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Math Help - Problem with vector question

  1. #1
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    Problem with vector question

    Hi
    Can someone tell me where i have gone wrong in the following question?

    Given that v=2i + 2j + k, w = 3i - j + k find a unit vector perpendicular to both v and w.

    unit vector = {a x b}/{|a x b|}

    | a x b |=  \sqrt{3^2+1^2+8^2}

    =\sqrt{74}


    a x b = i(2 x 1 - 1 x -1)-j(2 x 1 - 1 x 3)+k(2 x -1 - 2 x 3)
    =i(2+1)-j(2-3)+k(-2-6)
    =3i+j-8k

    therefore unit vector = \frac{1}{\sqrt{74}}(3i+j-8k)

    answer says its  \frac{1}{\sqrt{74}}(-3i-j+8k)

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone tell me where i have gone wrong in the following question?

    Given that v=2i + 2j + k, w = 3i - j + k find a unit vector perpendicular to both v and w.

    unit vector = {a x b}/{|a x b|}

    | a x b |=  \sqrt{3^2+1^2+8^2}

    =\sqrt{74}


    a x b = i(2 x 1 - 1 x -1)-j(2 x 1 - 1 x 3)+k(2 x -1 - 2 x 3)
    =i(2+1)-j(2-3)+k(-2-6)
    =3i+j-8k

    therefore unit vector = \frac{1}{\sqrt{74}}(3i+j-8k)

    answer says its  \frac{1}{\sqrt{74}}(-3i-j+8k)

    P.S
    \mathbf{v}\times\mathbf{w} = \left|\begin{matrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 1 \\ 3 & -1 & 1 \end{matrix}\right|

     = \mathbf{i}[2 \cdot 1 - 1\cdot (-1)] - \mathbf{j}[2 \cdot 1 - 1 \cdot 3] + \mathbf{k}[2 \cdot (-1) - 2 \cdot 3]

     = 3\mathbf{i} - \mathbf{j} - 8\mathbf{k}.

    I agree with the answer you have given for \mathbf{v} \times \mathbf{w}. Perhaps you copied the question down wrong or else the answer given is incorrect.


    Anyway, to find the unit vector in the direction of \mathbf{v} \times \mathbf{w}, divide it by its length.


    |\mathbf{v}\times\mathbf{w}| = \sqrt{3^2 + (-1)^2 + (-8)^2}

     = \sqrt{9 + 1 + 64}

     = \sqrt{74}.


    Therefore:

    \frac{\mathbf{v}\times\mathbf{w}}{|\mathbf{v}\time  s\mathbf{w}|} =  \frac{3\mathbf{i} - \mathbf{j} - 8\mathbf{k}}{\sqrt{74}}

     = \frac{3}{\sqrt{74}}\mathbf{i} - \frac{1}{\sqrt{74}}\mathbf{j} - \frac{8}{\sqrt{74}}\mathbf{k}

     = \frac{3\sqrt{74}}{74}\mathbf{i} - \frac{\sqrt{74}}{74}\mathbf{j} - \frac{4\sqrt{74}}{37}\mathbf{k}.
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  3. #3
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    Hello, Paymemoney!


    You've done nothing wrong . . .

    Their answer is the negative of yours.

    There are two vectors perpendicular to both \vec v\text{ and }\vec w,
    . . one "up" and one "down".

    They picked one, you picked the other.

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  4. #4
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    ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k
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