The perpendicular distance from the origin to the plane containing the points (1, -2, 1), (2, -1, -3) and (0, 1, 5) is...(in units)

Step-by-step solution

Let the given points be A(1, -2, 1) , B(2, -1, -3) and C(0, 1, 5) . The vectors in the plane are: AB = (2 - 1) i + (-1 - (-2)) j + (-3 - 1) k = i + j - 4 k AC = (0 - 1) i + (1 - (-2)) j + (5 - 1) k = - i + 3 j + 4 k The normal vector to the plane is given by the cross product n = AB AC : n = vmatrix i & j & k 1 & 1 & -4 -1 & 3 & 4 vmatrix = i (4 - (-12)) - j (4 - 4) + k (3 - (-1)) = 16 i + 4 k A simpler normal vector is n ' = 4 i + k . The equation of the plane passing through A(1, -2, 1) with normal vector 4 i + k