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Find Unit Vector perpendicular to the Surface

by dustydude
Tags: perpendicular, surface, unit, vector
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dustydude
#1
Dec4-10, 01:17 PM
P: 19
Find Unit Vector perpendicular to the Surface,

x3+zx=1 at the point P=(1,2,-1)

I figures that the perpendicular vector would be,

N(X)=grad(x3+zx)
= (3x2+z, 0, x)
N(P)= (3,0,1)
Then the unit vector would be,

n=N(P)/||N(P)||

n=(3/51/2,0,1/51/2)
The answer i get is not the right answer and i dont see where im going wrong.

Thanks,
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Dick
#2
Dec4-10, 01:59 PM
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P: 25,250
Is (3x^2+z,0,x) at P=(1,2,-1) really (3,0,1)?
dustydude
#3
Dec4-10, 02:28 PM
P: 19
Thanks for pointing that out Dick, minor error.

(2,0,1)

Its still not the right answer which is 1/271/2(5,1,1)

Dick
#4
Dec4-10, 02:44 PM
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P: 25,250
Find Unit Vector perpendicular to the Surface

Quote Quote by dustydude View Post
Thanks for pointing that out Dick, minor error.

(2,0,1)

Its still not the right answer which is 1/271/2(5,1,1)
Then there's probably a typo in the problem. The surface x^3+zx=1 equation doesn't have a 'y' in it. That means the y direction is tangent to the surface. The normal vector can't possibly have a nonzero y component.


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