JEE Mains · Maths · STD 12 - 11. three dimension geometry
If \(d_1\) is the shortest distance between the lines \(x+1=2 y=-12 z, x=y+2=6 z-6\) and \(d_2\) is the shortest distance between the lines \(\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}\), then the value of \(\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}\) is :
- A \(17\)
- B \(16\)
- C \(42\)
- D \(45\)
Answer & Solution
Correct Answer
(B) \(16\)
Step-by-step Solution
Detailed explanation
\(L_1: \frac{x+1}{1}=\frac{y}{1 / 2}=\frac{z}{-1 / 12}, L_2: \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}\) \(\mathrm{d}_1=\) shortest distance between \(\mathrm{L}_1 \& \mathrm{~L}_2\)…
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