The shortest distance between the lines $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ and $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ is

<p>Given two lines are represented as:</p> <p>$ \frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5} $</p> <p>and</p> <p>$ \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3} $</p> <p>The formula for the shortest distance between two lines is:</p> <p>$ d = \frac{|\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right)|}{|\vec{b}_1 \times \vec{b}_2|} $</p> <p>From the given lines:</p> <p>$ \vec{a}_1 = (3, -15, 9) $</p> <p>$ \vec{a}_2 = (-1, 1, 9) $</p> <p>$ \vec{b}_1 = (2, -7, 5) $</p> <p>$ \vec{b}_2 = (2, 1, -3) $</p> <p>Calculate the difference:</p> <p>$ \vec{a}_2 - \vec{a}_1 = (-1-3, 1+15, 9-9) = (-4, 16, 0) $</p> <p>Next, compute the cross product:</p> <p>$ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = (16\hat{i} + 16\hat{j} + 16\hat{k}) $</p> <p>The magnitude of the cross product is:</p> <p>$ |\vec{b}_1 \times \vec{b}_2| = |(16\hat{i} + 16\hat{j} + 16\hat{k})| = 16\sqrt{3} $</p> <p>Calculate the dot product:</p> <p>$ (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-4\hat{i} + 16\hat{j} + 0\hat{k}) \cdot (16\hat{i} + 16\hat{j} + 16\hat{k}) = -64 + 256 + 0 = 192 $</p> <p>Finally, find the shortest distance:</p> <p>$ d = \frac{|192|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = 4\sqrt{3} $</p> <p>Thus, the shortest distance is:</p> <p>$ \boxed{4 \sqrt{3}} $</p>