The angle between the lines vec{r}=(2 hat{i}- | vedclass

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Question

(A) The given lines are in the form $\vec{r} = \vec{a} + \lambda \vec{b}$.For the first line, the direction vector is $\vec{b_1} = \hat{i} + 4 \hat{j}

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$\frac{\pi}{2}$

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(A) The given lines are in the form $\vec{r} = \vec{a} + \lambda \vec{b}$.For the first line, the direction vector is $\vec{b_1} = \hat{i} + 4 \hat{j} + 3 \hat{k}$.For the second line, the direction vector is $\vec{b_2} = \hat{i} + 2 \hat{j} - 3 \hat{k}$.The angle $	heta$ between the two lines is given by $\cos 	heta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(1) + (4)(2) + (3)(-3) = 1 + 8 - 9 = 0$.Since the dot product is $0$, the angle between the lines is $	heta = \cos^{-1}(0) = \frac{\pi}{2}$.

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

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