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Unit Vector

  1. Feb 13, 2014 #1
    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]
  2. jcsd
  3. Feb 13, 2014 #2
    Check your definition of "unit vector." :wink:
  4. Feb 13, 2014 #3
    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
  5. Feb 13, 2014 #4


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    Derivation is correct.
  6. Feb 13, 2014 #5
    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]
  7. Feb 13, 2014 #6
    You're looking for "the" unit normal vector. Normal to what?
  8. Feb 14, 2014 #7
    Normal to the surface [itex]2x+3y+6z=12[/itex]
  9. Feb 14, 2014 #8
    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.
  10. Feb 14, 2014 #9
    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]

  11. Feb 14, 2014 #10
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