If the lines frac{x-1}{2} = frac{y+1}{3} = fr | vedclass

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(D) Let the given lines be $L_1: \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4} = \lambda$ and $L_2: \frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1} = \mu$

Vedclass · Accepted Answer

$\frac{9}{2}$

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(D) Let the given lines be $L_1: \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4} = \lambda$ and $L_2: \frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1} = \mu$. Any point on $L_1$ is $P(2\lambda+1, 3\lambda-1, 4\lambda+1)$ and any point on $L_2$ is $Q(\mu+3, 2\mu+k, \mu)$. For the lines to intersect,there must exist $\lambda$ and $\mu$ such that $P = Q$. Equating the coordinates: $2\lambda + 1 = \mu + 3 \implies 2\lambda - \mu = 2$  $(1)$ $3\lambda - 1 = 2\mu + k \implies 3\lambda - 2\mu = k + 1$  $(2)$ $4\lambda + 1 = \mu \implies 4\lambda - \mu = -1$  $(3)$ Subtracting $(1)$ from $(3)$: $(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2 \implies 2\lambda = -3 \implies \lambda = -\frac{3}{2}$. Substituting $\lambda = -\frac{3}{2}$ into $(1)$: $2(-\frac{3}{2}) - \mu = 2 \implies -3 - \mu = 2 \implies \mu = -5$. Now substitute $\lambda = -\frac{3}{2}$ and $\mu = -5$ into $(2)$: $3(-\frac{3}{2}) - 2(-5) = k + 1$ $-\frac{9}{2} + 10 = k + 1$ $\frac{11}{2} = k + 1$ $k = \frac{11}{2} - 1 = \frac{9}{2}$.

If the lines $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}$ and $\frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1}$ intersect,then find the value of $k$.

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