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    I don't have a clue about how to even start this question, would anyone be able to help? Thanks
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    (Original post by r3l3ntl3ss)
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    I don't have a clue about how to even start this question, would anyone be able to help? Thanks
    If a point lies on the intersection of both planes then it must satisfy both equations. Sub in (x,y,z) and solve.
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    (Original post by ghostwalker)
    If a point lies on the intersection of both planes then it must satisfy both equations. Sub in (x,y,z) and solve.
    I did the cross product of both of the direction vectors to get the direction vector of the line: \begin{pmatrix} sin(2\theta) \\ 0 \\ 0 \end{pmatrix}, and after subbing in I got to 2y\sin(\theta) = 0.

    Am I correct in saying the answer is \lambda\begin{pmatrix} sin(2\theta) \\ 0 \\ 0 \end{pmatrix}? It just seemed a bit odd.
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    (Original post by r3l3ntl3ss)
    I did the cross product of both of the direction vectors to get the direction vector of the line: \begin{pmatrix} sin(2\theta) \\ 0 \\ 0 \end{pmatrix},
    OK, that's useful - forgot about that

    and after subbing in I got to 2y\sin(\theta) = 0.
    Not sure how you got that

    Am I correct in saying the answer is \lambda\begin{pmatrix} sin(2\theta) \\ 0 \\ 0 \end{pmatrix}? It just seemed a bit odd.
    That seems odd to me too, as that is a line through the origin, but the planes don't even necessarily go through the origin.
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    (Original post by ghostwalker)
    OK, that's useful - forgot about that



    Not sure how you got that



    That seems odd to me too, as that is a line through the origin, but the planes don't even necessarily go through the origin.
    Since d was the same in both planes, I just equated one to the other and it all cancelled out leaving 2y\sin(\theta) = 0; i.e. y\sin(\theta) + z\cos(\theta) = -y\sin(\theta) + z\cos(\theta). I just thought it may be more difficult than that since it was part of a question which was allocated 45% of the marks.
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    (Original post by r3l3ntl3ss)
    Since d was the same in both planes, I just equated one to the other and it all cancelled out leaving 2y\sin(\theta) = 0; i.e. y\sin(\theta) + z\cos(\theta) = -y\sin(\theta) + z\cos(\theta). I just thought it may be more difficult than that since it was part of a question which was allocated 45% of the marks.
    Yep, sorry, my brain is starting to go.

    OK, y\sin\theta = 0, so for \sin\theta\not=0, we have y =0; but what about z? That's not zero.
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    (Original post by ghostwalker)
    Yep, sorry, my brain is starting to go.

    OK, y\sin\theta = 0, so for \sin\theta\not=0, we have y =0; but what about z? That's not zero.
    z can be any value can't it, since LHS cancels out RHS?
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    (Original post by r3l3ntl3ss)
    z can be any value can't it, since LHS cancels out RHS?
    It may do in that equation, but not in the original equations of the planes.
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    (Original post by r3l3ntl3ss)
    Am I correct in saying the answer is \lambda\begin{pmatrix} sin(2\theta) \\ 0 \\ 0 \end{pmatrix}? It just seemed a bit odd.
    Since theta is fixed, that can be reduced to

    \lambda\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}

    for a different lambda of couse, and \sin(2\theta)\not=0

    Note - this is the direction vector
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    (Original post by ghostwalker)
    Since theta is fixed, that can be reduced to

    \lambda\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}

    for a different lambda of couse, and \sin(2\theta)\not=0

    Note - this is the direction vector
    I see - the position vector is still confusing me a bit though; what were you implying in the previous post?
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    (Original post by r3l3ntl3ss)
    I see - the position vector is still confusing me a bit though; what were you implying in the previous post?
    Substitute your value of x,y into the equation of the plane, what do you get for z?
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    (Original post by ghostwalker)
    Substitute your value of x,y into the equation of the plane, what do you get for z?
    ah I see, z = d haha. Thank you!
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    (Original post by r3l3ntl3ss)
    ah I see, z = d haha. Thank you!
    Seriously?!
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    (Original post by ghostwalker)
    Seriously?!
    sorry - I'm really out of practice
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    (Original post by r3l3ntl3ss)
    sorry - I'm really out of practice
    Please explain how you got z=d from your equation.

    You have:

    ysin\theta + zcos\theta = d

    Correct?

    You've already said y=0
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    (Original post by Phichi)
    Please explain how you got z=d from your equation.

    You have:

    ysin\theta + zcos\theta = d

    Correct?

    You've already said y=0
    yeah sorry I meant to say z = \frac{d}{\cos(\theta)} - don't think I was thinking when I posted that
 
 
 
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