Question:
If the lines \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-\lambda}{3}\) and \(\frac{x}{1}=\frac{y+2}{2}=\frac{z}{4}\) intersect each other, then \(\lambda\) lies in the interval :
If the lines \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-\lambda}{3}\) and \(\frac{x}{1}=\frac{y+2}{2}=\frac{z}{4}\) intersect each other, then \(\lambda\) lies in the interval :
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Write a general point on each line using parameters \(t\) and \(s\), equate coordinates, and solve for \(\lambda\).
Updated On: Nov 12, 2026
- \((-5,-3)\)
- \((13,15)\)
- \((11,13)\)
- \((9,11)\)
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The Correct Option is C
Solution and Explanation
A point on the first line: \((4+t,\ 2+t,\ \lambda+3t)\). A point on the second line: \((s,\ -2+2s,\ 4s)\).
Equating: \(4+t=s\) and \(2+t=-2+2s\).
Substituting \(t=s-4\) in the second: \(s-2=-2+2s\Rightarrow s=0\), so \(t=-4\).
The third coordinates give \(\lambda+3t=4s\Rightarrow \lambda-12=0\Rightarrow \lambda=12\).
\(12\in(11,13)\).
Answer: (3) \((11,13)\)
Equating: \(4+t=s\) and \(2+t=-2+2s\).
Substituting \(t=s-4\) in the second: \(s-2=-2+2s\Rightarrow s=0\), so \(t=-4\).
The third coordinates give \(\lambda+3t=4s\Rightarrow \lambda-12=0\Rightarrow \lambda=12\).
\(12\in(11,13)\).
Answer: (3) \((11,13)\)
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