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    I'm stuck on this format of question:
    For what values of x and y is vector v = [6, 2, x]T orthogonal to the plane containing points
    A(1, 8, 2), B(1, −1, −4) and C(5, y, 15)? Record x then y on the same line.

    I'm not sure how to solve for x and y. With out them there I'd form two vectors AB and AC then use the cross product to get an equation for the plane and compare the co-efficients with the vector but I honestly have no idea how to get rid of the x and y values... would I set them to 0 and solve, or do I need simultaneous equations?
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    (Original post by matte30)
    I'm stuck on this format of question:
    For what values of x and y is vector v = [6, 2, x]T orthogonal to the plane containing points
    A(1, 8, 2), B(1, −1, −4) and C(5, y, 15)? Record x then y on the same line.

    I'm not sure how to solve for x and y. With out them there I'd form two vectors AB and AC then use the cross product to get an equation for the plane and compare the co-efficients with the vector but I honestly have no idea how to get rid of the x and y values... would I set them to 0 and solve, or do I need simultaneous equations?
    Proceed as normal(!). What do you get? It does work out.
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    (Original post by ghostwalker)
    Proceed as normal(!). What do you get? It does work out.
    AB = (1-1, -1-8, -4-2)

     = (0, -9, -6)

AC = (5-1, y-8, -4-15)

    = (4, y-8, -19)

\\

\\

    

AB\times AC

\[ \left| \begin{array}{ccccc} i & j & k & i & j \\ 0 & -9 & -6 & 0 & -9 \\ 4 & y-8 & -19 & 4 & y-8 \end{array} \right|\]
    

\\

\\

            

             = i(-9)(-19) + j(-6)(-4) + k(0)(y-8) - j(0)(-19) -i(-6)(y-8) - k(-9)(4)

             = -219i + 6yi + 24j + 36k
    (I think that's the best I can do for formatting; I hope it helps)
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    (Original post by ghostwalker)
    Proceed as normal(!).
    PRSOM
    matte30, it would be really helpful if you could LaTeX your working; you can literally just write [ latex] [/ latex] around it (without the spaces), and use \times for a cross if you want it.
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    (Original post by matte30)
    AB = (1-1, -1-8, -4-2)

     = (0, -9, -6)

AC = (5-1, y-8, -4-15)

    = (4, y-8, -19)
    (Matrix removed because I'm not sure how to format it; but am working on it now)
    

            AB\times AC

             = i(-9)(-19) + j(-6)(-4) + k(0)(y-8) - j(0)(-19) -i(-6)(y-8) - k(-9)(4)

             = -219i + 6yi + 24j + 36k
    Agree with your k component, but not the i, or j. j should be the negative of what you have.

    AC = (4,y-8,15-2)

    Then just compare the 2nd coordinate of your result with that of your vector v, to get the scaling factor, and ....

    Edit: Corrected.
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    (Original post by ghostwalker)
    Agree with your k component, but not the i, or j. j should be the negative of what you have.

    AC = (4,y-8,15-2)

    Then just compare the 1st coordinate of your result with that of your vector v, to get the scaling factor, and ....

    Edit: Corrected.
    AC = (4, y-8, 13)
    

AB \times AC = -144i + 6yi -48i -24j + 36k

= -192i +6yi -24j +36k
    I found why the j should be negative; thank you.
    To get the first co-ordinate do I set y to 0?
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    (Original post by matte30)
    AC = (4, y-8, 13)
    

AB \times AC = -144i + 6yi -48i -24j + 36k

= -192i +6yi -24j +36k
    I found why the j should be negative; thank you.
    To get the first co-ordinate do I set y to 0?
    Don't agree with the "-192i", I make it "-165i", otherwise we are in agreement.

    No, you don't set y to 0.

    This (corrected) vector is orthogonal to the plane, as is v.

    So, one must be a scalar multiple of the other.

    You want to compare the coefficients of the coordinates to work out the scaling factor. You need to check j - not i as I previously said (sorry about that) - to get the scaling factor.
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    (Original post by ghostwalker)
    Don't agree with the "-192i", I make it "-156i", otherwise we are in agreement.

    No, you don't set y to 0.

    This (corrected) vector is orthogonal to the plane, as is v.

    So, one must be a scalar multiple of the other.

    You want to compare the coefficients of the coordinates to work out the scaling factor. You need to check j - not i as I previously said (sorry about that) - to get the scaling factor.
    I cannot get -156i; the closet I have is -165i.
    The corrected k column finishes with +13.

    -9(13)i = -117i
    6(y-8)i = 6yi -48i
    -117 - 48 = 165

    ----------------------------------------------

    For the actual question I've been asked I've ended up comparing my j value with a 0 in the vector... once again, I'm not sure what to do now since there is no scale factor...
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    (Original post by matte30)
    I cannot get -156i; the closet I have is -165i.
    The corrected k column finishes with +13.

    -9(13)i = -117i
    6(y-8)i = 6yi -48i
    -117 - 48 = 165
    Yes, you're right -165i. Clearly, I can't multiply.

    So, (-165+6y)i-24j+36k

    I presume you can now finish off.
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    (Original post by ghostwalker)
    Yes, you're right -165i. Clearly, I can't multiply.

    So, (-165+6y)i-24j+36k

    I presume you can now finish off.
    For this question; yes. But what if the unit you're comparing with a 0? example v is changed from [6, 2, x]^T to [6, 0, x]^T?
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    (Original post by matte30)
    For this question; yes. But what if the unit you're comparing with a 0? example v is changed from [6, 2, x]^T to [6, 0, x]^T?
    Then it's not solveable. Your scalar multiple would have to be zero. And with that x would be 0, and there's no way you can get 6 by multiplying something by zero.
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    (Original post by ghostwalker)
    Then it's not solveable. Your scalar multiple would have to be zero. And with that x would be 0, and there's no way you can get 6 by multiplying something by zero.
    Thank you very much; you've been very helpful
 
 
 
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