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modphysnoob

  • 2 years ago

please look at my drawing @phi @KingGeorge

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  1. modphysnoob
    • 2 years ago
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    |dw:1360298829035:dw| find this vector

  2. itsmylife
    • 2 years ago
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    your x axis coordinate is incorrect

  3. itsmylife
    • 2 years ago
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    it should be (4,0)

  4. modphysnoob
    • 2 years ago
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    yes , thanks

  5. rebeccaskell94
    • 2 years ago
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    That is a beautiful triangle you have there.

  6. itsmylife
    • 2 years ago
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    then use mid point formula your gonna get, (2,1) as coordinates of bisector n then your vector is , V= 2i +j

  7. KingGeorge
    • 2 years ago
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    modphys doesn't want the mid point. He want the vector that's perpendicular.

  8. itsmylife
    • 2 years ago
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    well thats perpendicular bisector that means dividing equally that means mid point

  9. KingGeorge
    • 2 years ago
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    What I would do, is write the line in the form \(y=-x/2+2\). The slope of the intersecting line will be \[-\frac{1}{-\frac{1}{2}}=2.\] So you have a line that looks like \(y=2x\) as the line that intersects. Set them equal to each other to get \(2x=-x/2+2\implies 5x/2=2\implies x=1\).

  10. KingGeorge
    • 2 years ago
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    Plug that in again, \(x=1\implies y=2(1)=2\), so the point of intersection is \((1,2)\).

  11. KingGeorge
    • 2 years ago
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    Which in turn means the vector is \[\vec{i}+2\vec{j}\]

  12. modphysnoob
    • 2 years ago
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    I got same answer but here what my book did

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  13. modphysnoob
    • 2 years ago
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  14. KingGeorge
    • 2 years ago
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    If \(x=.8\), then the vector is \[.8\vec{i}+1.4\vec{j}\] Taking the dot product of that with the vector \[4\vec{i}-2\vec{j},\] we get \[.8\cdot4-2\cdot1.4=0.4\neq0\]so that solution is wrong.

  15. RadEn
    • 2 years ago
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    |dw:1360299909016:dw| just check the magnitude of vector A : (4*2)/sqrt(4^2+2^2) = 8/sqrt(20) = 8/2sqrt(5) = 4/5 * sqrt(5) 0.8 and 1.6 has the same magnitude

  16. phi
    • 2 years ago
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    Just as in your book, we can say the direction vector B (from pt (0,2) to (4,0) is (4,-2) let vector A be perpendicular to B. A dot B = 0 (x,y) dot (4,-2) =0 we get 4x = 2y pick a nice number for x, like 1. then y =2 vector A = (1,2) is perpendicular to vector B make A unit length, by dividing by its length. |A|= sqrt(5) Au = (1,2)/sqrt(5) now to get the length to scale Au so it reaches the line between (0,2) and (4,0), we notice that the vector (4,0) projected onto Au has the correct length Au dot (4,0) = 4/sqrt(5) The vector we want is 4/sqrt(5) * (1,2)/sqrt(5) = 4/5 * (1,2) or (0.8, 1.6)

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