Class 12 Three Dimensional Geometry · Long Answer3 Marksequations_of_lines
Class 12 Maths · Three Dimensional Geometry · Long Answer #42
Find the vector equation through (1,2,−1) parallel to 5x−25=14−7y=35z.
✅ Step-by-Step Solution
Final Answer: r=(1,2,-1)+t(7,-5,1), t∈ℝ; Cartesian: (x−(1))/(7)=(y−(2))/(-5)=(z−(-1))/(1).
- A line through A with nonzero direction d is r=A+td.
- Here A=(1,2,-1), d=(7,-5,1).
- Its components are x=7*t + 1, y=2 - 5*t, z=t - 1.
- r=(1,2,-1)+t(7,-5,1), t∈ℝ; Cartesian: (x−(1))/(7)=(y−(2))/(-5)=(z−(-1))/(1).
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] A line through A with nonzero direction d is r=A+td.
- [1 mark] Here A=(1,2,-1), d=(7,-5,1).
- [1 mark] Its components are x=7*t + 1, y=2 - 5*t, z=t - 1.
- [0.5 marks] r=(1,2,-1)+t(7,-5,1), t∈ℝ; Cartesian: (x−(1))/(7)=(y−(2))/(-5)=(z−(-1))/(1).
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