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amistre64

  • 5 years ago

What does the scalar result of dotting 2 vectors signify? I know a 0 means they are perps, but what is the scalar product actually mean?

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  1. amistre64
    • 5 years ago
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    <1,1>.<0,2sqrt(2)> = 0 + 2sqrt(2) = 2sqrt(2). That doesnt make sense to me a whole lot.

  2. amistre64
    • 5 years ago
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    that just seems to give the magnitude of the bottom vector

  3. anonymous
    • 5 years ago
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    <1,1>.<0,2sqrt(2)> = 0 + 2sqrt(2) = 2sqrt(2). how?

  4. nowhereman
    • 5 years ago
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    It gives the cosine of the angle between them. This is the way you can get angles in any vector space with an inner product.

  5. amistre64
    • 5 years ago
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    <1 , 1 > <0,2sqrt(2)> ------------ 0 + 2sqrt(2) = 2sqrt(2)

  6. amistre64
    • 5 years ago
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    yes, it gives the adjacent part of the cosine angle which is just the magnitude of the bottom vector right? the product of the magnitudes of the top and bottom vectors account for the 'hypotenuse' then

  7. anonymous
    • 5 years ago
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    <1 , 1 > <0,2sqrt(2)> ------------ 0 + 2sqrt(2) = 2sqrt(2). iam not geting it amistre

  8. amistre64
    • 5 years ago
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    see; it is how you multiply vectors; or at least one way to do it; the other being the cross product

  9. myininaya
    • 5 years ago
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    cross product can be found two ways if i remember correctly finding the determinant of a matrix or some kind of trig thing right?

  10. amistre64
    • 5 years ago
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    determinate of 2 vectors is usual

  11. amistre64
    • 5 years ago
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    the other way is just to equate the zero dot product of 2 vectors inthe plane and equat them

  12. amistre64
    • 5 years ago
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    v<x,y> n<a,b> ------- vx.a + vy.b = 0 u<x,y> n<a,b> -------- ux.a +uy.b = 0 equate them and solve the system of equations :)

  13. amistre64
    • 5 years ago
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    geterminate is mor emechanical tho

  14. amistre64
    • 5 years ago
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    lol... read thru the typos

  15. amistre64
    • 5 years ago
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    i now the scalar projection projects the top onto the bottom at |top|cos(a)

  16. amistre64
    • 5 years ago
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    i just dont understand if there is another way to use the scalar product other than associating it to the cos(a) between the 2 vectors :)

  17. myininaya
    • 5 years ago
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    example: find u*v if u=2i+3j and v=-2k | i j k | = i|3 0 | -j|2 0| +k|2 3| =i(-2(3)-0)-j(-2-0)+k(0-0)=-6i+2j | 2 3 0 | |0 -2| |0 -2| |0 0| | 0 0 -2|

  18. amistre64
    • 5 years ago
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    correct; just keep in mind to do +-+ :) that middle - is to account for being lazy i think lol

  19. myininaya
    • 5 years ago
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    =-6j+4k

  20. myininaya
    • 5 years ago
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    oops =-6i+4j

  21. myininaya
    • 5 years ago
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    i forgot about that othr two up there

  22. amistre64
    • 5 years ago
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    2a+3b+0c = 0 0a+0b-2c = 0 lol... thats would be hard to determine that way ; cross product is better in that case; and in most cases prolly

  23. myininaya
    • 5 years ago
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    yes that would be hard lol (or impossible) i dont see how we could do it that way

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