Class 12 Three Dimensional Geometry · Short Answer2 Marksequations_of_lines
Class 12 Maths · Three Dimensional Geometry · Short Answer #80
Find vector and Cartesian equations of the line through (4,3,-5) with direction (-2,-5,3).
✅ Step-by-Step Solution
Final Answer: r=(4,3,-5)+t(-2,-5,3), t∈ℝ; Cartesian: (x−(4))/(-2)=(y−(3))/(-5)=(z−(-5))/(3).
- A line through A with nonzero direction d is r=A+td.
- Here A=(4,3,-5), d=(-2,-5,3). Its components are x=4 - 2*t, y=3 - 5*t, z=3*t - 5.
- r=(4,3,-5)+t(-2,-5,3), t∈ℝ; Cartesian: (x−(4))/(-2)=(y−(3))/(-5)=(z−(-5))/(3).
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] A line through A with nonzero direction d is r=A+td.
- [1 mark] Here A=(4,3,-5), d=(-2,-5,3). Its components are x=4 - 2*t, y=3 - 5*t, z=3*t - 5.
- [0.5 marks] r=(4,3,-5)+t(-2,-5,3), t∈ℝ; Cartesian: (x−(4))/(-2)=(y−(3))/(-5)=(z−(-5))/(3).
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