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Anton [14]
3 years ago
15

A certain plane is described by 2x + 3y + 4z = 16. Find the unit vector normal to the surface in the direction away from the ori

gin.
Mathematics
1 answer:
Fynjy0 [20]3 years ago
8 0

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)

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