Unit Vector

by JasonHathaway
Tags: unit, vector
 P: 32 Hi everyone, Just wanna know how does the the unit vector become in that form: $\vec{n}=\frac{2x\vec{i}+2y\vec{j}}{\sqrt{(2x)^{2}+(2y)^{2}}}=\frac{x \vec{i}+y \vec{j}}{4}$
 P: 483 Check your definition of "unit vector."
 P: 32 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. $\vec{n}=\frac{2x\vec{i}+2y\vec{j}}{\sqrt{(2x)^{2}+(2y)^{2}}}=\frac{2(x \vec{i}+y\vec{j})}{\sqrt{4(x^{2}+y^{2}})}=\frac{2(x \vec{i}+y\vec{j})}{2\sqrt{(x^{2}+y^{2}})}=\frac{x \vec{i}+y\vec{j}}{\sqrt{x^{2}+y^{2}}}$ That's my best. :Z
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P: 5,891

Unit Vector

 Quote by JasonHathaway 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. $\vec{n}=\frac{2x\vec{i}+2y\vec{j}}{\sqrt{(2x)^{2}+(2y)^{2}}}=\frac{2(x \vec{i}+y\vec{j})}{\sqrt{4(x^{2}+y^{2}})}=\frac{2(x \vec{i}+y\vec{j})}{2\sqrt{(x^{2}+y^{2}})}=\frac{x \vec{i}+y\vec{j}}{\sqrt{x^{2}+y^{2}}}$ That's my best. :Z
Derivation is correct.
 P: 32 But how did it end up like this form: $\frac{x \vec{i}+y \vec{j}}{4}$ And I've found something similar in Thomas Calculus: Is $y^{2} + z^{2}$ equal to 1 or something? much like $sin^{2}\theta + cos^{2}\theta = 1$
 P: 483 You're looking for "the" unit normal vector. Normal to what?
 P: 32 Normal to the surface $2x+3y+6z=12$
 P: 483 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.
 P: 32 Sorry, that's not the correct surface, but the surface is $x^{2}+y^{2}=16$. But I think I've got the idea: $\vec{n}=\frac{x\vec{i}+y \vec{j}}{\sqrt{x^{2}+y^{2}}}=\frac{x\vec{i}+y\vec{j}}{\sqrt{16}}=\frac{ x\vec{i}+y\vec{j}}{4}$ right?
 P: 483

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