JEE Mains · Maths · STD 12 - 11. three dimension geometry
On which of the following lines lies the point of intersection of the line, \(\frac{{x - 4}}{2} = \frac{{y - 5}}{2} = \frac{{z - 3}}{1}\) and the plane, \(x + y + z = 2\) ?
- A \(\frac{{x + 3}}{3} = \frac{{4 - y}}{3} = \frac{{z + 1}}{{ - 2}}\)
- B \(\frac{{x - 4}}{1} = \frac{{y - 5}}{1} = \frac{{z - 5}}{{ - 1}}\)
- C \(\frac{{x - 1}}{1} = \frac{{y - 3}}{2} = \frac{{z + 4}}{{ - 5}}\)
- D \(\frac{{x - 2}}{2} = \frac{{y - 3}}{2} = \frac{{z + 3}}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{{x - 1}}{1} = \frac{{y - 3}}{2} = \frac{{z + 4}}{{ - 5}}\)
Step-by-step Solution
Detailed explanation
Put \((2 \lambda+4,2 \lambda+5, \lambda+3)\) in \(x+y+z=2\) \(2 \lambda+4+2 \lambda+5+\lambda+3=2\) \(5 \lambda=-10 \quad \lambda=-2\) \(P(0,1,1)\) Now put in options Answer is \(C\)
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