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Math Help - 3D Vector Problem

  1. #1
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    3D Vector Problem

    I have been give the following problem: the vector perpendicular to OA and OB where O, B, A are points with respective coordinates of (0,0,0), (-4,2,1) and (1,2,3) is of the form 4i+yj+zk. What are the values of y and z?

    How do i go about working this out???
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  2. #2
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    Quote Originally Posted by thermalwarrior View Post
    I have been give the following problem: the vector perpendicular to OA and OB where O, B, A are points with respective coordinates of (0,0,0), (-4,2,1) and (1,2,3) is of the form 4i+yj+zk. What are the values of y and z?

    How do i go about working this out???
    A vector perpendicular to two other vectors is given by the cross product:
    (-4, 2, 1) \times (1, 2, 3) = det \left ( \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} \right )

    This will either be parallel to your vector or antiparallel.

    -Dan
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    how does that lead to the values of y and z in the final form 4i + yj + zk?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by thermalwarrior View Post
    how does that lead to the values of y and z in the final form 4i + yj + zk?
    Quote Originally Posted by topsquark View Post
    A vector perpendicular to two other vectors is given by the cross product:
    (-4, 2, 1) \times (1, 2, 3) = det \left ( \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} \right )

    This will either be parallel to your vector or antiparallel.

    -Dan
    The result of the cross product will be of the form
    a(4i + yj + zk)
    where a is some constant. ie. they are parallel.

    So find the determinant, and set the x component equal to 4a. Solve for a. Then the y component will be ay, etc.

    -Dan
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    sorry im a bit confused - if i do that then a can only be 1 because the answer given in the questions has the x component as 4i. or have i misunderstood what you have wrote?
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  6. #6
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    Quote Originally Posted by thermalwarrior View Post
    sorry im a bit confused - if i do that then a can only be 1 because the answer given in the questions has the x component as 4i. or have i misunderstood what you have wrote?
    You are right. With your problem a = 1

    (Btw: I've got y = 13 and z = -10)
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  7. #7
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    can you show me how? that fits the question!
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  8. #8
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    As suggested take the cross product of OA and OB which will yield a vector perpendicular to both vectors. You are told that there is a vector of the form (4,y,z) that is also perpendicular to both OA and OB. So, (4,y,z) is some scalar multiple of the cross product between OA and OB:

    (-4, 2, 1) \times (1, 2, 3) = \left( \left| \begin{array}{cc} 2 & 1 \\ 2 & 3 \end{array} \right|,  -\left| \begin{array}{cc} -4 & 1 \\ 1 & 3 \end{array}  \right|, \left| \begin{array}{cc} -4 & 2 \\ 1 & 2 \end{array} \right| \right) = (a,  b, c) (leaving it up to you to find the cross product i.e. what a, b, and c are)

    So, (a, b, c) = k(4, y, z).

    Looking at the x component, you can see that a = 4k. Solve for k, then you should be able to find y and z (b = ky; c = kz).
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