Find Unit Vector perpendicular to the Surface


by dustydude
Tags: perpendicular, surface, unit, vector
dustydude
dustydude is offline
#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
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#2
Dec4-10, 01:59 PM
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P: 25,170
Is (3x^2+z,0,x) at P=(1,2,-1) really (3,0,1)?
dustydude
dustydude is offline
#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
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#4
Dec4-10, 02:44 PM
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P: 25,170

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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