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Class 12 Three Dimensional Geometry · Very Long Answer5 Marksorthogonal_directions

Class 12 Maths · Three Dimensional Geometry · Very Long Answer #59

Find vector and Cartesian equations of a line through (1,1,1) whose direction is orthogonal to (1,2,4) and (2,3,4). Also find the acute angle between the given directions.

✅ Step-by-Step Solution

Final Answer: r=(1,1,1)+t(-4,4,-1), t∈ℝ; (x−(1))/(-4)=(y−(1))/(4)=(z−(1))/(-1). Acute angle of the given lines=cos⁻¹(8*sqrt(609)/203).
  1. A common orthogonal direction is d×e.
  2. d=(1,2,4), e=(2,3,4), d×e=(-4,4,-1).
  3. Take the proportional nonzero direction n=(-4,4,-1); n·d=n·e=0.
  4. r=(1,1,1)+t(-4,4,-1), t∈ℝ; (x−(1))/(-4)=(y−(1))/(4)=(z−(1))/(-1).
  5. Use cos θ=|d·e|/(|d||e|)=8*sqrt(609)/203. r=(1,1,1)+t(-4,4,-1), t∈ℝ; (x−(1))/(-4)=(y−(1))/(4)=(z−(1))/(-1). Acute angle of the given lines=cos⁻¹(8*sqrt(609)/203).

📋 CBSE Stepwise Marking Scheme

  • [1 mark] A common orthogonal direction is d×e.
  • [1 mark] d=(1,2,4), e=(2,3,4), d×e=(-4,4,-1).
  • [1 mark] Take the proportional nonzero direction n=(-4,4,-1); n·d=n·e=0.
  • [1 mark] r=(1,1,1)+t(-4,4,-1), t∈ℝ; (x−(1))/(-4)=(y−(1))/(4)=(z−(1))/(-1).
  • [1 mark] Use cos θ=|d·e|/(|d||e|)=8*sqrt(609)/203. r=(1,1,1)+t(-4,4,-1), t∈ℝ; (x−(1))/(-4)=(y−(1))/(4)=(z−(1))/(-1). Acute angle of the given lines=cos⁻¹(8*sqrt(609)/203).

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