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Magnitude, Unit vectors,unknown variables

  1. Mar 4, 2014 #1
    1. The problem statement, all variables and given/known data
    a and b are vectors in R^3 s.t. a=(1,7,-4) and b= -3j-4k

    1. Find ||3a-3b|| (magnitude of 3a-3b)
    2. Find unit vector u in direction of 3a-3b, write answer in form (u1,u2,u3)
    3. Find vector of length 8 in direction of a (write answer in form "-")


    2. Relevant equations
    ||3a-3b|| = 3||a-b|| = 3*sqrt((a1-b1)^2 +.......(a3-b3)^2)
    3 is positve scalar and can be factored out I believe

    U = X/ ||X|| , X ≠ 0


    Length = sqrt(x^2 + y^2) (might be wrong, I am confused on finding length in same dir)



    3. The attempt at a solution

    Starting solving 1.
    3||a-b|| = 3 sqrt((1+3j+4k)^2 + (7+3j+4k)^2 + (-4+3j+4k)^2)
    I don't know what j and k represent

    2. I would need to know ||3a-3b|| first right?
    then do X/||X||?

    3. a= (1,7,-4) and L=8
    L is in same direction of a so
    8 = sqrt((u1-1)^2 + (u2-7)^2 + (u3 +4)^2) -> or maybe the a vals and u's are switched i don't know
     
  2. jcsd
  3. Mar 4, 2014 #2

    micromass

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    Not sure where this comes from.

    The formula is that if we are given ##a = (a_1,a_2,a_3)## and ##b = (b_1, b_2, b_3)##, then

    [tex]\|a-b\|= \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2 + (a_3 - b_3)^2}[/tex]

    Here you are given ##a= (1,7,-4)## and ##b=(0,-3,-4)##.
     
  4. Mar 4, 2014 #3
    Wow wait a minute, how do you know the values for b as (0,-3, -4)?
    What do j and k represent?

    Also, do you have any input regarding the other two questions I posted w/in this post?
     
  5. Mar 4, 2014 #4

    micromass

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    The notation ##-3j -4k## is just another notation for ##(0,-3,-4)##.

    The vector ##i## actually equals ##(1,0,0)##. The vector ##j=(0,1,0)## and ##k=(0,0,1)##. So

    [tex]-3-4k = -3(0,1,0) -4 (0,0,1) = (0,-3,0) + (0,0,-4) = (0,-3,-4)[/tex]

    Sure, but let's do (a) first.
     
  6. Mar 4, 2014 #5
    Okay gotcha, thanks! My prof did not cover that this lecture so I was unclear.

    Okay so with that in mind ||a-b|| = sqrt(101) right?
    and since 3 is a postive number it can be factored out thus,

    3*sqrt(101) = ||3a-3b|| right?
     
  7. Mar 4, 2014 #6

    micromass

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    Yes, that's correct.

    So, can you use this to solve (b) now?
     
  8. Mar 4, 2014 #7
    I think so but is my equation correct? Which values am I using
     
  9. Mar 4, 2014 #8
    actually scratch that I got the answer. I got:
    U = (1/sqrt 101, 10/sqrt 101, 0)

    How do I do the last part about length?
     
  10. Mar 5, 2014 #9

    micromass

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    Given your vector ##a##, then if ##\lambda>0## is a real number, then ##\lambda a## lies in the same direction of ##a##. So your answer will be ##\lambda a##. Your only job is to figure out what ##\lambda## is exactly.

    You want ##\|\lambda a\| = 8##. Can you figure out what ##\lambda## is now?
     
  11. Mar 5, 2014 #10
    yep got the answer as (8/sqrt 66 , 56/sqrt 66 , -32/sqrt 66)
     
  12. Mar 5, 2014 #11

    micromass

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    Right!
     
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