How Do Unit Vectors Achieve Their Form?

[JasonHathaway]

Hi everyone,


Just want to 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}

 

[olivermsun]

Check your definition of "unit vector."

 

[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

 

[mathman]

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.

 

[JasonHathaway]

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

 

[olivermsun]

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

 

[JasonHathaway]

Normal to the surface 2x+3y+6z=12

 

[olivermsun]

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]

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?

 

[olivermsun]

:thumbs: