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Perpendicular Unit Vectors

  1. Jul 30, 2015 #1
    1. The problem statement, all variables and given/known data
    From Kleppner and Kolenkow Chapter 1 (Just checking to see if I'm right)

    Given vector A=<3, 4, -4>
    a) Find a unit vector B that lies in the x-y plane and is perpendicular to A.
    b) Find a unit vector C that is perpendicular to both A and B.
    c)Show that A is perpendicular to the plane defined by B and C.

    2. Relevant equations
    Cross Product
    Dot Product
    Knowing how to find a plane from two vectors

    3. The attempt at a solution
    a) Using the following system...
    3x+4y=0 (using dot product and z=0)
    x+y=1 (For it to be a unit vector the sum of the two components must be 1)
    The perendicular vector is
    ##<4, -3, 0>##
    and the unit vector after dividing by the magnitude is...
    ##B=<4/5, -3/5, 0>##

    b)Finding the cross product of A and B and dividing by the magnitude...
    ##C=<\frac{-12,}{5\sqrt{41}}, \frac{-16}{5\sqrt{41}}, \frac{-5}{\sqrt{41}}>##

    c)##B \times C=<75, 100, -100>##
    Therefore the plane is defined by...
    ##3x+4y-4z=0##
    Therefore A is perpendicular since its components equal the constants that define the plane.

    I'm pretty sure I did everything right I just wanted to make sure and see if any of my wording is dumb or if I made a mistake. I don't know how to write the last part formally. Thanks all!
     
  2. jcsd
  3. Jul 30, 2015 #2
    You can show things are perpendicular by computing the dot product between two vectors.
     
  4. Jul 31, 2015 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    NO!! For it to be a unit vector we must have [itex]\sqrt{x^2+ y^2}= 1[/itex], not x+ y!

    So x+ y is NOT 1!

     
  5. Jul 31, 2015 #4
    I see, but since the unit vector I found follows that condition...
    ##\sqrt{(\frac{4}{5})^2 + (\frac{-3}{5})^2} = 1##
    is my solution still right?
     
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