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KatClaire

  • 3 years ago

Find a unit vector in the same direction as u= [3,0,-7,-4,2] (column)

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  1. KatClaire
    • 3 years ago
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    \[\left(\begin{matrix}3 \\ 0\\-7\\-4\\2\end{matrix}\right)\]

  2. amistre64
    • 3 years ago
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    what is the length of the given vector?

  3. amistre64
    • 3 years ago
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    suppose you have a vector that is 12 feet in length, how do you find out its "unit" length?

  4. KatClaire
    • 3 years ago
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    that's all that's given :\

  5. KatClaire
    • 3 years ago
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    12? lol

  6. amistre64
    • 3 years ago
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    yeah, and to find the length of a vector, you square the parts, add em up, and sqrt it all ....

  7. amistre64
    • 3 years ago
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    \[vector:<a,b,c,d,...,k>\]\[distance=\sqrt{a^2+b^2+c^2+d^2+...+k^2}\]

  8. amistre64
    • 3 years ago
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    parts: square 3: 9 0: 0 7:49 4:16 2: 4 --- sum: 78 ; so length must be sqrt(78) or did i do something wrong?

  9. KatClaire
    • 3 years ago
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    I did that and got sqrt 79, yes?

  10. amistre64
    • 3 years ago
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    vector = sqrt(78) units to find a vector of 1 unit, we divide both sides by sqrt(78) vector/ sqrt(78) = sqrt(78)/sqrt(78) units vector/ sqrt(78) = 1 unit therefore, a unit vector is created by dividing the components of the vector by its length

  11. amistre64
    • 3 years ago
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    18+10=28 .. not 29 :)

  12. bmelyk
    • 3 years ago
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    sorry my computer like died so I'm on my friends lol so it's 6/sqrt78 @amistre64 then you times that into each unit thingy

  13. amistre64
    • 3 years ago
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    each component then gets divided by the vectors length to obtain components of the unit vector. 3/sqrt(78) 0/sqrt(78) - 7/sqrt(78) - 4/sqrt(78) 2/sqrt(78)

  14. amistre64
    • 3 years ago
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    another way to notate it is just to scale the original vector by 1/length \[\frac{1}{\sqrt{78}}[3,0,-7,-4,2]\]

  15. KatClaire
    • 3 years ago
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    oh okay my problem was accepting the fact that my calculator wouldn't put my answers in fractions so I thought I must be wrong hahaha. Thanks!!

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