Unit Vector

1. Feb 13, 2014

JasonHathaway

Hi everyone,

Just wanna know how does the the unit vector become in that form:

$\vec{n}=\frac{2x\vec{i}+2y\vec{j}}{\sqrt{(2x)^{2}+(2y)^{2}}}=\frac{x \vec{i}+y \vec{j}}{4}$

2. Feb 13, 2014

olivermsun

Check your definition of "unit vector."

3. Feb 13, 2014

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

4. Feb 13, 2014

mathman

Derivation is correct.

5. Feb 13, 2014

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 $sin^{2}\theta + cos^{2}\theta = 1$

6. Feb 13, 2014

olivermsun

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

7. Feb 14, 2014

JasonHathaway

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

8. Feb 14, 2014

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.

9. Feb 14, 2014

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?

10. Feb 14, 2014

:thumbs: