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anonymous

  • 5 years ago

Find a vector that the same direction as −2i + 4j+ 2k but that has length 6.

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  1. anonymous
    • 5 years ago
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    desperately waiting for ur reply...so fed up with this problem right now !!! help

  2. anonymous
    • 5 years ago
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    hi guys

  3. anonymous
    • 5 years ago
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    do share ur views on this plss

  4. anonymous
    • 5 years ago
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    i am new here, on this site, looks fun

  5. anonymous
    • 5 years ago
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    surely it is !!! nothing lesser than a boon to the students !!! getting live help for free is just awesome

  6. anonymous
    • 5 years ago
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    feels like v r a community sharing its knowledge

  7. anonymous
    • 5 years ago
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    I am graduating highschool this spring

  8. anonymous
    • 5 years ago
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    congrats bro !!!

  9. anonymous
    • 5 years ago
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    the only bad thing is that me and my boyfriend are going to different universities :(

  10. anonymous
    • 5 years ago
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    which uni r u in ?

  11. anonymous
    • 5 years ago
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    Carnegie Mellon University

  12. anonymous
    • 5 years ago
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    what degree wer u in ?

  13. anonymous
    • 5 years ago
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    Computer Science, I know this sounds strange because I am a girl, but I LOVE computer science and programming

  14. anonymous
    • 5 years ago
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    thats good.....computers r not just monopoly of boyzzz

  15. anonymous
    • 5 years ago
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    :)

  16. anonymous
    • 5 years ago
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    may i ask ur name ?

  17. anonymous
    • 5 years ago
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    ankur, you still need an answer?

  18. anonymous
    • 5 years ago
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    You just need to find the unit vector of u and multiply it by 6

  19. anonymous
    • 5 years ago
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    ah got it !!! cheers !!!

  20. anonymous
    • 5 years ago
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    Cool

  21. nikvist
    • 5 years ago
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    \[|-2i+4j+2k|=\sqrt{(-2)^2+4^2+2^2}=2\sqrt{6}\] \[\vec{a}=\alpha(-2i+4j+2k)\quad,\quad\alpha=\frac{6}{2\sqrt{6}}=\frac{\sqrt{6}}{2}\] \[vec{a}=\frac{\sqrt{6}}{2}(-2i+4j+2k)=-\sqrt{6}i+2\sqrt{6}j+\sqrt{6}k\]

  22. anonymous
    • 5 years ago
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    hey one more short question

  23. anonymous
    • 5 years ago
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    ?

  24. anonymous
    • 5 years ago
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    unit vector that is perpendicular to the xz-plane

  25. anonymous
    • 5 years ago
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    (1,0,0) lies in x, (0,0,1) lies in z, so (0,1,0) will lie in y which is perpendicular to xz.

  26. anonymous
    • 5 years ago
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    @nikwist : could you please tell how do u type square roots like this on screen?

  27. anonymous
    • 5 years ago
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    and it's a unit vector.

  28. anonymous
    • 5 years ago
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    u r so intelligent !!! no wonder you have so many fans !!! thanx a lot again

  29. anonymous
    • 5 years ago
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    np...fan me if you haven't :)

  30. anonymous
    • 5 years ago
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    i have...do u have any idea how nikwist wrote his answers like that above?

  31. anonymous
    • 5 years ago
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    Yes, go into equation editor below, and there's a set of boxes to the right you can use. You can also type, sqrt{x} directly into the equation line to get\[\sqrt{x}\]

  32. anonymous
    • 5 years ago
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    cheers !!! u r really a lifesaver

  33. anonymous
    • 5 years ago
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    np :D

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