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anonymous

  • 4 years ago

the symbol\[A _{b}\] stands for the projection of vector A onto vector B. In other words, \[A _{b}\] represents the component of A that is parallel to B. Derive an expression for \[A _{b}\] in terms of the vectors A and B.

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  1. amistre64
    • 4 years ago
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    \[\frac{a.b}{|a|^2}a\]

  2. anonymous
    • 4 years ago
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    wow u did that really quickly

  3. amistre64
    • 4 years ago
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    just got out of calc3 that just went over it :)

  4. anonymous
    • 4 years ago
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    in the numerator...is that a times b?

  5. amistre64
    • 4 years ago
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    a dot b

  6. anonymous
    • 4 years ago
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    a dot b divided by magnitude of a squared all multiplied by a?

  7. amistre64
    • 4 years ago
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    yes, the left side is a scalar; and the right side is vector a scaled to that length

  8. anonymous
    • 4 years ago
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    would it be a bother to run me through how u got to this expression?

  9. amistre64
    • 4 years ago
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    |dw:1327527260794:dw|

  10. amistre64
    • 4 years ago
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    |b| cos(t) = the length of Ab, if we use your notation for proj{a} b

  11. amistre64
    • 4 years ago
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    \[|b|cos(t) =|b| \frac{a.b}{|a||b|}=\frac{a.b}{|a|}\] good so far?

  12. anonymous
    • 4 years ago
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    yea i get that

  13. anonymous
    • 4 years ago
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    i'm with u

  14. anonymous
    • 4 years ago
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    i'm also trying to decode your picture :)

  15. anonymous
    • 4 years ago
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    is that the x coordinate down there at the end?

  16. anonymous
    • 4 years ago
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    the long vector

  17. amistre64
    • 4 years ago
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    now, we need to scale that to a unit vector of a; since a is |a| long, lets divide it by |a| to get it to a length of 1

  18. amistre64
    • 4 years ago
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    \[\frac{a.b}{|a|}\ \frac{a}{|a|}=; unit\ a, scaled\ by\ needed\ length\]

  19. anonymous
    • 4 years ago
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    okay

  20. amistre64
    • 4 years ago
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    and since |a| |a| = |a|^2 i just condensed it alittle bit

  21. anonymous
    • 4 years ago
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    simple enough?

  22. amistre64
    • 4 years ago
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    |dw:1327527605851:dw|

  23. anonymous
    • 4 years ago
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    can u redraw that picture from above?

  24. anonymous
    • 4 years ago
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    oh okay now i see

  25. amistre64
    • 4 years ago
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    lol, i cant draw on this thing very well period :)

  26. anonymous
    • 4 years ago
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    okay....so that's it for the equation now?

  27. amistre64
    • 4 years ago
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    to find the the vector that is the proj of b onto a; yes; reduce a to ints unit equivalent; and scale it up by the magnitude of b*cos(t)

  28. anonymous
    • 4 years ago
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    and what u first initally gave me is that correct?

  29. amistre64
    • 4 years ago
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    yes, seeing how that is what we ended up with in the end i would say that its correct :)

  30. anonymous
    • 4 years ago
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    u are an absolute life saver

  31. amistre64
    • 4 years ago
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    thnx, good luck :)

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