Two parallel planes are given by, x+y+z=1 and x+y+z= 9 2 . A third plane that intersects them is given by 2 x-5 y+z=-5 , resulting in two parallel lines of intersection. If the distance ' d ' between these two parallel lines can be expressed as d= a b , where a and b are co-prime positive integers, then find the value of [d] . [Note: Where [k] denotes greatest integer function less than or equal to k .]

Step-by-step solution

P₁: x+y+z=1 ; P₂: x+y+z= 9 2 and P₃: 2 x-5 y+z=-5 Vector along the line of intersection of the planes P₁ / P₃ or P₂ / P₃ is = n = | array ccc i & j & k 1 & 1 & 1 2 & -5 & 1 array |=6 i + j -7 k Point on line L₁ which is the line of intersection of P₁ and P₃ say (x₁, y₁, 0 ) Hence, x₁+y₁=1 and 2 x₁-5 y₁=-5 Solving, we get y₁=1 and x₁=0 Hence, point on L₁ is (0,1,0) ......(A) ||ly point on line L₂ which is line of intersection of P₂ and P₃ say (x₂, y₂, 0 ) Hence x₂+y₂= 9 2 and 2 x₂-5 y₂=-5 Solving, we get y₂=2 and x₂