JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the foot of the perpendicular from point \((4,3,8)\) on the line \(L _{1}: \frac{ x - a }{l}=\frac{ y -2}{3}=\frac{ z - b }{4},\) \(l \neq 0\) is \((3,5,7),\) then the shortest distance between the line \(L _{1}\) and line \(L _{2}: \frac{ x -2}{3}=\frac{ y -4}{4}=\frac{ z -5}{5}\) is equal to:
- A \(\frac{1}{2}\)
- B \(\frac{1}{\sqrt{6}}\)
- C \(\sqrt{\frac{2}{3}}\)
- D \(\frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{6}}\)
Step-by-step Solution
Detailed explanation
\((3,5,7)\) satisfy the line \(L_{1}\) \(\frac{3-a}{\ell}=\frac{5-2}{3}=\frac{7-b}{4}\) \(\frac{3- a }{\ell}=1\) And \(\frac{7-b}{4}=1\) \(a+\ell=3 \quad \ldots(1)\) And \(b=3\) \(\overrightarrow{ v }_{1}=<4,3,8>-<3,5,7>\) \(\overrightarrow{ v }_{1}=<1,-2,1>\)…
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