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Class 12 Three Dimensional Geometry · Very Long Answer5 Marksprojection_and_reflection

Class 12 Maths · Three Dimensional Geometry · Very Long Answer #60

Find the foot and distance from P=(5,4,2) to r=(-1,3,1)+t(2,3,-1), t∈ℝ. Also find its reflection in this line.
Perpendicular projection and reflection
Perpendicular projection and reflection

✅ Step-by-Step Solution

Final Answer: Foot H=(1,6,0); distance=2*sqrt(6). Reflection=(-3,8,-2).
  1. Write H=A+td and impose (P−H)·d=0.
  2. A=(-1,3,1), d=(2,3,-1), P−A=(6,1,1); t=((P−A)·d)/(d·d)=1.
  3. H=A+td=(1,6,0). Check (P−H)·d=0.
  4. |P−H|=√(24)=2*sqrt(6). Reflection I=2H−P=(-3,8,-2).
  5. Foot H=(1,6,0); distance=2*sqrt(6). Reflection=(-3,8,-2).

📋 CBSE Stepwise Marking Scheme

  • [1 mark] Write H=A+td and impose (P−H)·d=0.
  • [1 mark] A=(-1,3,1), d=(2,3,-1), P−A=(6,1,1); t=((P−A)·d)/(d·d)=1.
  • [1 mark] H=A+td=(1,6,0). Check (P−H)·d=0.
  • [1 mark] |P−H|=√(24)=2*sqrt(6). Reflection I=2H−P=(-3,8,-2).
  • [1 mark] Foot H=(1,6,0); distance=2*sqrt(6). Reflection=(-3,8,-2).

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