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Math Help - 3D vectors finding perpindicular vector help!!!

  1. #1
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    3D vectors finding perpindicular vector help!!!

    hello everybody, first of all i would like to thank you for even trying to help me with my problem, I am doing my 3d vectors homework and I came upon a problem that I am having difficulties with, the problem is as follows:

    Determine a vector perpindicular to both the y-axis and the vector AB = [2,-1,2].

    any help will be appreciated.

    thanks
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  2. #2
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    Quote Originally Posted by dmitri View Post
    hello everybody, first of all i would like to thank you for even trying to help me with my problem, I am doing my 3d vectors homework and I came upon a problem that I am having difficulties with, the problem is as follows:

    Determine a vector perpindicular to both the y-axis and the vector AB = [2,-1,2].

    any help will be appreciated.

    thanks
    The y-axis has a unit vector of <0, 1, 0> and you want a vector mutually perpendicular to this and <2, -1, 2>. Take the cross product of these two vectors.

    -Dan
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    Hello, dmitri!

    Determine a vector perpendicular to both the y-axis and the vector AB = [2,-1,2].

    We have two vectors: . \begin{array}{ccc}\vec{u} &=&\langle 2,1,2\rangle \\ \vec{v} &=& \langle0,1,0\rangle \end{array}

    A vector perpendicular to both is their cross product.

    \vec{w} \;=\;\vec{u} \times \vec{v} \;=\;\left|\begin{array}{ccc}\bold i&\bold j&\bold k \\ 2&1&2\\0&1&0\end{array}\right| \;=\;\bold i(0-2) -\bold j(0-0) + \bold k(2-0) \;=\;-2\bold i + 2\bold k

    . . Therefore: . \vec{w} \;=\;\langle -2,0,2\rangle

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    thanks guys! i appreciate it
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