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

Correct Answer: 16

To find d₁, the shortest distance between the lines L₁ and L₂:

L₁ : \( \frac{x + 1}{2} = \frac{y - 1}{-12} = \frac{z}{1} \), L₂ : \( \frac{x - 1}{-7} = \frac{y + 8}{2} = \frac{z - 4}{5} \)

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 d₂ for lines L₃ and L₄:

L₃ : \( \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{-3} \), L₄ : \( \frac{x + 2}{1} = \frac{y + 2}{1} = \frac{z - 1}{6} \)

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

Finally,

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