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The given lines are: $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3\hat{i}-\hat{j})$ and $\vec{r}=(4\hat{i}-\hat{k})+\mu(2\hat{i}+3\hat{k})$. We can write these in parametric form as: Line 1: $x = 1 + 3\lambda$ $y = 1 - \lambda$ $z = -1$ Line 2: $x = 4 + 2\mu$ $y = 0$ $z = -1 + 3\mu$
For the lines to intersect, there must exist values of $\lambda$ and $\mu$ such that the coordinates $(x, y, z)$ are the same for both lines. Therefore, we set the corresponding coordinates equal to each other: $1 + 3\lambda = 4 + 2\mu$ (1) $1 - \lambda = 0$ (2) $-1 = -1 + 3\mu$ (3)
From equation (2), we have $\lambda = 1$. From equation (3), we have $3\mu = 0$, so $\mu = 0$. Substitute $\lambda = 1$ and $\mu = 0$ into equation (1): $1 + 3(1) = 4 + 2(0)$ $1 + 3 = 4$ $4 = 4$ Since the equation holds true, the lines intersect.
Using $\lambda = 1$ in the equation for Line 1: $x = 1 + 3(1) = 4$ $y = 1 - (1) = 0$ $z = -1$ So, the point of intersection is $(4, 0, -1)$. Using $\mu = 0$ in the equation for Line 2: $x = 4 + 2(0) = 4$ $y = 0$ $z = -1 + 3(0) = -1$ So, the point of intersection is $(4, 0, -1)$.
Final Answer: Yes, the lines intersect.<\/span>
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