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Find a unit vector for a particular vector

  1. Jan 13, 2016 #1
    1. The problem statement, all variables and given/known data

    Find two unit vectors in 2-space that make an angle of 45(deg) with 7i + 6j.

    2. Relevant equations

    Unit Vector = u = (1/||v||)*v

    Dot Product = u . v = ||u|| ||v|| cos(theta)

    3. The attempt at a solution

    I've been thinking about a way to do this. I originally thought that unit vectors could only be in the direction of the vector specified, because direction was the only thing that mattered.

    I was thinking, if I found the unit vector of 7i + 6j, and then I multiplied that by cos(45), and sin(45), it would give me two solutions of unit vectors 45 degrees from the direction of 7i + 6j.

    I'm not sure if this thinking is correct.
     
  2. jcsd
  3. Jan 13, 2016 #2

    LCKurtz

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    Hint: First find two vectors perpendicular to 7i+6j. What direction do you get if you add two equal length perpendicular vectors?
     
  4. Jan 14, 2016 #3
    Ahhh, okay I was thinking you would find the unit vectors first for some reason.

    So, we are just finding 2 vectors directed at 45 degrees from the original vector, and then finding the unit vectors along those directions.

    Thank you.
     
  5. Jan 15, 2016 #4
    Sorry to bring this question back up, but I just wanted to confirm my answer, as my math question online program I think might be reading my answer wrong.

    ##\vec{v} = 7 \hat{i} + 6\hat{j}##

    Let a vector 45 degrees from ##\vec{v}## be equal to ## \vec{w} = a\hat{i} + b\hat{j}##.

    ##cos(\theta) = \frac {\vec{v} \bullet \vec{w}} {||v|| ||w||}##

    ## \frac {\sqrt{2}} {2} = \frac {\langle 7, 6 \rangle \bullet \langle a, b \rangle} {\sqrt{85} \sqrt{a^{2} + b^{2}}} ##

    ## \frac {\sqrt{2}} {2} = \frac {7a + 6b} {\sqrt{85} \sqrt{a^{2} + b^{2}}} ##

    ## \sqrt{2} \sqrt{85} \sqrt{a^{2} + b^{2}} = 2(7a + 6b) ##

    ## (\sqrt{2} \sqrt{85} \sqrt{a^{2} + b^{2}})^{2} = (2(7a + 6b))^{2} ##

    ## (170)(a^{2} + b^{2}) = 4(7a + 6b)^{2} ##

    Let a = 1.

    ## (170)(1 + b^{2}) = 4(7 + 6b)^{2} ##
    ## 170 + 170b^{2} = 4(49 + 84b + 36b^{2})##
    ## 170 + 170b^{2} = 196 + 336b + 144b^{2}##
    ## 26b^{2} - 336b - 26 = 0 ##

    Quadratic equation gives:

    ## b = \frac {84 ± 85} {13} ##
    ## b = 13, b = - \frac {1} {13} ##

    Therefore, the two vectors that are 45 degrees from v, are:

    ## w_1 = \langle 1, 13 \rangle ##
    ## w_2 = \langle 1, -\frac {1} {13} \rangle ##

    The unit vectors are then:

    ## u_1 = \frac {\langle 1, 13 \rangle} {\sqrt{170}} ##
    ## u_2 = \frac {\langle 1, -\frac {1} {13} \rangle} {\sqrt{1 + \frac {1} {169}}} ##
     
  6. Jan 15, 2016 #5

    LCKurtz

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    Your answers are correct but I would expect your online software would expect you to simplify them, especially ##u_2##. You have certainly gone the long way around Robin Hood's barn to get it. Given the vector ##\langle 7,6\rangle##, isn't it obvious using the dot product that ##\langle -6,7\rangle## and its negative are perpendicular to it? Then you could have used my original hint. Still, you do have a solution and your answers are correct.
     
  7. Jan 16, 2016 #6

    ehild

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    @LCKurtz ingenious suggestion was to find a vector that is of equal length and perpendicular to the original one, and then add the two vectors. The resultant vector halves the angle between ##\vec a## and ##\vec b##. See figure.
    You get the blue vector if rotating ##\vec a = 7\vec i + 6 \vec j ## by 90°. ##\vec b = -6\vec i + 7 \vec j##. Adding them: ##\vec a + \vec b = 1\vec i + 13 \vec j##. Now you have to divide it by the norm. Do the same with the other perpendicular vector, negative of ##\vec b ##.

    grade45.jpg
     
  8. Jan 16, 2016 #7
    Ohhhh, thank you for that picture. I understand now. I knew that the 45 degree vector would be half the 90 degree, but I wasn't too sure on how to get it there. Thank you for the picture.
     
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