Register to reply

Unit Vector

by JasonHathaway
Tags: unit, vector
Share this thread:
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]
Phys.Org News Partner Mathematics news on Phys.org
'Moral victories' might spare you from losing again
Fair cake cutting gets its own algorithm
Effort to model Facebook yields key to famous math problem (and a prize)
olivermsun
#2
Feb13-14, 02:54 PM
P: 668
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: 6,040
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: 668
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: 668
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: 668


Register to reply

Related Discussions
Dot product of a vector with the derivative of its unit vector Calculus & Beyond Homework 7
How do you find the coordinate of a vector with the unit vector? Linear & Abstract Algebra 1
Ev(Unit Vector) and projection of a vector in a dot product Calculus 1
How do I calculate the binormal vector without using the principal unit normal vector Calculus & Beyond Homework 1
Symbolic represenation of the unit vector and the vector General Math 11