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Math Help - finding unit vector orthagonal with vector

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    finding unit vector orthagonal with vector

    Unsure how to do this.
    Find two different unit vectors, in 2dimensions, that make an angle of 90degress with 4i+3j
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Craka View Post
    Unsure how to do this.
    Find two different unit vectors, in 2dimensions, that make an angle of 90degress with 4i+3j
    note that if two vectors are orthogonal, their dot product is zero.

    so let one unit vector be <a,b> (note the other is <-a,-b>)

    then \left< a,b \right> \cdot \left< 4,3 \right> = 0

    then find the unit vector in the direction of <a,b>
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    Sorry but could please show the full solution. Thankyou
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    Quote Originally Posted by Craka View Post
    Sorry but could please show the full solution.
    You don't want to turn someone else's work do you?
    Solve this system.
    \begin{gathered}<br />
  4a + 3b = 0 \hfill \\<br />
  a^2  + b^2  = 1 \hfill \\ <br />
\end{gathered}
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    No i'm not trying to copy someone's work I am just trying to understand how to do the problem.
    The answer is in the form of <3/5, -4/5> and <-3/5 , 4/5>

    I can see where the 1/5 factor came in. It is from dividing by the magnitude of original vector ie the sqrt of [4^2 + 3^2], and yes I understand that if the dot product of two vector is zero than the two vectors must be orthagonal, however I still unsure how 4a+3b=0 and -4a-3b=0 and a^2+b^2=1 helps.

    I would really like to understand what to do here. It is not copying to do some homework. It is for study. Thankyou
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    I can see how the answer works. ie the dot product of <4,3> . <-3/5, 4/5> and dot product <4,3> . <3/5, -4/5 > work both out to be zero.
    What is the method of establishing this.

    To get a unit vector is just a matter of dividing the vector by its length which is the sqrt of the squared sums of the vector components.

    Guess what I'm also trying to ask here is in 2 dimensional vectors, to find a orthogonal vector is it just a matter of swapping say the x and y component of vector around and making one component negative , and how would you do so for 3 dimensional vectors?
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    Quote Originally Posted by Craka View Post
    I can see how the answer works. ie the dot product of <4,3> . <-3/5, 4/5> and dot product <4,3> . <3/5, -4/5 > work both out to be zero.
    What is the method of establishing this.

    To get a unit vector is just a matter of dividing the vector by its length which is the sqrt of the squared sums of the vector components.

    Guess what I'm also trying to ask here is in 2 dimensional vectors, to find a orthogonal vector is it just a matter of swapping say the x and y component of vector around and making one component negative , and how would you do so for 3 dimensional vectors?
    <a,b>.<4,3> = 0

    => 4a + 3b = 0

    pick a = 1. what does b have to be?

    thus, you can find values a and b that work. infinitely many values will work, in fact. you just need 1 of them. once you find the unit vector in the direction <a,b>, take its negative to find the another unit vector
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    How would I go about doing this with a 3 dimensional vector ?
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    Quote Originally Posted by Craka View Post
    How would I go about doing this with a 3 dimensional vector ?
    same thing. pick random values for two of the components, and solve for the third
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    Quote Originally Posted by Craka View Post
    How would I go about doing this with a 3 dimensional vector ?
    One important point: in the plane there exist a single line perpendicular to a given line at a given point. That means that the set of vectors perpendicular to a given vector must all lie on that line. Given, for example, the vector <a, b> it is very easy to see that <a, b>.<-b, a>= 0 and so any vector perpendicular to <a, b> must b a multiple of <-b, a>. In particular, there are only two unit vectors perpendicular to a given vector, pointing in opposite directions.

    In three dimensions, there exist an entire PLANE of line perpendicular to a given line at a given point. There are now an infinite number of unit vectors perpendicular to a given vector.
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