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Math Help - surface of a parallelogram - vectors

  1. #1
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    surface of a parallelogram - vectors

    I have to determine the surface of a parallelogram whose vectors of diagonals are \vec{c}=5\vec{m}+5\vec{n}, \vec{d}=\vec{m}-\vec{n}, and \vec{m} and \vec{n} are unit vectors and \angle (\vec{m}, \vec{n})=\frac{\pi}{6}.

    Please help, I have to idea where to start.
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    Quote Originally Posted by harriette View Post
    I have to determine the surface of a parallelogram whose vectors of diagonals are \vec{c}=5\vec{m}+5\vec{n}, \vec{d}=\vec{m}-\vec{n}, and \vec{m} and \vec{n} are unit vectors and \angle (\vec{m}, \vec{n})=\frac{\pi}{6}.

    Please help, I have to idea where to start.
    Dear harriette,

    The vector equation of a plane is given by, (\underline{a}-\underline{b}).\underline n=0 where \underline{a},\underline{b} are the position vectors of two points on the plane and \underline n is a vector perpendicular to the plane.

    Hope you would be able to continue along these lines.
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    I suppose I should express \vec{a}, \vec{b} of the parallelogram (the base) using \vec{m}, \vec{n} and then use cross product on that? Is that correct?
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    Quote Originally Posted by harriette View Post
    I suppose I should express \vec{a}, \vec{b} of the parallelogram (the base) using \vec{m}, \vec{n} and then use cross product on that? Is that correct?
    Dear harriette,

    We are given two vectors of the plane that we have to find out. Those are the vectors of the diagonals. What we dont know is a perpendicular vector to the plane. Can you think of a way to find a perpendicular vector using the two given vectors?
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  5. #5
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    Their cross product will be perpendicular to the plane, but I don't see why do I even need a perpendicular vetor.
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    Anyone?
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    Quote Originally Posted by harriette View Post
    Their cross product will be perpendicular to the plane, but I don't see why do I even need a perpendicular vetor.
    Correct. Therefore, you can take, \vec{n}=\vec{c}\times\vec{d}. The equation of the plane will be, \vec{c}.(\vec{c}\times\vec{d})=0 because in this case \vec{c} lies on the plane. Alternatively you can write the equation of the plane as \vec{d}.(\vec{c}\times\vec{d})=0.

    Hope you will be able to continue.
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  8. #8
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    I'm so embarassed, but I really don't see where you're going with this.

    Here's what I wanted to do. Let ABCD be the parallelogram in question, \vec{a}=AB, \vec{b}=AD. Surface area of that parallelogram is S=|\vec{a}|h, where h denotes the height. That equals to |\vec{a}| |\vec{b}|\sin \alpha which is | \vec{a} \times \vec{b}|.
    So I was hoping I could express \vec{a}, \vec{b} using \vec{n}, \vec{m} and then calculate it using the above equation.

    Does that make any sense?
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  9. #9
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    Quote Originally Posted by harriette View Post
    I'm so embarassed, but I really don't see where you're going with this.

    Here's what I wanted to do. Let ABCD be the parallelogram in question, \vec{a}=AB, \vec{b}=AD. Surface area of that parallelogram is S=|\vec{a}|h, where h denotes the height. That equals to |\vec{a}| |\vec{b}|\sin \alpha which is | \vec{a} \times \vec{b}|.
    So I was hoping I could express \vec{a}, \vec{b} using \vec{n}, \vec{m} and then calculate it using the above equation.

    Does that make any sense?
    Ok. If you could find \vec{a} and \vec{b} using \vec{m} and \vec{n} that would be great (Can you please tell me how?). But I dont think that would be easy. Instead you could use the given vectors of the diagonals and get the equation of the plane as I had mentioned in my last post.
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  10. #10
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    I can get the equation of the plane but then what? I have no idea how will I get the surface area of a parallelogram using the equation of a plane.
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  11. #11
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    Quote Originally Posted by harriette View Post
    I can get the equation of the plane but then what? I have no idea how will I get the surface area of a parallelogram using the equation of a plane.
    Let \vec{a} and \vec{b} be two sides of the parallelogram.

    \vec{a}+\vec{b}=\vec{c}

    \vec{a}-\vec{b}=\vec{d}

    We get \vec{a}=\frac{1}{2}(\vec{c}+\vec{d}) and \vec{b}=\frac{1}{2}(\vec{c}-\vec{d}).

    Find \vec{a} and \vec{b} in terms of \vec{m} and \vec{n}. Compute \vec{a}\times\vec{b}.
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  12. #12
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    Quote Originally Posted by alexmahone View Post
    Let \vec{a} and \vec{b} be two sides of the parallelogram.

    \vec{a}+\vec{b}=\vec{c}

    \vec{a}-\vec{b}=\vec{d}

    We get \vec{a}=\frac{1}{2}(\vec{c}+\vec{d}) and \vec{b}=\frac{1}{2}(\vec{c}-\vec{d}).

    Find \vec{a} and \vec{b} in terms of \vec{m} and \vec{n}. Compute \vec{a}\times\vec{b}.

    Thank you, both of you for your replies. This is what I wanted to do (I'm glad to see the idea was okay.)


    I still don't understand the part with the perpendicular vector Sudharaka mentioned.
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  13. #13
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    Quote Originally Posted by harriette View Post
    I still don't understand the part with the perpendicular vector Sudharaka mentioned.
    Nor do I.
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  14. #14
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    Quote Originally Posted by harriette View Post
    I can get the equation of the plane but then what? I have no idea how will I get the surface area of a parallelogram using the equation of a plane.
    In your first post.....

    I have to determine the surface of a parallelogram.............
    So I did not know you have to find the area of the paralleogram.........I told you the method of finding the plane (surface) which the parallelogram lies. Anyway if you want to find the area of the parallelogram you will have to find the sides of the parallelogram (as vectors). See the attached figure.

    \vec{x}+\vec{y}=5\vec{m}+5\vec{n}

    \vec{x}-\vec{y}=\vec{m}-\vec{n}

    Solving the above equations you can obtain the vectors of the sides of the parallelogram. Then \vec{x}\times\vec{y} would give you the area.

    Sorry if I confused you in my previous posts..........I hope this helps to clarify things.....just forget what I told before.
    Attached Thumbnails Attached Thumbnails surface of a parallelogram - vectors-sp.pdf  
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  15. #15
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    Oh, now I see where the mixup was.
    Thank you.
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