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Thread: vectors orthogonal etc.

  1. #1
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    vectors orthogonal etc.

    Which of the vectors u1=(1,2), u2=(0,1), u3=(-2,-4), u4=(-2,1), u5=(2,4), u6=(-6,3) are
    a. orthogonal?
    b. in the same direction?
    c. in opposite directions?

    my answers
    a. u1 & u4, u1 & u6, u3 & u4, u3 & u6, u4 & u5, u5 & u6 because their dot products are 0
    b. u1, u2 & u5 AND u4 & u6
    c. u3 & u5

    does anyone know if I did this right? did I miss anything?
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  2. #2
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    Hello, antman!

    Given the vectors: .$\displaystyle \begin{array}{ccc}u_1&=&\langle1,2\rangle \\ u_2&=&\langle0,1\rangle \\u_3&=&\langle\text{-}2,\text{-}4\rangle \\u_4&=& \langle\text{-}2,1\rangle \\u_5&=&\langle2,4\rangle \\ u_6&=&\langle\text{-}6,3\rangle \end{array}$

    which are:

    a) orthogonal?
    b) in the same direction?
    c) in opposite directions?

    My answers

    $\displaystyle a)\;\;u_1, u_4,\;\;u_1, u_6,\;\;u_3,u_4,\;\;u_3,u_6,\;\;u_4,u_5,\;\;u_5,u_ 6$ . . . . Right!

    $\displaystyle b)\;\;u_1, {\color{red}\rlap{//}}u_2,u_5,\;\;u_4,u_6$

    $\displaystyle c)\;\;u_3, u_5,\;\;{\color{blue}u_1,u_3} $

    Pretty good work . . .

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  3. #3
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    I understand where I went wrong and what I missed from your help, but is there a specific equation to use as a proof? For example, the vectors in the same direction seem to be multiples of one another.
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  4. #4
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    Quote Originally Posted by antman View Post
    For example, the vectors in the same direction seem to be multiples of one another.
    If $\displaystyle u_1$ and $\displaystyle u_2$ are multiples of one another (and thus parallel), then you should be able to find a scalar $\displaystyle c$ such that $\displaystyle u_1=cu_2.$ Can you find such a $\displaystyle c?$

    If you setup a system of equations to find this value, you will see that the system is inconsistent. Therefore, $\displaystyle \langle1,2\rangle$ and $\displaystyle \langle0,1\rangle$ are not parallel. And if they are not parallel, they cannot be in the same direction.
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