JEE MainMathematicsThree Dimensional Geometry
A line L₁ passes through the origin and its direction cosines satisfy the relations l + m - n = 0 and l^2 + 2m^2 - n^2 = 0 . It is given that L₁ is perpendicular to the y-axis. If L₂ is another line given by the equation x-1 2 = y-2 1 = z+1 2 , then the square of the shortest distance between L₁ and L₂ is
Options
- A72 13
- B2
- C4
- D0
Correct answer
B. 2
Step-by-step solution
Given relations for the direction cosines of L₁ are: l + m - n = 0 n = l + m ... (i) l^2 + 2m^2 - n^2 = 0 ... (ii) Substituting (i) into (ii): l^2 + 2m^2 - (l + m)^2 = 0 l^2 + 2m^2 - (l^2 + m^2 + 2lm) = 0 m^2 - 2lm = 0 m(m - 2l) = 0 This gives m = 0 or m = 2l . Since L₁ is perpendicular to the y-axis, its direction cosine along the y-axis must be zero, so m = 0 . For m = 0 , n = l . Thus, the direction ratios of L₁ are (l, 0, l) (1, 0, 1) . Since L₁ passes through the origin (0,0,0) , its vector equation is: r = (