JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(l_{1}\) be the line in \(xy\)-plane with \(x\) and \(y\) intercepts \(\frac{1}{8}\) and \(\frac{1}{4 \sqrt{2}}\) respectively, and \(l_{2}\) be the line in \(zx\)-plane with \(x\) and \(z\) intercepts \(-\frac{1}{8}\) and \(-\frac{1}{6 \sqrt{3}}\) respectively. If \(d\) is the shortest distance between the line \(l_{1}\) and \(l_{2}\), then \(d ^{-2}\) is equal to
- A \(52\)
- B \(51\)
- C \(46\)
- D \(59\)
Answer & Solution
Correct Answer
(B) \(51\)
Step-by-step Solution
Detailed explanation
\(8 x+4 \sqrt{2} y=1, z=0\) \(\Rightarrow \frac{x-\frac{1}{8}}{1}=\frac{y-0}{-\sqrt{2}}=\frac{z-0}{0}=\lambda\) \(-8 x-6 \sqrt{3} z=1, y=0\) \(\Rightarrow \frac{x+\frac{1}{8}}{3 \sqrt{3}}=\frac{y-0}{0}=\frac{z-0}{-4}\)…
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