The vector equation of the plane which is at a distance of 5 units from the origin and normal to the vector 2 i + j - 2 k is

Step-by-step solution

The normal vector to the plane is n = 2 i + j - 2 k . The magnitude of the normal vector is | n | = 2^2 + 1^2 + (-2)^2 = 9 = 3 . The unit normal vector is n = n | n | = 2 i + j - 2 k 3 . The vector equation of a plane at a perpendicular distance d from the origin is given by r n = d . Substituting d = 5 and n = 2 i + j - 2 k 3 , we get: r ( 2 i + j - 2 k 3 ) = 5 r (2 i + j - 2 k ) = 15 Answer: r (2 i + j - 2 k ) = 15