JEE Mains · Maths · STD 12 - 11. three dimension geometry
The image of the line \(\frac{{x - 1}}{3} = \frac{{y - 3}}{1} = \frac{{z - 4}}{{ - 5}}\) in the plane \(2x - y + z + 3 = 0\) is the line :
- A \(\;\frac{{x - 3}}{3} = \frac{{y + 5}}{1} = \frac{{z - 2}}{{ - 5}}\)
- B \(\;\frac{{x - 3}}{{ - 3}} = \frac{{y + 5}}{{ - 1}} = \frac{{z - 4}}{5}\)
- C \(\;\frac{{x + 3}}{3} = \frac{{y - 5}}{1} = \frac{{z - 2}}{{ - 5}}\)
- D \(\;\frac{{x + 3}}{{ - 3}} = \frac{{y - 5}}{{ - 1}} = \frac{{z + 2}}{5}\)
Answer & Solution
Correct Answer
(C) \(\;\frac{{x + 3}}{3} = \frac{{y - 5}}{1} = \frac{{z - 2}}{{ - 5}}\)
Step-by-step Solution
Detailed explanation
Equation of \(AB\) is \(\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\lambda\) Co-ordinate of point \(\mathrm{B}\) is \(\Rightarrow x=1+2 \lambda\) point satisfy the equation of plane \(y=3-\lambda\) \(z = 4 + \lambda \)…
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