anonymous
  • anonymous
Determine all the real numbers for which the vector <1-c,c/3> is a unit vector.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
The vector is a unit vector if its magnitude is 1: \[\sqrt{(1-c)^2+\left(\frac{c}{3}\right)^2}=1\]
anonymous
  • anonymous
yea so i was thinking then it could be everything except 0 and 1 , also mayb am overthinking it
anonymous
  • anonymous
Actually if \(c=0\), the magnitude is 1, making it a unit vector for this \(c\).

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anonymous
  • anonymous
lol am really over thinking it lmao
anonymous
  • anonymous
All you have to do is solve for \(c\): \[\sqrt{(1-c)^2+\left(\frac{c}{3}\right)^2}=1\]
anonymous
  • anonymous
so it would jus be 0 ?
anonymous
  • anonymous
Well, let's see: \[(1-c)^2+\left(\frac{c}{3}\right)^2=1\] \[1-2c+c^2+\frac{c^2}{9}=1\] \[9-18c+9c^2+c^2=9\] \[10c^2-18c=0\] \[2c\left(5c-9\right)=0\] \[c=0~~\text{and}~~c=\frac{9}{5}\]
anonymous
  • anonymous
i see, i guess am just getting mixed up with the formula of u/|u|
anonymous
  • anonymous
Ah, that's the unit vector in the *direction* of u. All you need here is the magnitude
anonymous
  • anonymous
oooohh
anonymous
  • anonymous
thx

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