Find the perpendicular distance from the point P(9,11,−8) ft to a plane defined by three points A(1,9,−4) ft, B(−4,−8,6) ft, and C(−1,−2,2) ft

Distance = ______ ft

Respuesta :

Answer:

  0 ft

Explanation:

The equation of the plane can be found from the cross product AC×BC. That vector is ...

  N = (2, 11, -6) × (-3, -6, 4) = (8, 10, 21)

Then the equation of the plane is ...

  8x +10y +21z = 14 . . . . . 14 = N·A

Point P satisfies this equation, so is on the plane. The distance is 0 feet.

  8(9) +10(11) -8(21) = 72 +110 -168 = 14