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  1. #1
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    vectors

    how to get, two vectors of length 26 units and slope 5/12?
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  2. #2
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    Quote Originally Posted by rabih2011 View Post
    how to get, two vectors of length 26 units and slope 5/12?
    I assume that your vectors are in \mathbb{R}^2.

    1. Then \vec {v_1} = (26,0)

    2. The 2nd vector could be \vec{v_2}=k\cdot (12,5)

    3. Determine k such that |\vec {v_2}| = 26

    Spoiler:
    You should come out with k = 2
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  3. #3
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    Quote Originally Posted by rabih2011 View Post
    how to get, two vectors of length 26 units and slope 5/12?
    Call the vector <x, y>. Saying it has length 26 means that \sqrt{x^2+ y^2}= 26. Saying that it has slope 5/12 means that x/y= 12. That gives you two equations to solve for x and y. From x/y= 12, x= 12y so the first equation becomes \sqrt{144y^2+ y^2}= \sqrt{145}y= 26. You may assume that x and y are positive: if x and y are both positive or both negative, the sign disappears in both formulas. That is, if you find <x, y> which satisfies those equations, so does <-x, -y>. (I don't think it is necessary to "assume" your vectors are in \mathbb{R}^2- if they were not they would not have "slope".)
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  4. #4
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    Hello, rabih2011!

    Find two vectors of length 26 units and slope \frac{5}{12}

    Let the vector be \langle x,y\rangle

    Its length is 26: . \sqrt{x^2+y^2} \:=\:26 \quad\Rightarrow\quad x^2 + y^2 \:=\:676 .[1]

    Its slope is \frac{5}{12}:\;\;\frac{y}{x} \:=\:\frac{5}{12} \quad\Rightarrow\quad y \:=\:\frac{5}{12}x .[2]

    Substitute [2] into [1]: . x^2 + \left(\frac{5}{12}x\right)^2 \:=\:676 \quad\Rightarrow\quad x^2 + \frac{25}{144}x^2 \:=\:676

    . . \frac{169}{144}x^2 \:=\:676 \quad\Rightarrow\quad x^2 \:=\:576 \quad\Rightarrow\quad x \:=\:\pm24

    Substitute into [1]: . y \:=\:\frac{5}{12}(\pm24) \:=\:\pm10


    The two vectors are: . \begin{array}{c}\langle 24,\;10\rangle \\ \langle \text{-}24,\text{-}10\rangle \end{array}
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  5. #5
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    Arghh! I forgot to square the 5/12! Thanks, Soroban.
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