Given the lines vec{r} = (3+t)hat{i} + (1-t)h | vedclass

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(B) The first line is given by $\vec{r} = (3\hat{i} + \hat{j} - 2\hat{k}) + t(\hat{i} - \hat{j} - 2\hat{k})$. The direction vector is $\vec{b_1} = \ha

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Coincident

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(B) The first line is given by $\vec{r} = (3\hat{i} + \hat{j} - 2\hat{k}) + t(\hat{i} - \hat{j} - 2\hat{k})$. The direction vector is $\vec{b_1} = \hat{i} - \hat{j} - 2\hat{k}$ and it passes through $P_1(3, 1, -2)$.The second line is given by $x = 4+k, y = -k, z = -4-2k$,which can be written as $\vec{r} = (4\hat{i} - 4\hat{k}) + k(\hat{i} - \hat{j} - 2\hat{k})$. The direction vector is $\vec{b_2} = \hat{i} - \hat{j} - 2\hat{k}$ and it passes through $P_2(4, 0, -4)$.Since $\vec{b_1} = \vec{b_2}$,the lines are parallel.To check if they are coincident,we check if $P_1$ lies on the second line. Substituting $P_1(3, 1, -2)$ into the second line equations:$3 = 4+k \implies k = -1$$1 = -k \implies k = -1$$-2 = -4-2k \implies -2 = -4-2(-1) = -4+2 = -2$Since $k = -1$ satisfies all equations,the point $P_1$ lies on the second line. Therefore,the lines are coincident.

Given the lines $\vec{r} = (3+t)\hat{i} + (1-t)\hat{j} + (-2-2t)\hat{k}$,$t \in R$ and $x = 4+k, y = -k, z = -4-2k$,$k \in R$. What is the relationship between these two lines?

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