A line L_1 passes through the point with posi | vedclass

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(B) The equation of line $L_1$ is given by $\vec{r} = 3 \hat{i} + \lambda(-\hat{i} + \hat{j} + \hat{k})$.The equation of line $L_2$ is given by $\vec{

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$2 \hat{i}+\hat{j}+\hat{k}$

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(B) The equation of line $L_1$ is given by $\vec{r} = 3 \hat{i} + \lambda(-\hat{i} + \hat{j} + \hat{k})$.The equation of line $L_2$ is given by $\vec{r} = (\hat{i} + \hat{j}) + \mu(\hat{i} + \hat{k})$.For the point of intersection,the position vectors must be equal:$3 \hat{i} - \lambda \hat{i} + \lambda \hat{j} + \lambda \hat{k} = \hat{i} + \hat{j} + \mu \hat{i} + \mu \hat{k}$Comparing the coefficients of $\hat{i}, \hat{j}, \hat{k}$:For $\hat{j}$: $\lambda = 1$.For $\hat{k}$: $\lambda = \mu$,so $\mu = 1$.For $\hat{i}$: $3 - \lambda = 1 + \mu \Rightarrow 3 - 1 = 1 + 1 \Rightarrow 2 = 2$,which is consistent.Substituting $\lambda = 1$ into the equation of $L_1$:$\vec{r} = 3 \hat{i} + 1(-\hat{i} + \hat{j} + \hat{k}) = 2 \hat{i} + \hat{j} + \hat{k}$.Thus,the position vector of the point of intersection is $2 \hat{i} + \hat{j} + \hat{k}$.

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