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Math Help - Unit Outward normal

  1. #1
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    Unit Outward normal

    ** See attachment**

    I'm having a hard time trying to figure out how to get outward unit normals for each surface of this shape. Here I have the question and the answer but still don't seem to be able to solve it..

    I understand that to get a unit normal I divide by the length of n but I cant see how the answer is

    S1: x = 0 and n (hat) = (-1,0,0)
    S2: y = 0 and n (hat) = (0,-1,0)
    S3: z = 0 and n (hat) = (0,0,-1)

    is it..

    for S1: nabla ( 3y + 2z -6 )

    so is the unit normal

    1/(sqrt(11))*(0,3,2) ????

    If so then the answers I have are wrong!

    Please can someone help. I have included the problem in the attachment.
    Attached Thumbnails Attached Thumbnails Unit Outward normal-unitv.jpg  
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  2. #2
    Member Abu-Khalil's Avatar
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    You have 4 planes in \mathbb{R}^3:

    1. x=0
    2. y=0
    3. z=0
    4. 6x+3y+2z=6

    So you must have 4 normals, one for each:

    1. \hat n_1=(-1,0,0)
    2. \hat n_2=(0,-1,0)
    3. \hat n_3=(0,0,-1)
    4. \hat n_4=\frac{1}{\sqrt{36+9+4}}(6,3,2)=\frac{1}{7}(6,3  ,2)

    And remember that they must be exteriors to the enclosed region.
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  3. #3
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    Quote Originally Posted by dankelly07 View Post
    ** See attachment**

    I'm having a hard time trying to figure out how to get outward unit normals for each surface of this shape. Here I have the question and the answer but still don't seem to be able to solve it..

    I understand that to get a unit normal I divide by the length of n but I cant see how the answer is

    S1: x = 0 and n (hat) = (-1,0,0)
    S2: y = 0 and n (hat) = (0,-1,0)
    S3: z = 0 and n (hat) = (0,0,-1)

    is it..

    for S1: nabla ( 3y + 2z -6 )
    No, what did you do, solve for -6x?

    S1 is the plane x= 0. \nabla x= i= (1, 0, 0)
    S2 is the plane y= 0. \nabla y= j= (0, 1, 0)
    S3 is the plane z= 0. \nabla z= k= (0, 0, 1)
    those already have unit length.

    S4 is the plane 6x+ 3y+ 2z= 6 so \nabla 6x+ 3y+ 2z= 6i+ 3j+ 2k= (6, 3, 2). Its length is \sqrt{36+ 9+ 4}= 7.

    so is the unit normal

    1/(sqrt(11))*(0,3,2) ????

    If so then the answers I have are wrong!

    Please can someone help. I have included the problem in the attachment.
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