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Magnitude of vector CD?

  1. Mar 4, 2007 #1
    C, D are points defined by position vectors c and d. Magnitude of c is 5, mag of d is 7, c dotproduct d is 4 ie c.d = 4, find the magnitude of vector CD.

    So i started this way

    c.d = magc*magdcos@
    = 35cos@, @ = 83.4 degrees

    But still no idea how do get magnitude of vector cd. Thank you!
    Last edited: Mar 4, 2007
  2. jcsd
  3. Mar 4, 2007 #2


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    Are you sure it isnt the magnitude of vector CD? (i.e. the vector from point C to point D)? If so, use the cosine rule on the triangle, since you know two vectors and an angle.
    Last edited: Mar 4, 2007
  4. Mar 4, 2007 #3
    oops yes you are right, ok i use the cos rule and do get an answer(which is correct), but I am after it as an exact value (root66). How would i get that? Thanks!
    Last edited: Mar 4, 2007
  5. Mar 4, 2007 #4


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    ?? You say you got an answer but you want an exact value? Did you use a calculator to get [itex]cos(\theta)[/itex]. Since you want to use [itex]cos(\theta)[/itex] in the cosine rule, why not just use [itex]cos(\theta)= \frac{4}{35}[/itex] rather than finding [itex]\theta[/itex] itself?
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