If L_1 is a line through the point 5 hat{i}+8 | vedclass

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(B) Line $L_1$ passes through $5 \hat{i}+8 \hat{j}+11 \hat{k}$ and is parallel to $2 \hat{i}+3 \hat{j}+4 \hat{k}$.Thus,$L_1: \vec{r} = (5 \hat{i}+8 \h

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$\hat{i}+2 \hat{j}+3 \hat{k}$

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(B) Line $L_1$ passes through $5 \hat{i}+8 \hat{j}+11 \hat{k}$ and is parallel to $2 \hat{i}+3 \hat{j}+4 \hat{k}$.Thus,$L_1: \vec{r} = (5 \hat{i}+8 \hat{j}+11 \hat{k}) + \lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$.Line $L_2$ passes through $4 \hat{i}+6 \hat{j}+8 \hat{k}$ and is parallel to $3 \hat{i}+4 \hat{j}+5 \hat{k}$.Thus,$L_2: \vec{r} = (4 \hat{i}+6 \hat{j}+8 \hat{k}) + \mu(3 \hat{i}+4 \hat{j}+5 \hat{k})$.For intersection,equate the two lines:$(5+2\lambda) \hat{i} + (8+3\lambda) \hat{j} + (11+4\lambda) \hat{k} = (4+3\mu) \hat{i} + (6+4\mu) \hat{j} + (8+5\mu) \hat{k}$.Comparing components:$5+2\lambda = 4+3\mu \Rightarrow 2\lambda - 3\mu = -1$ $(i)$$8+3\lambda = 6+4\mu \Rightarrow 3\lambda - 4\mu = -2$ (ii)Solving $(i)$ and (ii): Multiply $(i)$ by $3$ and (ii) by $2$:$6\lambda - 9\mu = -3$$6\lambda - 8\mu = -4$Subtracting gives $\mu = -1$. Substituting $\mu = -1$ into $(i)$: $2\lambda - 3(-1) = -1 \Rightarrow 2\lambda = -4 \Rightarrow \lambda = -2$.Substituting $\lambda = -2$ into $L_1$: $\vec{r} = (5-4) \hat{i} + (8-6) \hat{j} + (11-8) \hat{k} = \hat{i} + 2 \hat{j} + 3 \hat{k}$.

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