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Vectors and Matrices Watch

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    Hi guys, my answer to 1a is:

    \hat{n} = \frac{\vec{u} \times \vec{v}}{\sqrt{|\vec{u}| + |\vec{v}|}}

    d = \frac{\vec{a}.(\vec{u} \times \vec{v})}{\sqrt{|\vec{u}| + |\vec{v}|}}

    Is that correct?

    I'm not sure how to do 1b, would someone be able to explain it? Thanks.
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    I'm getting "Invalid Attachment".
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    (Original post by ghostwalker)
    I'm getting "Invalid Attachment".
    Sorry, TSR is giving me problems and won't allow me to attach to the original post, here it is:

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    (Original post by r3l3ntl3ss)
    Sorry, TSR is giving me problems and won't allow me to attach to the original post, here it is:

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    For part b, do you know how to do row reduction?
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    (Original post by rayquaza17)
    For part b, do you know how to do row reduction?
    using Gaussian Elimination right? yeah I do
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    (Original post by r3l3ntl3ss)
    Hi guys, my answer to 1a is:

    \hat{n} = \frac{\vec{u} \times \vec{v}}{\sqrt{|\vec{u}| + |\vec{v}|}}

    d = \frac{\vec{a}.(\vec{u} \times \vec{v})}{\sqrt{|\vec{u}| + |\vec{v}|}}

    Is that correct?
    Don't think your denominator is correct there. What's your thinking on that?
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    (Original post by ghostwalker)
    Don't think your denominator is correct there. What's your thinking on that?
    Doesn't the unit vector \hat{n} represent the cross product between the two direction vectors divided by the square root of the magnitude of both of the direction vectors?
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    (Original post by r3l3ntl3ss)
    Doesn't the unit vector \hat{n} represent the cross product between the two direction vectors divided by the square root of the magnitude of both of the direction vectors?
    Wondering where you got the bit in bold from.

    It's easy to construct a counterexample. If the two vectors are i and j, then the normal is k, but that formula gives k/sqrt(2), which isn't a unit vector.
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    (Original post by r3l3ntl3ss)
    using Gaussian Elimination right? yeah I do
    Actually are you familiar with the rank of a matrix?
    You get at least one solution when the rank is 4, so you need to try different values of alpha and beta to get the rank as 4.
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    (Original post by ghostwalker)
    Wondering where you got the bit in bold from.

    It's easy to construct a counterexample. If the two vectors are i and j, then the normal is k, but that formula gives k/sqrt(2), which isn't a unit vector.
    ah I see, I guess I was getting confused.

    The denominator of both should be |\vec{u}| \times |\vec{v}|, right?

    (Original post by rayquaza17)
    Actually are you familiar with the rank of a matrix?
    You get at least one solution when the rank is 4, so you need to try different values of alpha and beta to get the rank as 4.
    yup I'm familiar with rank, but shouldn't the rank be a maximum of 3 since the matrix has 3 rows?
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    (Original post by r3l3ntl3ss)
    yup I'm familiar with rank, but shouldn't the rank be a maximum of 3 since the matrix has 3 rows?
    Oops, yeah! I pressed the wrong button.
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    (Original post by rayquaza17)
    Oops, yeah! I pressed the wrong button.
    Haha fair enough - what I'm unsure of is the question saying at least one solution. If the rank is less than 3, doesn't that mean there's infinite solutions? I'd understand how to do it if it said just one solution.
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    (Original post by r3l3ntl3ss)
    ah I see, I guess I was getting confused.

    The denominator of both should be |\vec{u}| \times |\vec{v}|, right?
    Nay.

    You want |\vec{u} \times \vec{v}|,
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    (Original post by r3l3ntl3ss)
    Haha fair enough - what I'm unsure of is the question saying at least one solution. If the rank is less than 3, doesn't that mean there's infinite solutions? I'd understand how to do it if it said just one solution.
    Erm I think if the rank is 3 then you have one unique solution.
    If the rank is less than 3, you have infinitely many solutions.

    I think you might have to consider like rank 3, rank 2, etc then for the answer.
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    (Original post by ghostwalker)
    Nay.

    You want |\vec{u} \times \vec{v}|,
    That's the same isn't it?

    (Original post by rayquaza17)
    Erm I think if the rank is 3 then you have one unique solution.
    If the rank is less than 3, you have infinitely many solutions.

    I think you might have to consider like rank 3, rank 2, etc then for the answer.
    ah yeah that's what I suspected, thanks!
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    (Original post by r3l3ntl3ss)
    That's the same isn't it?
    I hope that's a joke.
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    (Original post by ghostwalker)
    I hope that's a joke.
    hah don't worry it was
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    (Original post by rayquaza17)
    Erm I think if the rank is 3 then you have one unique solution.
    If the rank is less than 3, you have infinitely many solutions.

    I think you might have to consider like rank 3, rank 2, etc then for the answer.
    I'm still a bit unsure on how to solve this question, I've reduced the matrix to this:

    \begin{pmatrix} 1 & 0 & {\frac{3}{4}} \\ 0 & 1 & -{\frac{5}{4}} \\ \alpha-1 & 0 & 0\end{vmatrix} \begin{matrix} \frac{\beta}{4} \\ -{\frac{3\beta}{4}} \\ 2+2\beta \end{pmatrix}
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    (Original post by r3l3ntl3ss)
    I'm still a bit unsure on how to solve this question, I've reduced the matrix to this:

    \begin{pmatrix} 1 & 0 & {\frac{3}{4}} \\ 0 & 1 & -{\frac{5}{4}} \\ \alpha-1 & 0 & 0\end{vmatrix} \begin{matrix} \frac{\beta}{4} \\ -{\frac{3\beta}{4}} \\ 2+2\beta \end{pmatrix}
    Maybe I was right the first time and you didn't need to consider ranks, sorry!

    You need to consider the bottom row of the matrix. To get solutions, we need to have a-1=2+2b.
    If a=1, what is b? If a isn't 1, what's the relation between a and b?
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    (Original post by rayquaza17)
    Maybe I was right the first time and you didn't need to consider ranks, sorry!

    You need to consider the bottom row of the matrix. To get solutions, we need to have a-1=2+2b.
    If a=1, what is b? If a isn't 1, what's the relation between a and b?
    I'm still really confused, so if \alpha = 1 then \beta = -1; would you just say that x = \frac{2 + 2\beta}{\alpha - 1} as the relation?
 
 
 
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