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Math Help - Quick vector question

  1. #1
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    Quick vector question

    Hi
    I need help on the following question:

    Find a unit vector perpendicular to the plane of the two vectors (2,-1,1) and (3,4,-1). What is the sine of the angle between these two vectors?

    I have found that the unit vector perpendicular to the plane is:
    \frac{-3i+5j+11k}{\sqrt{155}}

    However how would i work out the angle. i used the a x b = absin\theta equation but cannot get the right answer.

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following question:

    Find a unit vector perpendicular to the plane of the two vectors (2,-1,1) and (3,4,-1). What is the sine of the angle between these two vectors?

    I have found that the unit vector perpendicular to the plane is:
    \frac{-3i+5j+11k}{\sqrt{155}}

    However how would i work out the angle. i used the a x b = absin\theta equation but cannot get the right answer.

    P.S
    Use the definition of the scalar product:

    \vec a \cdot \vec b = |\vec a| \cdot |\vec b| \cdot \cos(\angle(\vec a, \vec b))

    Solve for \cos(\angle(\vec a, \vec b)) and afterwards determine \angle(\vec a, \vec b)
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  3. #3
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    Quote Originally Posted by earboth View Post
    Use the definition of the scalar product:

    \vec a \cdot \vec b = |\vec a| \cdot |\vec b| \cdot \cos(\angle(\vec a, \vec b))

    Solve for \cos(\angle(\vec a, \vec b)) and afterwards determine \angle(\vec a, \vec b)
    why would i use the scalar form? It is asking to find the sine angle.
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  4. #4
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    Why not? You do know that sin x= \sqrt{1- cos^2 x}, don't you?

    Either way will give the correct answer. Since you don't show what you did or what answer you got, no one can tell what you did wrong.
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  5. #5
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    ok i used the formula which i said and i get the answer:
    a x b = (absin\theta) * unit vector of a,b

    \frac{a * b}{unit vector of a,b} = absin\theta

    \frac{-3i+5j+11k}{1} * \frac{\sqrt{155}}{-3i+5j+11k} = absin\theta

    -3i+5j+11k cancels so i am left with:

    \frac{\sqrt{155}}{\sqrt{156}} = sin\theta

    therefore sin\theta = 0.9968
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  6. #6
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    so i used the way you said and i get the same answer:

    a \cdot b = abcos\theta

    cos\theta = \frac{1}{\sqrt{156}}

    sub into sin\theta = \sqrt{1-cos^{2}\theta}

    so sin\theta = \sqrt{1-0.08006^2}

    sin\theta = 0.9968<br />
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