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Vector Length

  1. Nov 7, 2006 #1
    Find the vector that satisfies the stated conditions: Length (17^1/2) and same direction as v = <7,0,-6>.

    The book gives the answer as being u = (1/5)<7,0,-6>. Having some trouble getting this answer... here is how I went about attempting to solve it:

    1. Finding the vector length of the given vector.
    ||v|| = (7^2 + 0^2 + -6^2)^(1/2)
    = (49+0+36)^(1/2)
    = (85)^(1/2)

    2. Setting the two vectors equal...
    ||u|| = (17)^(1/2) = k||v|| Since multiples of the vector will give same direction
    (17)^(1/2)/||v|| = k

    So k = (17)^(1/2)/(85)^(1/2)

    So then my answer would be u = (17)^(1/2)/(85)^(1/2) <7,0,-6>.


    Can anyone help point out where I am messing up?? Thank you
     
  2. jcsd
  3. Nov 7, 2006 #2

    radou

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    Work with vectors. Your first condition is u = k v. The second one is ||u||=||kv|| = 17^(1/2). This should work just fine.
     
  4. Nov 8, 2006 #3
    I'd recommend first finding the unit vector [tex]v = \frac{u}{||u||}[/tex], because you know that vector has a magnitude of 1. Then, you know that [tex]||kv||=\sqrt{17}[/tex], and you know that [tex]||v||=1[/tex], so [tex]k=\sqrt{17}[/tex]. So, your new vector [tex]w=kv=<\frac{7\sqrt{17}}{\sqrt{85}},0,\frac{-6\sqrt{17}}{\sqrt{85}}>=<7\sqrt{\frac{1}{5}},0,-6\sqrt{\frac{1}{5}}>[/tex] has a magnitude of [tex]\sqrt{17}[/tex]. So, your answer should be correct... I'd check the answer in your book with your instructor.
     
    Last edited: Nov 8, 2006
  5. Nov 8, 2006 #4

    NateTG

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    Are you sure that the book doesn't have the following?
    [tex]\sqrt{\frac{1}{5}} \left<7,0-6\right>[/tex]

    (Which is one of many possible expressions equivalent to the answer you've given.)
     
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