Find the point of intersection of the lines v | vedclass

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(A) The position vectors of arbitrary points on the given lines are:Line $1$: $\vec{r} = (3\lambda + 1)\hat{i} + (1 - \lambda)\hat{j} - \hat{k}$Line $

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$(4, 0, -1)$

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(A) The position vectors of arbitrary points on the given lines are:Line $1$: $\vec{r} = (3\lambda + 1)\hat{i} + (1 - \lambda)\hat{j} - \hat{k}$Line $2$: $\vec{r} = (2\mu + 4)\hat{i} + 0\hat{j} + (3\mu - 1)\hat{k}$If the lines intersect,they must have a common point for some values of $\lambda$ and $\mu$:$(3\lambda + 1)\hat{i} + (1 - \lambda)\hat{j} - \hat{k} = (2\mu + 4)\hat{i} + 0\hat{j} + (3\mu - 1)\hat{k}$Equating the components,we get:$3\lambda + 1 = 2\mu + 4$  $(i)$$1 - \lambda = 0$  $(ii)$$-1 = 3\mu - 1$  $(iii)$From $(ii)$,we get $\lambda = 1$.From $(iii)$,$3\mu = 0 \Rightarrow \mu = 0$.Substituting $\lambda = 1$ and $\mu = 0$ into $(i)$:$3(1) + 1 = 4$ and $2(0) + 4 = 4$. Since $4 = 4$,the lines intersect.To find the point of intersection,substitute $\lambda = 1$ into the first line equation:$\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + 1(3\hat{i} - \hat{j}) = 4\hat{i} + 0\hat{j} - \hat{k}$.Thus,the coordinates of the point of intersection are $(4, 0, -1)$.

Find the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$.

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