JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the line \(\mathrm{L}\) intersect the lines \(\mathrm{x}-2=-\mathrm{y}=\mathrm{z}-1,2(\mathrm{x}+1)=2(\mathrm{y}-1)=\mathrm{z}+1\) and be parallel to the line \(\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}\). Then which of the following points lies on \(\mathrm{L}\) ?
- A \(\left(-\frac{1}{3}, 1,1\right)\)
- B \(\left(-\frac{1}{3}, 1,-1\right)\)
- C \(\left(-\frac{1}{3},-1,-1\right)\)
- D \(\left(-\frac{1}{3},-1,1\right)\)
Answer & Solution
Correct Answer
(B) \(\left(-\frac{1}{3}, 1,-1\right)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{L}_1: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\lambda\) \(\mathrm{L}_2: \frac{\mathrm{x}+1}{\frac{1}{2}}=\frac{\mathrm{y}-1}{\frac{1}{2}}=\frac{\mathrm{z}+1}{1}=\mu\) dr of line \(MN\) will be…
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