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Math Help - Find the unit vector perpendicular to the plane of 2 vector

  1. #1
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    Red face Find the unit vector perpendicular to the plane of 2 vector

    Question : Find the unit vector perpendicular to the plane of two vectors \bar{a} and \bar{b} ,

    \bar{a} = \hat{i} - \hat{j} + 2\hat{k} and \bar{b} = 2\hat{i} + 3\hat{j} - \hat{k}
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    Quote Originally Posted by zorro View Post
    Question : Find the unit vector perpendicular to the plane of two vectors \bar{a} and \bar{b} ,

    \bar{a} = \hat{i} - \hat{j} + 2\hat{k} and \bar{b} = 2\hat{i} + 3\hat{j} - \hat{k}
    Use the formula: \overrightarrow{u_n}=\dfrac{\vec a \times \vec b}{|\vec a \times \vec b|}
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    Is this correct?

    Quote Originally Posted by earboth View Post
    Use the formula: \overrightarrow{u_n}=\dfrac{\vec a \times \vec b}{|\vec a \times \vec b|}


    | \bar a \times \bar b| = \begin{vmatrix}i & j & k \\ 1 & -1  & 2 \\ 2 & 3 & -1 \end{vmatrix} = -5i + 5j +5k

    | \bar a \times \bar b| = \sqrt{ (- 5)^2 + 5^2 + 5^2 } = \sqrt{75} = 5 \sqrt{3}

    u_n =  \frac{\bar a \times \bar b}{|\bar a \times \bar b|} = -i + j + k..............Is this correct ??
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    Quote Originally Posted by zorro View Post
    | \bar a \times \bar b| = \begin{vmatrix}i & j & k \\ 1 & -1 & 2 \\ 2 & 3 & -1 \end{vmatrix} = -5i + 5j +5k Mr F says: Correct.

    | \bar a \times \bar b| = \sqrt{ (- 5)^2 + 5^2 + 5^2 } = \sqrt{75} = 5 \sqrt{3} Mr F says: Correct.

    u_n =  \frac{\bar a \times \bar b}{|\bar a \times \bar b|} = -i + j + k..............Is this correct ?? Mr F says: No.
    You divided by 5. You're meant to divide by 5 \sqrt{3}.
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    okay

    Quote Originally Posted by mr fantastic View Post
    You divided by 5. You're meant to divide by 5 \sqrt{3}.

    The result should be then

    \frac{-5i+5j+5k}{5 \sqrt{3}}
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    Quote Originally Posted by zorro View Post
    The result should be then

    \frac{-5i+5j+5k}{5 \sqrt{3}}
    Well it should be written as:  - \frac{{\sqrt 3 }}<br />
{3}i + \frac{{\sqrt 3 }}<br />
{3}j + \frac{{\sqrt 3 }}<br />
{3}k
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    Could u please explain

    Quote Originally Posted by Plato View Post
    Well it should be written as:  - \frac{{\sqrt 3 }}<br />
{3}i + \frac{{\sqrt 3 }}<br />
{3}j + \frac{{\sqrt 3 }}<br />
{3}k

    Could u please explain how u got this answer.......
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    Quote Originally Posted by zorro View Post
    Could u please explain how u got this answer.......
    Basic arithmetic. Cancel the common factor of 5.
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    how did u get \sqrt{3} at the denominator
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    Quote Originally Posted by zorro View Post
    how did u get \sqrt{3} at the denominator
    \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} (you must have been taught how to rationalise the denominator of surds ....?)
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    Quote Originally Posted by mr fantastic View Post
    \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} (you must have been taught how to rationalise the denominator of surds ....?)
    okay but is that neccessary ..........my answer is correct isnt it??
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  12. #12
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    Quote Originally Posted by zorro View Post
    okay but is that neccessary ..........my answer is correct isnt it??
    Yes, it is. But whether or not it got full marks would depend on whether the person marking it wanted a simplified answer. Certainly at this level the common factor of 5 should be cancelled. It is personal taste whether or not to rationalise the denominator. You should discuss this further with your instructor.
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