Class 12 Three Dimensional Geometry · Very Long Answer5 Marksprojection_and_reflection

Class 12 Maths · Three Dimensional Geometry · Very Long Answer #88

Find the foot and distance from P=(1,6,3) to r=(0,1,2)+t(1,2,3), t∈ℝ. Also find its reflection in this line.

✅ Step-by-Step Solution

Final Answer: Foot H=(1,3,5); distance=sqrt(13). Reflection=(1,0,7). Joining line: r=(1,6,3)+t(0,-6,4), t∈ℝ; (y−(6))/(-6)=(z−(3))/(4); x=1.

Write H=A+td and impose (P−H)·d=0.
A=(0,1,2), d=(1,2,3), P−A=(1,5,1); t=((P−A)·d)/(d·d)=1.
H=A+td=(1,3,5). Check (P−H)·d=0.
|P−H|=√(13)=sqrt(13). Reflection I=2H−P=(1,0,7).
Foot H=(1,3,5); distance=sqrt(13). Reflection=(1,0,7). Joining line: r=(1,6,3)+t(0,-6,4), t∈ℝ; (y−(6))/(-6)=(z−(3))/(4); x=1.

[1 mark] Write H=A+td and impose (P−H)·d=0.
[1 mark] A=(0,1,2), d=(1,2,3), P−A=(1,5,1); t=((P−A)·d)/(d·d)=1.
[1 mark] H=A+td=(1,3,5). Check (P−H)·d=0.
[1 mark] |P−H|=√(13)=sqrt(13). Reflection I=2H−P=(1,0,7).
[1 mark] Foot H=(1,3,5); distance=sqrt(13). Reflection=(1,0,7). Joining line: r=(1,6,3)+t(0,-6,4), t∈ℝ; (y−(6))/(-6)=(z−(3))/(4); x=1.

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