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Class 12 Three Dimensional Geometry · Very Long Answer5 Marksorthogonal_directions

Class 12 Maths · Three Dimensional Geometry · Very Long Answer #55

Find vector and Cartesian equations of a line through (-1,3,-2) whose direction is orthogonal to (1,2,3) and (-3,2,5). Also find its distance from the origin.

✅ Step-by-Step Solution

Final Answer: r=(-1,3,-2)+t(2,-7,4), t∈ℝ; (x−(-1))/(2)=(y−(3))/(-7)=(z−(-2))/(4). Distance from origin=sqrt(345)/69.
  1. A common orthogonal direction is d×e.
  2. d=(1,2,3), e=(-3,2,5), d×e=(4,-14,8).
  3. Take the proportional nonzero direction n=(2,-7,4); n·d=n·e=0.
  4. r=(-1,3,-2)+t(2,-7,4), t∈ℝ; (x−(-1))/(2)=(y−(3))/(-7)=(z−(-2))/(4).
  5. Distance from origin is |P×n|/|n|=sqrt(345)/69. r=(-1,3,-2)+t(2,-7,4), t∈ℝ; (x−(-1))/(2)=(y−(3))/(-7)=(z−(-2))/(4). Distance from origin=sqrt(345)/69.

📋 CBSE Stepwise Marking Scheme

  • [1 mark] A common orthogonal direction is d×e.
  • [1 mark] d=(1,2,3), e=(-3,2,5), d×e=(4,-14,8).
  • [1 mark] Take the proportional nonzero direction n=(2,-7,4); n·d=n·e=0.
  • [1 mark] r=(-1,3,-2)+t(2,-7,4), t∈ℝ; (x−(-1))/(2)=(y−(3))/(-7)=(z−(-2))/(4).
  • [1 mark] Distance from origin is |P×n|/|n|=sqrt(345)/69. r=(-1,3,-2)+t(2,-7,4), t∈ℝ; (x−(-1))/(2)=(y−(3))/(-7)=(z−(-2))/(4). Distance from origin=sqrt(345)/69.

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