Answer:
the unit vector is u=(2/√29 ,3/√29,4/√29)
Step-by-step explanation:
From the plane equation
2x+3y+4z = 16
2x+3y+4*(z-4) =0
(2,3,4)*(x-0,y-0,z-4) =0
then the vector n=(2,3,4) is normal to the plane that contains the point (0,0,4)
the modulus of n will be
|n|= √(2²+3²+4²) = √29
then the unit vector will be
u=n/|n| = (2,3,4)/√29= (2/√29 ,3/√29,4/√29)
since there are 2 possible choices
(2/√29 ,3/√29,4/√29) or (-2/√29 ,-3/√29,-4/√29)
then the vector that points away from the origin is
(2/√29 ,3/√29,4/√29)
