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Math Help - Questions on finding k and normal vectors

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    Questions on finding k and normal vectors

    Okay, so the end goal of what I need help with is to find the normal vector. But I also don't understand how we are setting up a matrix to find it. Here is a specific example -

    I have a vector e=[1,1] and normal vector = e x k

    So my matrix is

    [ i j k ]
    [ 1 1 0 ]
    [ 0 0 1 ]

    How are we getting 0 in the top row for k?
    How are we getting the bottom row?
    How are we finding k in general?

    My thoughts -
    k=ixj, which here equals 0, so is that why there is a 0 for k in the middle row? And if k=0, why do we have a 1 in the bottom row? What does the right hand rule give us when the two vectors are the same?

    Another example -

    e=[0,6]

    [ i j k ]
    [ 0 6 0 ]
    [ 0 0 1 ]

    The right hand rule for i and j would make k go into the paper, which means it should be -1 right? So why do we have a positive one right there? Is that matrix wrong? If not, what am I missing?

    Thank you for any help. I know I may be missing some details so just let me know and I'll respond asap.
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    Re: Questions on finding k and normal vectors

    Quote Originally Posted by SterlingM View Post
    Okay, so the end goal of what I need help with is to find the normal vector. But I also don't understand how we are setting up a matrix to find it. Here is a specific example -

    I have a vector e=[1,1] and normal vector = e x k

    So my matrix is

    [ i j k ]
    [ 1 1 0 ]
    [ 0 0 1 ]

    How are we getting 0 in the top row for k?
    How are we getting the bottom row?
    How are we finding k in general?

    My thoughts -
    k=ixj, which here equals 0, so is that why there is a 0 for k in the middle row? And if k=0, why do we have a 1 in the bottom row? What does the right hand rule give us when the two vectors are the same?

    Another example -

    e=[0,6]

    [ i j k ]
    [ 0 6 0 ]
    [ 0 0 1 ]

    The right hand rule for i and j would make k go into the paper, which means it should be -1 right? So why do we have a positive one right there? Is that matrix wrong? If not, what am I missing?

    Thank you for any help. I know I may be missing some details so just let me know and I'll respond asap.
    Your original vectors are only in the x,y plane, so have a 0 component in the z axis.

    You also need a 3D vector which is orthogonal to it. An obvious second vector to use is < 0, 0, 1 > because, being on the z axis, it is already orthogonal to your e vectors which don't have any z component, and their cross product will give you a third vector which is orthogonal to both.

    Incidentally, you are not setting up a MATRIX to evaluate the cross products, you are setting up a DETERMINANT. So it SHOULD be

    \displaystyle \begin{align*} \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right|&= \mathbf{i}\left| \begin{matrix}  1 & 0 \\ 0 & 1 \end{matrix} \right| - \mathbf{j} \left| \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right| + \mathbf{k} \left| \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right| \end{align*}

    Go from here.
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    Re: Questions on finding k and normal vectors

    Quote Originally Posted by Prove It View Post
    Your original vectors are only in the x,y plane, so have a 0 component in the z axis.

    You also need a 3D vector which is orthogonal to it. An obvious second vector to use is < 0, 0, 1 > because, being on the z axis, it is already orthogonal to your e vectors which don't have any z component, and their cross product will give you a third vector which is orthogonal to both.

    Incidentally, you are not setting up a MATRIX to evaluate the cross products, you are setting up a DETERMINANT. So it SHOULD be

    \displaystyle \begin{align*} \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right|&= \mathbf{i}\left| \begin{matrix}  1 & 0 \\ 0 & 1 \end{matrix} \right| - \mathbf{j} \left| \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right| + \mathbf{k} \left| \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right| \end{align*}

    Go from here.
    Okay that makes a lot of sense. So on using the orthogonal vector <0,0,1> - when would that 1 ever be different? For the second example I gave above, why wouldn't it be <0,0,-1>?
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    Re: Questions on finding k and normal vectors

    with any plane, there are 2 normal vectors (if you imagine the plane as being "horizontal", one points up, one points down). so to decide "which one we want", we need to decide how our space is "oriented" (we have two choices). the usual way (but not, the ONLY way) is the "right-hand rule", if the first two vectors are your index and middle fingers (at right angles to each other), the normal vector is your thumb.

    this is an arbitrary convention, there's no mathematical reason why we couldn't choose "the left-hand rule" instead (this is much like the convention of choosing "counter-clockwise" as positive rotation).

    in terms of determinants, this boils down to changing the ORDER of the rows 2 and 3 of your matrix (which flips the sign). perhaps you see why this means u x v = -(v x u) (and thus why u x u = 0). so yes, you COULD use (0,0,-1)...and then you'd get "the other normal" k x e = -(e x k) = e x (-k).
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    Re: Questions on finding k and normal vectors

    Quote Originally Posted by Deveno View Post
    with any plane, there are 2 normal vectors (if you imagine the plane as being "horizontal", one points up, one points down). so to decide "which one we want", we need to decide how our space is "oriented" (we have two choices). the usual way (but not, the ONLY way) is the "right-hand rule", if the first two vectors are your index and middle fingers (at right angles to each other), the normal vector is your thumb.

    this is an arbitrary convention, there's no mathematical reason why we couldn't choose "the left-hand rule" instead (this is much like the convention of choosing "counter-clockwise" as positive rotation).

    in terms of determinants, this boils down to changing the ORDER of the rows 2 and 3 of your matrix (which flips the sign). perhaps you see why this means u x v = -(v x u) (and thus why u x u = 0). so yes, you COULD use (0,0,-1)...and then you'd get "the other normal" k x e = -(e x k) = e x (-k).
    In fact, you could use any scalar multiple of < 0, 0, 1 >.
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    Re: Questions on finding k and normal vectors

    Quote Originally Posted by Deveno View Post
    with any plane, there are 2 normal vectors (if you imagine the plane as being "horizontal", one points up, one points down). so to decide "which one we want", we need to decide how our space is "oriented" (we have two choices). the usual way (but not, the ONLY way) is the "right-hand rule", if the first two vectors are your index and middle fingers (at right angles to each other), the normal vector is your thumb.

    this is an arbitrary convention, there's no mathematical reason why we couldn't choose "the left-hand rule" instead (this is much like the convention of choosing "counter-clockwise" as positive rotation).

    in terms of determinants, this boils down to changing the ORDER of the rows 2 and 3 of your matrix (which flips the sign). perhaps you see why this means u x v = -(v x u) (and thus why u x u = 0). so yes, you COULD use (0,0,-1)...and then you'd get "the other normal" k x e = -(e x k) = e x (-k).
    Okay I followed you until the last part. How do I know which normal vector to choose? If I use the right hand rule and my thumb is pointing up towards the sky, would I use <0,0,1> and if its pointing down I would use <0,0,-1>? If so, for

    [ i j k ]
    [ 0 6 0 ]
    [ 0 0 1 ]

    Would the 1 be -1 instead? Someone told me that 1 is correct, but I don't see how because when I use the right hand rule my thumb points down.

    What do you mean by the ORDER? Using the matrix above for example (not saying its right, but just an example), do you mean that changing the order would be
    [ i j k ]
    [ 0 0 1]
    [ 0 6 0]
    ?

    Thanks for the help so far. Sorry if I'm making you repeat yourself. Stick with me, I feel like I'm very close to understanding.
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    Re: Questions on finding k and normal vectors

    The determinant form always gives the entire vector- you don't need the 'right hand rule'. You are confusing that with a different formula that also gives the cross product: "the cross product, u x v, is the vector whose length is given by |u||v|sin(\theta), where \theta is the angle between u and v, and whose direction is given by the 'right hand rule' (if you curl the fingers of your right hand from u to v, then your thumb points in the direction of u x v").

    For example, the vectors i and j are the unit vectors in the positive x and y directions. They have length 1 and the angle between them is a right angle so |i||j|sin(\theta)= (1)(1)(1)= 1. That, is the length of either i x j or j x i is 1. If I curl the fingers of my right hand from the positive x axis to the positive y axis, my thumb points in the direction of the positive z axis: i x j= k, the unit vector in the positive z direction. If I curl the fingers of my right hand from the positive y axis to the positive x axis, my thumb points in the direction of the negative z axis: j x i= -k. Both of those are given automatically, without worrying about the right hand rule, by the "determinant" formula:
    i\times j= \left|\begin{array}{ccc}i & j & k \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right|= i\left|\begin{array}{cc} 0 & 0 \\ 1 & 0\end{array}\right|- j\left|\begin{array}{cc}1 &  0 \\ 0 & 0 \end{array}\right|+ k\left|\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right|= 0i- 0j+ k

    while
    j  \times i=\left|\begin{array}{ccc}i & j & k \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right|= i\left|\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right|- j\left|\begin{array}{cc}0 &  0 \\ 1 & 0 \end{array}\right|+ k\left|\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right|= 0i- 0j- k
    Last edited by HallsofIvy; September 3rd 2012 at 08:12 AM.
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    Re: Questions on finding k and normal vectors

    Quote Originally Posted by HallsofIvy View Post
    The determinant form always gives the entire vector- you don't need the 'right hand rule'. You are confusing that with a different formula that also gives the cross product: "the cross product, u x v, is the vector whose length is given by |u||v|sin(\theta), where \theta is the angle between u and v, and whose direction is given by the 'right hand rule' (if you curl the fingers of your right hand from u to v, then your thumb points in the direction of u x v").

    For example, the vectors i and j are the unit vectors in the positive x and y directions. They have length 1 and the angle between them is a right angle so |i||j|sin(\theta)= (1)(1)(1)= 1. That, is the length of either i x j or j x i is 1. If I curl the fingers of my right hand from the positive x axis to the positive y axis, my thumb points in the direction of the positive z axis: i x j= k, the unit vector in the positive z direction. If I curl the fingers of my right hand from the positive y axis to the positive x axis, my thumb points in the direction of the negative z axis: j x i= -k. Both of those are given automatically, without worrying about the right hand rule, by the "determinant" formula:
    i\times j= \left|\begin{array}{ccc}i & j & k \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right|= i\left|\begin{array}{cc} 0 & 0 \\ 1 & 0\end{array}\right|- j\left|\begin{array}{cc}1 &  0 \\ 0 & 0 \end{array}\right|+ k\left|\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right|= 0i- 0j+ k

    while
    j  \times i=\left|\begin{array}{ccc}i & j & k \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right|= i\left|\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right|- j\left|\begin{array}{cc}0 &  0 \\ 1 & 0 \end{array}\right|+ k\left|\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right|= 0i- 0j- k
    So for the i x j and j x i matrices you have, how do you know that k is a vector of 0's?
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    Re: Questions on finding k and normal vectors

    Quote Originally Posted by HallsofIvy View Post
    The determinant form always gives the entire vector- you don't need the 'right hand rule'. You are confusing that with a different formula that also gives the cross product: "the cross product, u x v, is the vector whose length is given by |u||v|sin(\theta), where \theta is the angle between u and v, and whose direction is given by the 'right hand rule' (if you curl the fingers of your right hand from u to v, then your thumb points in the direction of u x v").

    For example, the vectors i and j are the unit vectors in the positive x and y directions. They have length 1 and the angle between them is a right angle so |i||j|sin(\theta)= (1)(1)(1)= 1. That, is the length of either i x j or j x i is 1. If I curl the fingers of my right hand from the positive x axis to the positive y axis, my thumb points in the direction of the positive z axis: i x j= k, the unit vector in the positive z direction. If I curl the fingers of my right hand from the positive y axis to the positive x axis, my thumb points in the direction of the negative z axis: j x i= -k. Both of those are given automatically, without worrying about the right hand rule, by the "determinant" formula:
    i\times j= \left|\begin{array}{ccc}i & j & k \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right|= i\left|\begin{array}{cc} 0 & 0 \\ 1 & 0\end{array}\right|- j\left|\begin{array}{cc}1 &  0 \\ 0 & 0 \end{array}\right|+ k\left|\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right|= 0i- 0j+ k

    while
    j  \times i=\left|\begin{array}{ccc}i & j & k \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right|= i\left|\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right|- j\left|\begin{array}{cc}0 &  0 \\ 1 & 0 \end{array}\right|+ k\left|\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right|= 0i- 0j- k
    but, you see, the "right-hand rule" is "built-in" to that matrix formula. ok, the "labelling" of the unit basis vectors i ,j and k makes it seem "natural" that "i comes first." but that's just another way of defining the cross-product. in general, when given two vectors (in 3-space) a and b, the plane spanned by the two has a one-dimensional "normal othogonal space" (a line). namely L = {t(axb): t in R}. t can be positive OR negative, there's no way to tell L and -L apart.

    the choice of a formula for the determinant IS the choice of an orientation. "xyz"-space is "the mirror-image" of "xzy"-space. we tend to think of the ordered basis {(1,0,0),(0,1,0),(0,0,1)} as natural, but it's not. abstractly, any ordered basis will do. we adopt a "standard orientation" as a convention.

    and why this matters, is the OP wants to find "the" normal vector. which pre-supposes a "standard orientation", which has to invoke "some kind of rule" to distinguish the two different directions we can travel on the line L. which vector is put "on the second row" and which vector is put "on the third row" matters (it will affect the sign of the resultant determinant). both exk and kxe are BOTH orthogonal to (1,1,0) and (0,0,1), they just point "in opposite directions".

    let me put it this way: if i ask you to find uxv, no problem, right? if i ask for "the cross-product of two vectors"...now there's a problem...which one comes FIRST? you have to put "the right vector on the right row to get the right answer" with the determinant formula. if you are finding axb, you have to put a on the first row, and b on the second row...if you get it backwards you get bxa = -(axb). if you're trying to "see" what happens "in actual space" you need some sort of "road-map" for how to label your axes (so you don't label them "out of order").

    this matters much more in "physical interpretation" of vector math, than it does in the abstract. in physics, they speak of THE normal, because it's the direction in which some actual force is applied. in the case at hand, the vector being sought-after is:

    (1)(1)i + (0)(0)j + (1)(0)k - (0)(0)i - (1)(1)j - (0)(1)k = i - j = (1,0,0) - (0,1,0) = (1,-1,0).

    since this lies in the xy-plane, we can just call it: (1,-1). using the "right-hand rule": our index finger is (1,1), our middle finger is k (pointing straight up) which means our thumb is 90 degrees clockwise from (1,1), that is: (1,-1).

    we could just use the rule for the cross-product:

    (1,1,0) x (0,0,1) = (1*1 - 0*0, 0*0 - 1*1, 1*0 - 1*0) = (1,-1,0) as well.
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    Re: Questions on finding k and normal vectors

    I think this might help.

    Your vector e = [1,1] can be written as i + j. A generalization is if your vector e = [a,b] then it can be written as ai + bj.
    Your vector e is written as if it was in 2D. But, you can write this as if it was in 3D (or more).

    So, in 3D, your e would be [1,1,0] or i + j + 0k. In general, if your e = [a,b] then it would be [a,b,0] in 3D or ai +bj +0k.

    Now, as mentioned before, i,j, and k are unit vectors pointing in the positive x-,y-, and z-axis respectively. This means that your i,j, and k already satisfy the right-hand rule. For emphasis, i,j, and k are vectors, meaning that i = [1,0,0], j = [0,1,0], and k = [0,0,1]. I emphasize this because this basically answers your question, e x k is actually e x [0,0,1] (in this case [1,1,0] x [0,0,1]), and this also shows that the i,j, and k were chosen. If you go to more advanced math, you can actually define your own i,j, and k (which will most likely roughly satisfy the right hand rule). Sometimes the i,j,k are referred to as the components of your vector.

    So, your top row is basically your components (which has to be listed in the order i,j,k, otherwise you mess around with the orientation).
    Since you are doing exk, your second row is the components of e (listed in the i,j,k order), and your third row is the components of k (again, listed in i,j,k order).
    If you were told to do kxe, then your second row would be the components of k, and the third row would be the components of e, and this would be the same as ex(-k) (i.e. e x [0,0,-1])

    A general example:
    If s = [a,b,c] and t = [d,e,f] then

    s x t =
    i j k
    a b c
    d e f

    and t x s =
    i j k
    d e f
    a b c

    That's how you would set up the determinant
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    Re: Questions on finding k and normal vectors

    Quote Originally Posted by Bingk View Post
    I think this might help.

    Your vector e = [1,1] can be written as i + j. A generalization is if your vector e = [a,b] then it can be written as ai + bj.
    Your vector e is written as if it was in 2D. But, you can write this as if it was in 3D (or more).

    So, in 3D, your e would be [1,1,0] or i + j + 0k. In general, if your e = [a,b] then it would be [a,b,0] in 3D or ai +bj +0k.

    Now, as mentioned before, i,j, and k are unit vectors pointing in the positive x-,y-, and z-axis respectively. This means that your i,j, and k already satisfy the right-hand rule. For emphasis, i,j, and k are vectors, meaning that i = [1,0,0], j = [0,1,0], and k = [0,0,1]. I emphasize this because this basically answers your question, e x k is actually e x [0,0,1] (in this case [1,1,0] x [0,0,1]), and this also shows that the i,j, and k were chosen. If you go to more advanced math, you can actually define your own i,j, and k (which will most likely roughly satisfy the right hand rule). Sometimes the i,j,k are referred to as the components of your vector.

    So, your top row is basically your components (which has to be listed in the order i,j,k, otherwise you mess around with the orientation).
    Since you are doing exk, your second row is the components of e (listed in the i,j,k order), and your third row is the components of k (again, listed in i,j,k order).
    If you were told to do kxe, then your second row would be the components of k, and the third row would be the components of e, and this would be the same as ex(-k) (i.e. e x [0,0,-1])

    A general example:
    If s = [a,b,c] and t = [d,e,f] then

    s x t =
    i j k
    a b c
    d e f

    and t x s =
    i j k
    d e f
    a b c

    That's how you would set up the determinant
    Okay awesome that makes a lot of sense! So is k being [0,0,1] something like...universal unless otherwise noted? Could (and if so, when?) k be [0,0,-1] or [0,0,0]?
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    Re: Questions on finding k and normal vectors

    Yeah, you may say that it's "universal" unless otherwise noted.

    k would most likely never be [0,0,0] ( a very unusual vector ), because the resulting determinant would be 0i+0j+0k (you may verify this)

    I can't think of good examples where you would need to use k = [0,0,-1], but it might help some computations simpler (i.e. if k = [0,0,1] then -k = [0,0,-1], and e x k = (-k) x e).
    Recall that k = [0,0,1] could be interpreted as pointing in the positive z-axis, so if k = [0,0,-1], it would be pointing in the negative z-axis. Basic example of where you would change i,j, and k is if you want to "change" your x,y,z axes. So, if you really wanted your new positive z-axis to be the standard negative z-axis (i.e. k=[0,0,-1]), then if you want to satisfy the right hand rule, you would have to change your i and j, so i = [0,1,0] and j = [1,0,0] (i.e. i points in the positive y axis, and j points in the positive x axis). It can get complicated
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    Re: Questions on finding k and normal vectors

    Quote Originally Posted by SterlingM View Post
    So for the i x j and j x i matrices you have, how do you know that k is a vector of 0's?
    I have no idea what you are talking about. k is NOT "a vector of 0's". In a given coordinate system, k, by definition, is the unit vector in the direction of the z-axis- <0, 0, 1> as I said.
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