If the angle between the planes x-2 y+3 z-5=0 and x+ y+2 z+7=0 is ⁻¹ ( 1 14 ) then the difference between the values of is

Step-by-step solution

The angle between two planes with normal vectors n₁ and n₂ is determined by = | n₁ n₂ | | n₁ || n₂ | . For planes P₁: x-2y+3z-5=0 and P₂: x+ y+2z+7=0 , their normals are n₁ = 1,-2,3 and n₂ = 1, ,2 . Given = 1 14 , the dot product and magnitudes yield: n₁ n₂ = 7 - 2 | n₁ | = 14 | n₂ | = 5+ ^2 Substituting gives 1 14 = |7-2 | 14 5+ ^2 , which simplifies to 1 14 = (7-2 )^2 5+ ^2 after squaring. Solving 5+ ^2 = 14(7-2 )^2 produces the quadratic 55 ^2 - 392 + 681 = 0 . Using the quadratic formula, = 392 62 110 , giving