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Class 12 Three Dimensional Geometry · Long Answer3 Marksprojection_and_reflection

Class 12 Maths · Three Dimensional Geometry · Long Answer #41

Find the foot and distance from P=(0,2,3) to r=(-3,1,-4)+t(5,2,3), t∈ℝ.
Perpendicular projection
Perpendicular projection

✅ Step-by-Step Solution

Final Answer: Foot H=(2,3,-1); distance=sqrt(21).
  1. Write H=A+td and impose (P−H)·d=0.
  2. A=(-3,1,-4), d=(5,2,3), P−A=(3,1,7); t=((P−A)·d)/(d·d)=1.
  3. H=A+td=(2,3,-1). Check (P−H)·d=0.
  4. |P−H|=√(21)=sqrt(21). Foot H=(2,3,-1); distance=sqrt(21).

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Write H=A+td and impose (P−H)·d=0.
  • [1 mark] A=(-3,1,-4), d=(5,2,3), P−A=(3,1,7); t=((P−A)·d)/(d·d)=1.
  • [1 mark] H=A+td=(2,3,-1). Check (P−H)·d=0.
  • [0.5 marks] |P−H|=√(21)=sqrt(21). Foot H=(2,3,-1); distance=sqrt(21).

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