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anonymous

  • 4 years ago

consider the vector I =[ 1/Square root (2) , 1/Square root (2)] and J= [ -1/Square root (2), 1/Square root (2)]. Write the vector (2,-6) in terms of I and J.

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  1. phi
    • 4 years ago
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    james gave you a good starting point. I= [1 1]/sqrt(2) J= [-1 1]/sqrt(2) you want to find two scalars (two simple numbers) that you multiply I and J by to get (2,-6) call the numbers a and b then you must solve (2,-6)= a*I + b*J

  2. phi
    • 4 years ago
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    First rule: multiplying a vector by a scalar uses this rule: a (x,y)= (ax, ay)

  3. phi
    • 4 years ago
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    so, start with (2,-6)= a*I + b*J (2, -6) = a*(1,1)/sqrt(2) + b*(-1,1)/sqrt(2) multiply both sides by sqrt(2) . This means multiply every term by sqrt(2) you get sqrt(2) * (2, -6) = a*(1,1) + b*(-1,1)

  4. anonymous
    • 4 years ago
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    hmm this one doesn't make sense to me at all

  5. anonymous
    • 4 years ago
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    ohh ok

  6. phi
    • 4 years ago
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    Now use the rule First rule: multiplying a vector by a scalar uses this rule: a (x,y)= (ax, ay) on all the terms

  7. anonymous
    • 4 years ago
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    so its 2 sqrt(2) , -6 sqrt(2) = a (1,1) + b (-1,1)

  8. phi
    • 4 years ago
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    yes, but apply the rule to the two terms on the right hand side also

  9. anonymous
    • 4 years ago
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    how do you do that?

  10. phi
    • 4 years ago
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    a (x,y)= (ax, ay)

  11. anonymous
    • 4 years ago
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    so its 2sqrt(2) , -6sqrt(2) = (a1, a1) + (-b, b1)

  12. phi
    • 4 years ago
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    well, I would not write a1, because a*1 is just a.... also, put the left side in parens because we do not want to forget is is a vector

  13. anonymous
    • 4 years ago
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    ok. its [2sqrt(2), -6sqrt(2)] = (a,a) + (-b,b) ?

  14. phi
    • 4 years ago
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    yes, now add the two vectors on the right hand side

  15. anonymous
    • 4 years ago
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    how do you do that?

  16. phi
    • 4 years ago
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    You were able to subtract two vectors, for example P-Q in the last problem. Adding is the same, except you add instead of subtract.

  17. anonymous
    • 4 years ago
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    so its like (1+-1, 1+1) = (0,2) ?

  18. phi
    • 4 years ago
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    yes, but now we are adding variables instead of numbers

  19. anonymous
    • 4 years ago
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    what do you mean?

  20. phi
    • 4 years ago
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    (a,a) + (-b,b) = ?? add corresponding entries

  21. anonymous
    • 4 years ago
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    (-ab, ab)

  22. phi
    • 4 years ago
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    add (not multiply)

  23. anonymous
    • 4 years ago
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    (a+-b, a+b)

  24. phi
    • 4 years ago
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    yes, but normally you would write a + -b as just a-b .

  25. anonymous
    • 4 years ago
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    oh ok. now what do i do?

  26. phi
    • 4 years ago
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    You have (2sqrt(2), -6sqrt(2)) = (a-b, a+b)

  27. phi
    • 4 years ago
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    this means you have two vectors that equal each other. If they equal each other, what does that mean about each entry in the vector?

  28. anonymous
    • 4 years ago
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    umm no clue

  29. phi
    • 4 years ago
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    if I told you ( 1, 2) = (x, y) what is x and what is y how do you know?

  30. anonymous
    • 4 years ago
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    that x = 1 and y =2

  31. phi
    • 4 years ago
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    Now use the same idea on our problem (2sqrt(2), -6sqrt(2)) = (a-b, a+b)

  32. anonymous
    • 4 years ago
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    that a-b = 2sqrt(2) and a+b = -6sqrt(2)

  33. phi
    • 4 years ago
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    yes. we have 2 equations and two unknowns. The way to solve these types of equations is write them down like this: a-b = 2sqrt(2) a+b = -6sqrt(2) add the two equations together (we are adding equal things to equal things, so the sum of them is still equal things) 2a-b+b= 2sqrt(2)-6sqrt(2) can you solve for a?

  34. anonymous
    • 4 years ago
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    lol thats confusing

  35. anonymous
    • 4 years ago
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    would it be a = 2sqrt(2) -6sqrt(2) / 2 ??

  36. phi
    • 4 years ago
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    first, you should put parens around (2sqrt(2) -6sqrt(2)) because you are dividing *both* terms by 2 (you divide the whole right hand side by 2) next, you should simplify

  37. anonymous
    • 4 years ago
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    so its [sqrt(2), -3sqrt(2)]?

  38. phi
    • 4 years ago
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    it is not a vector.

  39. anonymous
    • 4 years ago
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    what do you mean?

  40. anonymous
    • 4 years ago
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    oh sqrt(2)i , -3sqrt(2) j

  41. phi
    • 4 years ago
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    no. I mean you start with (2*sqrt(2)-6*sqrt(2))/2 which is just a (ugly) number: divide both terms by 2 to get (sqrt(2)-3*sqrt(2)) factor out sqrt(2) to get (1-3)*sqrt(2) simplify 1-3 to get a= -2*sqrt(2)

  42. phi
    • 4 years ago
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    btw, if you want info on how to solve simultaneous equations like this one, watch http://www.khanacademy.org/video/solving-systems-by-elimination?topic=worked-examples-4 and his other examples

  43. anonymous
    • 4 years ago
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    ok

  44. anonymous
    • 4 years ago
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    but what would be the answer for this?

  45. phi
    • 4 years ago
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    we are getting there.... use a+b = -6sqrt(2) plus your value for a that we just found to get b

  46. anonymous
    • 4 years ago
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    so b = -6sqrt(2) - 2sqrt(2) ?

  47. phi
    • 4 years ago
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    well let's check that a + b= -6sqrt(2) -2*sqrt(2) + b= -6sqrt(2) add +2sqrt(2) to both sides to get b = -6sqrt(2)+2sqrt(2) so close, but not exactly correct. can you simplify this?

  48. anonymous
    • 4 years ago
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    lol i am so confused

  49. phi
    • 4 years ago
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    can you simplify b = -6sqrt(2)+2sqrt(2)

  50. anonymous
    • 4 years ago
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    umm -4sqrt(2) ?

  51. phi
    • 4 years ago
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    yes. we are almost done.

  52. phi
    • 4 years ago
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    we have to answer Write the vector (2,-6) in terms of I and J. which we decided meant: (2,-6)= a*I + b*J we now know a and b, so the final answer is?

  53. anonymous
    • 4 years ago
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    2sqrt(2) i + -4sqrt(2) j ?

  54. phi
    • 4 years ago
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    and what is I and J ? It does not hurt to write out the whole solution

  55. anonymous
    • 4 years ago
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    what do you mean?

  56. phi
    • 4 years ago
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    I =[ 1/Square root (2) , 1/Square root (2)] J= [ -1/Square root (2), 1/Square root (2)] [2,-6]=2sqrt(2) I + -4sqrt(2) J

  57. phi
    • 4 years ago
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    You can check your answer by multiplying the scalars times the vectors, and then adding the vectors to see if you get [2, -6] btw, if you have time, you might want to brush up on your algebra using http://www.khanacademy.org/#core-algebra

  58. anonymous
    • 4 years ago
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    ohh k

  59. anonymous
    • 4 years ago
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    thanks a lot

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