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Math Help - Angle between plane and xy-plane

  1. #1
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    Angle between plane and xy-plane

    I have a plane with the equation -900y+165z=0 and I want to find the angle between this plane and the xy-plane. What do I do?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    I'm not sure that I understand your question. It looks like all you have is the equation to a line in the zy plane, which is, by definition, perpindicular to the xy plane. Now, we could perhaps find the angle of the line that goes through the zy plane by anylizing its slope which is 900/165 (I don't feel like reducing). Does this help you?
    Last edited by VonNemo19; May 2nd 2009 at 07:31 AM. Reason: 1
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  3. #3
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    The normal to the given plane is <0,-900,165> and the normal to the xy-plane is <0,0,1>.
    What is the angle between those two vectors?
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  4. #4
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    Hello, No Logic Sense!

    Find the angle between the plane -900y+165z\:=\:0 and the xy-plane.
    Given vectors \vec u and \vec v, the angle \theta between them is given by: . \cos\theta \:=\:\frac{|\vec u \cdot\vec v|}{|\vec v||\vec v|}

    The angle between their normal vectors is the angle between the planes.


    The first plane has normal vector: . \vec u \:=\:\langle 0,\text{-}900,165\rangle \:=\:\langle0,\text{-}60,11\rangle
    . . Its magnitude is: . |\vec u| \:=\:\sqrt{0^2 + (\text{-}60)^2+9^2}  \:=\:\sqrt{3721} \:=\:61

    The xy-plane has normal vector: . \vec v \:=\:\langle0,0,1\rangle
    . . It is a unit vector: . |\vec v| \:=\:1

    Hence: . \cos\theta \;=\;\frac{|\langle0,\text{-}60,11\rangle\cdot\langle0,0,1\rangle|}{61\cdot1} \;=\;\frac{0+0+11}{61} \;=\;\frac{11}{61}


    Therefore: . \theta \;=\;\arccos\left(\tfrac{11}{61}\right) \;\approx\;79.6^o

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  5. #5
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    Ahaaa, but my normal vector for the plane is:

    (0;-900;165)
    and then the xy-plane has the (0;0;1)

    I get the angle to 88,24 degrees, is that wrong?
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  6. #6
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    Quote Originally Posted by No Logic Sense View Post
    Ahaaa, but my normal vector for the plane is:

    (0;-900;165)
    and then the xy-plane has the (0;0;1)

    I get the angle to 88,24 degrees, is that wrong? .....Yes
    \cos(\alpha)=\dfrac{(0,-900,165) \cdot (0,0,1)}{915 \cdot 1} = \dfrac{11}{61}

    which is exactly the same value as posted by Soroban.

    But your calculator is switched to the very special mode where a right angle is measured 100
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  7. #7
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    Quote Originally Posted by earboth View Post
    \cos(\alpha)=\dfrac{(0,-900,165) \cdot (0,0,1)}{915 \cdot 1} = \dfrac{11}{61}

    which is exactly the same value as posted by Soroban.

    But your calculator is switched to the very special mode where a right angle is measured 100
    "grads"- a system used primarily to measure slope of roads has 100 grads per right angle. I once upon a time had a calculator that could be set to "degrees", "radians", or "grads" but I haven't seen one in years.
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