If the angle between the line x= y -1 2 = z -3 and the plane x+2 y+3 z=4 is ⁻¹ 5 14 , then the value of is

Step-by-step solution

The line L has direction vector d = (1, 2, ) and the plane P: x + 2y + 3z = 4 has normal vector n = (1, 2, 3) . The given angle between the line and plane is ⁻¹ 5 14 , so = 5 14 and ^2 = 1 - ^2 = 9 14 , yielding = 3 14 . Using the angle formula = | d n | | d | | n | , we have d n = 5 + 3 , | d | = 5 + ^2 , and | n | = 14 . Substituting gives 3 14 = |5 + 3 | 5 + ^2 14 , which simplifies to 3 = |5 + 3 | 5 + ^2 . Squaring both sides yields 9 = (5 + 3 )^2 5 + ^2 , and solving the resulting equation gives = 20 30 = 2 3