JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the shortest distance between the lines \(\frac{{x - 1}}{\alpha } = \frac{{y + 1}}{{ - 1}} = \frac{z}{1},(\alpha \ne \, - 1)\) and \(x+y+z +1=0= 2x - y + z + 3\) is \(\frac {1}{\sqrt 3},\) then a value \(\alpha \) is
- A \(-\frac {16}{19}\)
- B \(-\frac {19}{16}\)
- C \(\frac {32}{19}\)
- D \(\frac {19}{32}\)
Answer & Solution
Correct Answer
(C) \(\frac {32}{19}\)
Step-by-step Solution
Detailed explanation
Plane passing through \(x+y+z+1=0\) and \(2 x-y+z+3=0\) is \(x+y+z+1+\lambda\) \((2 x-y+z+3)=0\) \(\Rightarrow(2 \lambda+1) x+(1-\lambda) y+(1+\lambda) z+3 \lambda+1=0\) Parallel to the given line if \(\alpha(2 \lambda+1)-1(1-\lambda)+1(1+\lambda)=0\)…
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