{aligned} \]
The position vectors \(3\hat{i} + 2\hat{j} - 4\hat{k}\) and \(5\hat{i} - 2\hat{j}\) only fix where each line sits, not its direction. The other options have denominators 13, 19 and 17, which cannot arise, since \(|\vec{b}_{1}||\vec{b}_{2}| = 21\).
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