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unicwaan

  • one year ago

Can someone help me answer a precalculus question please?

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  1. unicwaan
    • one year ago
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    Find the angle between the given vectors to the nearest tenth of a degree. u = <2, -4>, v = <3, -8>

  2. unicwaan
    • one year ago
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    @EducationsFinest

  3. unicwaan
    • one year ago
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    @Vocaloid

  4. anikhalder
    • one year ago
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    You can write u as 2i -4j and v as 3i -8j. Then you can apply the dot product rule to find the angle. Take a look at this page: http://www.wikihow.com/Find-the-Angle-Between-Two-Vectors Or take a look at these videos. It will really make your base strong. https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces

  5. IrishBoy123
    • one year ago
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    you don't have to re-write them, this is really good compact notation you are using: \(\large \vec u = <2, -4>, \vec v = <3, -8>\) for \(\large \vec u = <u_x, u_y>, \vec v = <v_x, v_y>\), we have \(\large \vec u \bullet \vec v = u_x \times v_x + u_y \times v_y = |\vec u||\vec v| cos \theta\). \(\large \theta \) is the angle between these vectors. \(\large |\vec u|\) is the magnitude of \(\large \vec u\) and equals \(\large \sqrt {u_x^2 + u_y^2}\), ditto for \(\large |\vec v|\). that's all you need to answer this, but you should always check first for some trickery in the question, eg where they are clearly parallel.

  6. IrishBoy123
    • one year ago
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    @anikhalder thanks for the medal! have one back :p

  7. anikhalder
    • one year ago
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    Thank you very much. I really appreciate your explanation and I hope rather than the medals our mutual friend here won't be having troubles finding the angle between vectors anymore :)))

  8. IrishBoy123
    • one year ago
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    cool!

  9. unicwaan
    • one year ago
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    Thank you all!

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