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JEE Mains · Maths · STD 12 - 11. three dimension geometry

Two lines \(\frac{{x - 3}}{1} = \frac{{y + 1}}{3} = \frac{{z - 6}}{{ - 1}}\) and \(\frac{{x + 5}}{7} = \frac{{y - 2}}{{ - 6}} = \frac{{z - 3}}{4}\) intersect at the point \(R\). The reflection of \(R\) in the \(xy -\) plane has coordinates

  1. A \((2, -4, -7)\)
  2. B \((2, 4, 7)\)
  3. C \((2, -4, 7)\)
  4. D \((-2, 4, 7)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \((2, -4, -7)\)

Step-by-step Solution

Detailed explanation

Points on the given lines are \((\lambda+3,3 \lambda-1,-\lambda+6)\) and \((7 \alpha-5,-6 \alpha+2,4 \alpha+3)\) \(\Rightarrow \lambda+3=7 \alpha-5\) \(3 \lambda-1=-6 \alpha+2\) \(\Rightarrow \alpha=1, \lambda=-1\) Point \(R\) is \((2,-4,7)\) Image of \(\mathrm{R}\) under…
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