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Unit Vector

by JasonHathaway
Tags: unit, vector
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JasonHathaway
#1
Feb13-14, 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]
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olivermsun
#2
Feb13-14, 02:54 PM
P: 786
Check your definition of "unit vector."
JasonHathaway
#3
Feb13-14, 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

mathman
#4
Feb13-14, 04:09 PM
Sci Advisor
P: 6,059
Unit Vector

Quote Quote by JasonHathaway View Post
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
Derivation is correct.
JasonHathaway
#5
Feb13-14, 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]
olivermsun
#6
Feb13-14, 06:01 PM
P: 786
You're looking for "the" unit normal vector. Normal to what?
JasonHathaway
#7
Feb14-14, 03:35 AM
P: 77
Normal to the surface [itex]2x+3y+6z=12[/itex]
olivermsun
#8
Feb14-14, 06:40 AM
P: 786
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 x2 + y2 = 1) then you need to state the original problem.
JasonHathaway
#9
Feb14-14, 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?
olivermsun
#10
Feb14-14, 08:06 AM
P: 786


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