The distance of the point P(3,8,2) from the line x-1 2 = y-3 4 = z-2 3 measured parallel to the plane 3 x+2 y-2 z+15=0 is

Step-by-step solution

Let P(3,8,2) be the given point and L: x-1 2 = y-3 4 = z-2 3 the line. The parametric form of L gives Q(1 + 2t, 3 + 4t, 2 + 3t) for some t R . The plane : 3x + 2y - 2z + 15 = 0 has normal vector n = (3, 2, -2) . The distance from P to L measured parallel to is the length of the segment PQ where PQ is parallel to , meaning PQ n = 0 . The vector PQ = Q - P = (2t - 2, 4t - 5, 3t) . Setting the dot product to zero: (2t - 2)(3) + (4t - 5)(2) + (3t)(-2) = 0 Simplifying: 6t - 6 + 8t - 10 - 6t = 0 8t - 16 = 0 t = 2 Substit