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Class 12 Three Dimensional Geometry · Long Answer3 Marksprojection_and_reflection

Class 12 Maths · Three Dimensional Geometry · Long Answer #83

Find the foot and distance from P=(-1,8,4) to r=(0,-1,3)+t(2,-2,-4), 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=(-2,1,7); distance=sqrt(59). Reflection=(-3,-6,10).
  1. Write H=A+td and impose (P−H)·d=0.
  2. A=(0,-1,3), d=(2,-2,-4), P−A=(-1,9,1); t=((P−A)·d)/(d·d)=-1.
  3. H=A+td=(-2,1,7). Check (P−H)·d=0.
  4. |P−H|=√(59)=sqrt(59). Reflection I=2H−P=(-3,-6,10). Foot H=(-2,1,7); distance=sqrt(59). Reflection=(-3,-6,10).

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Write H=A+td and impose (P−H)·d=0.
  • [1 mark] A=(0,-1,3), d=(2,-2,-4), P−A=(-1,9,1); t=((P−A)·d)/(d·d)=-1.
  • [1 mark] H=A+td=(-2,1,7). Check (P−H)·d=0.
  • [0.5 marks] |P−H|=√(59)=sqrt(59). Reflection I=2H−P=(-3,-6,10). Foot H=(-2,1,7); distance=sqrt(59). Reflection=(-3,-6,10).

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