Consider the lines L_1 and L_2 given byL_1: f | vedclass

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(A) Let the point $P$ on $L_1$ be $(2\lambda+1, \lambda+3, 2\lambda+2)$ and the point $Q$ on $L_2$ be $(\mu+2, 2\mu+2, 3\mu+3)$.The direction ratios o

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$2\sqrt{6}$

Vedclass · Answer

(A) Let the point $P$ on $L_1$ be $(2\lambda+1, \lambda+3, 2\lambda+2)$ and the point $Q$ on $L_2$ be $(\mu+2, 2\mu+2, 3\mu+3)$.The direction ratios of the line $PQ$ are given by $((\mu+2)-(2\lambda+1), (2\mu+2)-(\lambda+3), (3\mu+3)-(2\lambda+2)) = (\mu-2\lambda+1, 2\mu-\lambda-1, 3\mu-2\lambda+1)$.Since the direction ratios of $L_3$ are $1, -1, -2$,we have:$\frac{\mu-2\lambda+1}{1} = \frac{2\mu-\lambda-1}{-1} = \frac{3\mu-2\lambda+1}{-2}$.From the first two parts: $-\mu+2\lambda-1 = 2\mu-\lambda-1 \Rightarrow 3\lambda = 3\mu \Rightarrow \lambda = \mu$.Substituting $\lambda = \mu$ into the equality $\frac{\mu-2\lambda+1}{1} = \frac{3\mu-2\lambda+1}{-2}$:$\frac{-\lambda+1}{1} = \frac{\lambda+1}{-2} \Rightarrow 2\lambda - 2 = \lambda + 1 \Rightarrow \lambda = 3$.Thus,$\lambda = 3$ and $\mu = 3$. The points are $P(7, 6, 8)$ and $Q(5, 8, 12)$.The length $PQ = \sqrt{(5-7)^2 + (8-6)^2 + (12-8)^2} = \sqrt{(-2)^2 + 2^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.

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