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Vectors: Cross Product

  1. Feb 19, 2017 #1
    Hello, I apologize in advance for the way this post looks. I am new to this forum and I've never used LaTeX Primer. I noticed that someone has prevoiusly asked the same question, but I still do not understand how to get to the answer. Also, I tried posting an image but I could not; and this other post contains the image so I will leave a link to the other forum.
    https://www.physicsforums.com/threads/cross-product-homework.352545/




    1. The problem statement, all variables and given/known data

    The figure shows two vectors $$ \vec T$$ and $$\vec U$$ separated by an angle $$θ_{TU}$$ You are given that $$\vec T =(3,1,0), \vec U =(2,4,0), ~and~ \vec T × \vec U = \vec V$$ Find the sine of the angle between $$\vec T ~and~ \vec U $$


    2. Relevant equations

    $$| \vec T × \vec U |= | \vec T || \vec V |\sinθ$$

    3. The attempt at a solution

    I know that the answer is $$\frac {\sqrt{2}} 2$$ The magnitude of vector V is 10 and you square root that, 10 is also opposite of the angle theta. However, I don't understand how the hypotenuse is 20 therefore you would simplify $$\frac {\sqrt{10}} {\sqrt20}$$ to get the answer
     
  2. jcsd
  3. Feb 19, 2017 #2

    Charles Link

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    You didn't compute ## T \times U ## as far as I can see. This is a necessary computation. The easiest way that I know is by the determinant method. Have they shown you the determinant method in your coursework? ## \\ ## Editing... Looking over your work, the correct equation is ## | T \times U|=|T||U| sin(\theta) ##. (The equation you have written is incorrect.) The correct result is thereby ## \frac{|V|}{|T||U|}=sin(\theta) ## where ## V=T \times U ##.
     
    Last edited: Feb 19, 2017
  4. Feb 20, 2017 #3
    Isn't the magnitude of T ##\sqrt{3^2+1^2+0^2}## = ##\sqrt{10}## and the magnitude of U ##\sqrt{2^2+4^2+0^2}## = ##\sqrt{20}## ? But if you plug that into ## \frac{|V|}{|T||U|}=sin(\theta) ## , the answer is not correct. Where am I going wrong?
     
  5. Feb 20, 2017 #4

    Student100

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    It gives you the correct answer.

    What is ##\vec{T} \times \vec{U}##?
     
  6. Feb 20, 2017 #5
    It is (0,0,10), so the magnitude of that is 10.
     
  7. Feb 20, 2017 #6

    Student100

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    Yes, sooo...

    $$\frac{|V|}{|T||U|} = \sin(\theta)$$
    $$\arcsin(\frac{|V|}{|T||U|}) = \theta$$
    $$\arcsin(\frac{\sqrt{100}}{\sqrt{10}\sqrt{20}}) = \theta$$
    $$\arcsin(\sqrt{\frac{100}{200}}) = \theta$$

    Can you finish it from there?
     
  8. Feb 20, 2017 #7
    Yes, I just forgot to take the arcsin of that. Thank you!
     
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