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anonymous

  • one year ago

Let v1 = (2, -6) and v2 = (-4, 7). Compute the unit vectors in the direction of |v1| and |v2|. And can anyone double check if this graph is right? Draw and label v1, v2, and v1+v2. https://gyazo.com/ca330d1301b8dd28e0cdfa3e72f6443c

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  1. Alex_Mattucci
    • one year ago
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    Your graph is looking great!

  2. Alex_Mattucci
    • one year ago
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    is that all?

  3. anonymous
    • one year ago
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    Compute the unit vectors in the direction of |v1| and |v2|. What exactly is this question trying to find?

  4. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{\vec{V_1} }{\left|\left| \vec{V_1}\right|\right|} }\)

  5. SolomonZelman
    • one year ago
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    This is the unit vector (with magnitude 1) in direction of \(\vec{V_1}\)

  6. anonymous
    • one year ago
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    How would you plug in the vector v1 into this equation to find a value?

  7. SolomonZelman
    • one year ago
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    Note: V\(_1\) with two bars on each side, means "magnitude of V\(_1\).

  8. SolomonZelman
    • one year ago
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    You want units vecotrs with the same directions as \(\vec{V_1}\) and \(\vec{V_2}\), right?

  9. SolomonZelman
    • one year ago
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    |dw:1442448091015:dw|

  10. SolomonZelman
    • one year ago
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    that means that if you take each component and divide by this magnitude, you get a unit vector in same direction.

  11. SolomonZelman
    • one year ago
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    \(\left|\vec{V_1}\right|=\sqrt{(-2)^2+(6)^2{\color{white}{\large|}}}\)

  12. SolomonZelman
    • one year ago
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    I mixed it up, 2, and -6. For magnitude that doesn't matter though. you still get 2√10

  13. SolomonZelman
    • one year ago
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    |dw:1442448470003:dw|

  14. SolomonZelman
    • one year ago
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    Now, you need to simplify this and you are done with the unit vector for V\(\large _1\)

  15. anonymous
    • one year ago
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    Oh, so the value just ends up being the x value of the vector/ magnitude and the y value/magnitude?

  16. SolomonZelman
    • one year ago
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    yes

  17. SolomonZelman
    • one year ago
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    And same for 3 D vector: < x-component/magnitude, y-component/magnitude, z-component > (and same for any N-dimensional vector (only physics stops at 3 D))

  18. SolomonZelman
    • one year ago
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    Then you need to find the magnitude of \(\vec{V_2}\) and divide each of the \(\vec{V_2}\)'s components by this magnitude. And lets recall that: \(\vec{V_2}=<4,-7>\)

  19. SolomonZelman
    • one year ago
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    So the rest is yours:)

  20. anonymous
    • one year ago
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    So magnitude of v2 ends up being √65, then the unit vector would be 4/√65 and -7/√65?

  21. SolomonZelman
    • one year ago
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    let me see: √{4²+(-7)²}=√{16+49}=√65 Yes, then it would be: < 4/√65 , -7/√65 >

  22. SolomonZelman
    • one year ago
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    You can use a taylor polynomial approximation of some nth degree near a=64, of f(x)=√x. (If you want.... :D)

  23. SolomonZelman
    • one year ago
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    Imagine you didn't have a calculator, then you would need one ... jk

  24. anonymous
    • one year ago
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    Oh jeez, I have no idea about 3d vectors or taylor polynomials at all....But thanks so much for all the help!

  25. SolomonZelman
    • one year ago
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    3D is just more complicated to draw, that is all. And tailor polynomial just comes from standard theorem of calculus and integration of parts, starting from: \(\displaystyle \int_{0}^{x}f'(t)dt\)

  26. SolomonZelman
    • one year ago
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    You will learn it pretty soon I am sure.... as for now, if you don't have any questions regarding your problem, then good luck:)

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