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anonymous

  • one year ago

Could somebody explain how to do this: Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3> Answer choices: A. -9.1 Degrees B. 1.8 Degrees C. 0.9 Degrees D. 11.8 Degrees Not sure how to find an angle between vectors.

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  1. anikhalder
    • one year ago
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    Do you know how to find the magnitude or length of a vector? And also so you know about scalar or dot product of 2 vectors?

  2. anonymous
    • one year ago
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    I know Scalar and Dot Product. @anikhalder I sort of know magnitude...

  3. anikhalder
    • one year ago
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    That's perfect! So let's go through this: What is the length of u vector?

  4. anonymous
    • one year ago
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    I'm not sure how to solve for length...

  5. anikhalder
    • one year ago
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    so its like, suppose i have a vector a = <3, 4> Then the length of the vector a is : \[\left| a \right|=\sqrt{3^{2} + {4^{2}}}\] \[\left| a \right|=\sqrt{25}\] That is the length of vector a is 5 So using the same way can you solve for the length of vector u?

  6. anonymous
    • one year ago
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    Oh! Is the magnitude |u| = SqRt. 41 ?

  7. anikhalder
    • one year ago
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    Perfect! You rock! What about the magnitude (or length) of vector v?

  8. anonymous
    • one year ago
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    |v| = 5 ?

  9. anikhalder
    • one year ago
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    Awesome! Now do you know the formula for dot (or scalar) product of 2 vectors a and b?

  10. anonymous
    • one year ago
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    Yes, would the product be: a x b = 32 ?

  11. anikhalder
    • one year ago
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    Yes that's correct! But, I will suggest when writing dot product don't use 'x' sign. You will learn that there is another type of product called the vector or cross product which uses this sign...but we don't need to worry about right now. Suppose I have a vector a and vector b Then the dot product of vectors a and b is also defined as: \[a.b=\left| a \right|\left| b \right|\cos \theta\] Where theta is the angle between the vectors a and b. So for our question at hand. Let's replace a by u and b by v. So, we get: \[u.v=\left| u\right|\left| v\right|\cos \theta\] We can replace u.v as 32 and the magnitudes of u and v on the right hand side of the equation to get: \[\cos \theta=\frac{ u.v }{ \left| u \right|\left| v \right|}\] Can you solve for costheta now?

  12. anonymous
    • one year ago
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    Is costheta = .99 ?

  13. anonymous
    • one year ago
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    @anikhalder

  14. anikhalder
    • one year ago
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    Yes! I find so as well. Now just find the cosine inverse of 0.99 and you'll get it in degrees

  15. anonymous
    • one year ago
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    I got 8.1 degrees, approximately, but that is not one of my answers.

  16. anikhalder
    • one year ago
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    That's what I get and now I must confess that I am confused as well

  17. anikhalder
    • one year ago
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    Let's verify

  18. anonymous
    • one year ago
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    Ok

  19. anikhalder
    • one year ago
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    http://onlinemschool.com/math/assistance/vector/angl/ Can you type the values in and see what you get for costheta because I get 0.99 in this website as well

  20. anonymous
    • one year ago
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    I am getting .99

  21. anonymous
    • one year ago
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    @anikhalder Can you help me with another problem, and then maybe go back to this one?

  22. anikhalder
    • one year ago
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    oh wait

  23. anonymous
    • one year ago
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    Express the complex number in trigonometric form. -3 + 3 square root of three i

  24. anikhalder
    • one year ago
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    http://www.rapidtables.com/calc/math/Arccos_Calculator.htm

  25. anonymous
    • one year ago
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    And I'm confused as to what that calculator meant.

  26. anikhalder
    • one year ago
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    Check this, we were just using 0.99, we cant do that we have to use the full result which was 0.9995120761 plug this and find the cos inverse. You will see that we get our answer :)

  27. anonymous
    • one year ago
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    Oh. I see. So the answer would be 1.8 degrees?

  28. anikhalder
    • one year ago
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    Yep! And yes just take 4 minutes and see this video on complex numbers! he explains better than me :)))

  29. anonymous
    • one year ago
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    What video?

  30. anikhalder
    • one year ago
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    Sorry...my bad i forgot to give the link: https://www.youtube.com/watch?v=6z6fzPXUbSQ

  31. anonymous
    • one year ago
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    Thank you so much. (: I watched that video, though, and am still confused on my example. He helped me with a couple others, but the one I posted about is a bit confusing after watching the video.

  32. anikhalder
    • one year ago
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    You mean its like : \[3+3\sqrt{3}i\] So, \[\left| z \right| = \sqrt{3^{2}+(3\sqrt{3})^{2}} = 6\] and \[\tan \theta = \frac{ 3\sqrt{3} }{ 3}\] i.e. \[\theta = 60 degrees\] Now, in polar or trigonometric form, z = \[\left| z \right|(\cos \theta + i \sin \theta)\] Just substitute the values and find the answer :)))

  33. anonymous
    • one year ago
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    What is the Z value?

  34. anonymous
    • one year ago
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    And is the theta value 120 degrees?

  35. anikhalder
    • one year ago
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    you can say \[\left| z \right|\] is like the absolute value of the complex number (like the|dw:1437768487456:dw|

  36. anonymous
    • one year ago
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    So the z value is 3SqRt3. But you have in 3 as positive, the 3 is negative. So would that make the equation (in radians): 3SqRt2(Cos2pi/3 + i sin2pi/3)

  37. anikhalder
    • one year ago
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    oho...i forgot the minus sign...yep!

  38. anonymous
    • one year ago
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    Ok, so the theta value is 120 degrees? Not 150?

  39. anikhalder
    • one year ago
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    Wait hold on, how is r coming to 3sqrt3 it should be 6

  40. anonymous
    • one year ago
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    Oh. Ok, how do you get 6?

  41. anikhalder
    • one year ago
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    |dw:1437768935393:dw|

  42. anonymous
    • one year ago
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    Oh ok thanks. So it would be the same answer, but 6 as r vs. 3sqrt3 as r?

  43. anikhalder
    • one year ago
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    yes it would 6(costheta + i sintheta)

  44. anonymous
    • one year ago
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    Ok thanks! Theta = 2pi/3 right?

  45. anikhalder
    • one year ago
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    yes. because tantheta in this case is 3sqrt3 / -3 which will give theta as 2pi/3

  46. anikhalder
    • one year ago
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    :) I guess so. Please verify the answer with your book and your teacher. If I made any mistake please tell me :)))

  47. anonymous
    • one year ago
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    Thanks so much!

  48. anikhalder
    • one year ago
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    Thank you very much for bearing with me! My pleasure! Have a nice night! I will go to sleep!

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