Question:
Let l1 be the line in xy-plane with x and y intercepts ⅛ and \(\frac{1}{4√2} \) respectively and l2 be the line in zx-plane with x and z intercepts -⅛ and \(−\frac{1}{6√3}\) respectively. If d is the shortest distance between the line l1 and l2, then d–2 is equal to ______.
Let l1 be the line in xy-plane with x and y intercepts ⅛ and \(\frac{1}{4√2} \) respectively and l2 be the line in zx-plane with x and z intercepts -⅛ and \(−\frac{1}{6√3}\) respectively. If d is the shortest distance between the line l1 and l2, then d–2 is equal to ______.
Updated On: Sep 24, 2024
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Correct Answer: 51
Solution and Explanation
The correct answer is: 51.
\(\frac{x-\frac{1}{8}}{\frac{1}{8}}=\frac{y}{-\frac{1}{4\sqrt2}}=\frac{z}{0}....L_1\)
or
\(\frac{x-\frac{1}{8}}{1}=\frac{y}{-\sqrt2}=\frac{z}{0}....(i)\)
\(=\frac{\sqrt2}{\sqrt{32+16+54}}=\frac{!}{\sqrt51}\)
d–2 = 51
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