JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the shortest distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}\) and \(\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}\) is \(\frac{1}{\sqrt{3}}\), then the sum of all possible values of \(\lambda\) is
- A \(16\)
- B \(6\)
- C \(12\)
- D \(15\)
Answer & Solution
Correct Answer
(A) \(16\)
Step-by-step Solution
Detailed explanation
SHORTEST distance \(\frac{\left|\left(a_{2}-a_{1}\right) \cdot\left(b_{1} \times b_{2}\right)\right|}{\left|b_{1} \times b_{2}\right|}\) \(a _{1}=(1,2,3)\) \(a _{2}=(2,4,5)\) \(\overrightarrow{ b }_{2}=2 \hat{ i }+3 \hat{ j }+\lambda \hat{ k }\)…
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