# Unit Outward normal

• Dec 29th 2009, 11:26 AM
dankelly07
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.
• Dec 29th 2009, 02:54 PM
Abu-Khalil
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.
• Dec 30th 2009, 08:11 AM
HallsofIvy
Quote:

Originally Posted by dankelly07
** 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. $\nabla z= k= (0, 0, 1)$
S4 is the plane 6x+ 3y+ 2z= 6 so $\nabla 6x+ 3y+ 2z= 6i+ 3j+ 2k= (6, 3, 2)$. Its length is $\sqrt{36+ 9+ 4}= 7$.