Question

If $d_1$ is the shortest distance between the lines $\frac{x+1}{2} = \frac{y-1}{-12} = \frac{z-1}{-7} \quad \text{and} \quad \frac{x-1}{7} = \frac{y+8}{2} = \frac{z-4}{5},$ and $d_2$ is the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3} \quad \text{and} \quad \frac{x+2}{1} = \frac{y+2}{2} = \frac{z-1}{6},$ then the value of $\frac{32\sqrt{3}d_1}{d_2}$ is:

Answer 1

To find d 1, the shortest distance between the lines L 1 and L 2:

L 1 : $\frac{x + 1}{2} = \frac{y - 1}{-12} = \frac{z}{1}$, L 2 : $\frac{x - 1}{-7} = \frac{y + 8}{2} = \frac{z - 4}{5}$

Using the formula for the distance between two skew lines $d = \frac{|(\vec{a_2} - \vec{a_1}) \times (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$, we calculate:

$d_1 = 2$

Similarly, to find d 2 for lines L 3 and L 4:

L 3 : $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{-3}$, L 4 : $\frac{x + 2}{1} = \frac{y + 2}{1} = \frac{z - 1}{6}$

we get: $d_2 = \frac{12}{\sqrt{3}}$

Finally,

$\frac{32 \sqrt{3} d_1}{d_2} = \frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}} = 16$