Class 12 Three Dimensional Geometry · Short Answer2 Marksequations_of_lines
Class 12 Maths · Three Dimensional Geometry · Short Answer #79
Find vector and Cartesian equations of the line through (3,4,-7) with direction (-2,-5,13).
✅ Step-by-Step Solution
Final Answer: r=(3,4,-7)+t(-2,-5,13), t∈ℝ; Cartesian: (x−(3))/(-2)=(y−(4))/(-5)=(z−(-7))/(13).
- A line through A with nonzero direction d is r=A+td.
- Here A=(3,4,-7), d=(-2,-5,13). Its components are x=3 - 2*t, y=4 - 5*t, z=13*t - 7.
- r=(3,4,-7)+t(-2,-5,13), t∈ℝ; Cartesian: (x−(3))/(-2)=(y−(4))/(-5)=(z−(-7))/(13).
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] A line through A with nonzero direction d is r=A+td.
- [1 mark] Here A=(3,4,-7), d=(-2,-5,13). Its components are x=3 - 2*t, y=4 - 5*t, z=13*t - 7.
- [0.5 marks] r=(3,4,-7)+t(-2,-5,13), t∈ℝ; Cartesian: (x−(3))/(-2)=(y−(4))/(-5)=(z−(-7))/(13).
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