The equation of the plane containing the line x-2 3 = y+1 2 = z-4 -2 and the point (0,5,0) is

Step-by-step solution

A plane containing a given line and a point not on the line is determined by the direction vector of the line and the vector from a point on the line to the given point. Using the symmetric form x-2 3 = y+1 2 = z-4 -2 , the point P₁ = (2, -1, 4) lies on the line, and the direction vector is d = (3, 2, -2) . The given point is P₂ = (0, 5, 0) , so P₁P₂ = (-2, 6, -4) . The normal vector n to the plane is the cross product d P₁P₂ . n_x = (2)(-4) - (-2)(6) = 4 , n_y = -[(3)(-4) - (-2)(-2)] = -[-12 - 4] = 16 , n_z = (3)(