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

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(D) The direction vector of the first line is $\vec{b_1} = \hat{i} - \hat{j} + \hat{k}$.The direction vector of the second line is $\vec{b_2} = \hat{i

Vedclass · Accepted Answer

$\frac{\pi}{2}$

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(D) The direction vector of the first line is $\vec{b_1} = \hat{i} - \hat{j} + \hat{k}$.The direction vector of the second line is $\vec{b_2} = \hat{i} + 3\hat{j} + 2\hat{k}$.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: $\vec{b_1} \cdot \vec{b_2} = (1)(1) + (-1)(3) + (1)(2) = 1 - 3 + 2 = 0$.Since the dot product is $0$,$\cos 	heta = 0$,which implies $	heta = \frac{\pi}{2}$.

The angle between the lines $\vec{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-3}{2}$ is

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