The equation of the line of shortest distance | vedclass

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(D) Let the points on the two lines be $P(3\lambda + 6, -\lambda + 7, \lambda + 4)$ and $Q(-3\mu, 2\mu - 9, 4\mu + 2)$.The vector $\vec{PQ} = (-3\mu -

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$\frac{x - 3}{2} = \frac{y - 8}{5} = \frac{z - 3}{-1}$

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(D) Let the points on the two lines be $P(3\lambda + 6, -\lambda + 7, \lambda + 4)$ and $Q(-3\mu, 2\mu - 9, 4\mu + 2)$.The vector $\vec{PQ} = (-3\mu - 3\lambda - 6)\hat{i} + (2\mu + \lambda - 16)\hat{j} + (4\mu - \lambda - 2)\hat{k}$.Since $\vec{PQ}$ is perpendicular to both lines,its dot product with the direction vectors of the lines must be zero.Direction vectors are $\vec{d_1} = 3\hat{i} - \hat{j} + \hat{k}$ and $\vec{d_2} = -3\hat{i} + 2\hat{j} + 4\hat{k}$.$\vec{PQ} \cdot \vec{d_1} = 0 \Rightarrow 3(-3\mu - 3\lambda - 6) - 1(2\mu + \lambda - 16) + 1(4\mu - \lambda - 2) = 0 \Rightarrow -7\mu - 11\lambda = 4$.$\vec{PQ} \cdot \vec{d_2} = 0 \Rightarrow -3(-3\mu - 3\lambda - 6) + 2(2\mu + \lambda - 16) + 4(4\mu - \lambda - 2) = 0 \Rightarrow 29\mu + 7\lambda = 22$.Solving the system of equations,we get $\lambda = -1$ and $\mu = 1$.Substituting these values,we get $P(3, 8, 3)$ and $Q(-3, -7, 6)$.The direction ratios of line $PQ$ are $(-3 - 3, -7 - 8, 6 - 3) = (-6, -15, 3)$,which simplifies to $(2, 5, -1)$.The equation of the line passing through $P(3, 8, 3)$ with direction ratios $(2, 5, -1)$ is $\frac{x - 3}{2} = \frac{y - 8}{5} = \frac{z - 3}{-1}$.

The equation of the line of shortest distance between the lines $\frac{x - 6}{3} = \frac{y - 7}{-1} = \frac{z - 4}{1}$ and $\frac{x}{-3} = \frac{y + 9}{2} = \frac{z - 2}{4}$ is:

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