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Unit Vector 
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#1
Feb1314, 01:36 PM

P: 77

Hi everyone,
Just wanna know how does the the unit vector become in that form: [itex]\vec{n}=\frac{2x\vec{i}+2y\vec{j}}{\sqrt{(2x)^{2}+(2y)^{2}}}=\frac{x \vec{i}+y \vec{j}}{4}[/itex] 


#2
Feb1314, 02:54 PM

P: 601

Check your definition of "unit vector."



#3
Feb1314, 04:05 PM

P: 77

As far as I know, the unit vector or the normal vector is the vector divided by its magnitude.
But that's not what I need to know, what I need to know is the manipulation that occurred. [itex]\vec{n}=\frac{2x\vec{i}+2y\vec{j}}{\sqrt{(2x)^{2}+(2y)^{2}}}=\frac{2(x \vec{i}+y\vec{j})}{\sqrt{4(x^{2}+y^{2}})}=\frac{2(x \vec{i}+y\vec{j})}{2\sqrt{(x^{2}+y^{2}})}=\frac{x \vec{i}+y\vec{j}}{\sqrt{x^{2}+y^{2}}}[/itex] That's my best. :Z 


#4
Feb1314, 04:09 PM

Sci Advisor
P: 6,031

Unit Vector



#5
Feb1314, 04:44 PM

P: 77

But how did it end up like this form: [itex]\frac{x \vec{i}+y \vec{j}}{4}[/itex]
And I've found something similar in Thomas Calculus: Is [itex]y^{2} + z^{2}[/itex] equal to 1 or something? much like [itex]sin^{2}\theta + cos^{2}\theta = 1 [/itex] 


#6
Feb1314, 06:01 PM

P: 601

You're looking for "the" unit normal vector. Normal to what?



#7
Feb1414, 03:35 AM

P: 77

Normal to the surface [itex]2x+3y+6z=12[/itex]



#8
Feb1414, 06:40 AM

P: 601

Okay, but clearly that isn't where the gradient in the original post came from. So if you want to know what happened in post #3 (why x^{2} + y^{2} = 1) then you need to state the original problem.



#9
Feb1414, 07:41 AM

P: 77

Sorry, that's not the correct surface, but the surface is [itex]x^{2}+y^{2}=16[/itex].
But I think I've got the idea: [itex]\vec{n}=\frac{x\vec{i}+y \vec{j}}{\sqrt{x^{2}+y^{2}}}=\frac{x\vec{i}+y\vec{j}}{\sqrt{16}}=\frac{ x\vec{i}+y\vec{j}}{4}[/itex] right? 


#10
Feb1414, 08:06 AM

P: 601




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