The lines bar{r} = (hat{i} + hat{j} - hat{k}) | vedclass

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(A) The given lines are in the form $\bar{r} = \bar{a}_1 + \lambda \bar{b}_1$ and $\bar{r} = \bar{a}_2 + \mu \bar{b}_2$. Here,$\bar{b}_1 = 3 \hat{i} -

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intersecting but not perpendicular

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(A) The given lines are in the form $\bar{r} = \bar{a}_1 + \lambda \bar{b}_1$ and $\bar{r} = \bar{a}_2 + \mu \bar{b}_2$. Here,$\bar{b}_1 = 3 \hat{i} - \hat{j}$ and $\bar{b}_2 = 2 \hat{i} + 3 \hat{k}$. First,check if the lines are parallel: $\bar{b}_1$ is not a scalar multiple of $\bar{b}_2$,so they are not parallel. Next,check for intersection by equating the lines: $(\hat{i} + \hat{j} - \hat{k}) + \lambda(3 \hat{i} - \hat{j}) = (4 \hat{i} - \hat{k}) + \mu(2 \hat{i} + 3 \hat{k})$. Equating components: $1 + 3\lambda = 4 + 2\mu \implies 3\lambda - 2\mu = 3$ $1 - \lambda = 0 \implies \lambda = 1$ $-1 = -1 + 3\mu \implies 3\mu = 0 \implies \mu = 0$ Substituting $\lambda = 1$ and $\mu = 0$ into the first equation: $3(1) - 2(0) = 3$,which is $3 = 3$. Since the equations are consistent,the lines intersect. Check for perpendicularity: $\bar{b}_1 \cdot \bar{b}_2 = (3)(2) + (-1)(0) + (0)(3) = 6 
eq 0$. Thus,the lines are intersecting but not perpendicular.

The lines $\bar{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3 \hat{i} - \hat{j})$ and $\bar{r} = (4 \hat{i} - \hat{k}) + \mu(2 \hat{i} + 3 \hat{k})$ are

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