Class 12 Three Dimensional Geometry · Short Answer2 Marksequations_of_lines
Class 12 Maths · Three Dimensional Geometry · Short Answer #77
Find vector and Cartesian equations of the line through (2,1,3) with direction (1,1,-2).
✅ Step-by-Step Solution
Final Answer: r=(2,1,3)+t(1,1,-2), t∈ℝ; Cartesian: (x−(2))/(1)=(y−(1))/(1)=(z−(3))/(-2).
- A line through A with nonzero direction d is r=A+td.
- Here A=(2,1,3), d=(1,1,-2). Its components are x=t + 2, y=t + 1, z=3 - 2*t.
- r=(2,1,3)+t(1,1,-2), t∈ℝ; Cartesian: (x−(2))/(1)=(y−(1))/(1)=(z−(3))/(-2).
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] A line through A with nonzero direction d is r=A+td.
- [1 mark] Here A=(2,1,3), d=(1,1,-2). Its components are x=t + 2, y=t + 1, z=3 - 2*t.
- [0.5 marks] r=(2,1,3)+t(1,1,-2), t∈ℝ; Cartesian: (x−(2))/(1)=(y−(1))/(1)=(z−(3))/(-2).
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