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Math Help - another vector question

  1. #1
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    another vector question

    Find a unit vector that is perpendicular to the vector 6i+8j
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  2. #2
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    Quote Originally Posted by hooke View Post
    Find a unit vector that is perpendicular to the vector 6i+8j
    If this vector belongs to 2-D space then you only have to reverse the order of the components and make one component negantive:

    \vec v = (6,8)~\implies~ \vec n = (-8,6)~\vee~ \vec n = (8, -6)

    The unit vector of \vec n is calculated by: \vec n^0 = \dfrac{\vec n}{|\vec n|}

    In 3-D space the normal unit vector is simply \vec n = (0,0,1)
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  3. #3
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    Hello, hooke!

    Find a unit vector that is perpendicular to the vector 6i+8j
    Two vectors are perpendicular if their dot product is zero.


    Let the vector be: . \vec v \:=\:ai+bj

    Then we have: . (6i + 8j)\cdot(ai + bj) \:=\:0 \quad\Rightarrow\quad 6a + 8b \:=\:0 \quad\Rightarrow\quad b \:=\:-\frac{3a}{4}

    Let a = 4\quad\hdots then: . b = -3

    . . Hence, a perpendicular vector is: . \vec v \:=\:4i - 3j


    For unit vector, divide by its magnitude: . |\vec v| \:=\:\sqrt{4^2 + (\text{-}3)^2} \:=\:5


    Therefore: . \vec u \:=\:\frac{\vec v}{|\vec v|} \:=\:\frac{4i-3j}{5} \:=\:\tfrac{4}{5}i - \tfrac{3}{5}j

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