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Math Help - Scalar Product/Norm proof

  1. #1
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    Scalar Product/Norm proof

    Hello, I am having trouble. I would appreciate if someone could help me with part b of this problem, I have solved part a.

    Problem:
    Let u = (a,b) and v = (c, d) be nonzero points in the plane R^n and let \theta be the radian measure of the angle with vertex 0 formed by 0, u, and v.
    **notations: ||u|| is the norm and <u,v> is the scalar/dot product

    a. Prove that \left\| u \right\|^2 \left\| v \right\|^2 - \left( <u,v> \right)^2 = \left( ||u|| ||v|| sin \theta \right) ^2

    (1) LHS = \left\| u \right\|^2 \left\| v \right\|^2 - \left( ||u|| ||v|| cos \theta \right)^2 by definition of dot product
    (2) LHS = <u,u><v,v>-<u,u><v,v>cos^2 by distribution and the property that <u,u> = ||u||^2
    (3) LHS = [<u,u><v,v>](1-cos^2 \theta) by factoring
    (4) LHS = RHS by sine cosine rule that sin^2 \theta + cos^2 \theta = 1

    b. I am given that |ad-bc|= \left\| u \right\| \left\| v \right\| |sin \theta |
    I have to use this to verify that |ad-bc|/2 is the area of the triangle with vertices 0, u, and v and that that as a consequence |ad-bc| is the area of the parallelogram with vertices 0, u, u+v, and v.

    I know that the area of a triangle is \frac {1}{2}bh, so if the area of the triangle with vertices 0, u, v. Then shouldn't the area be |ac-bd|/2 instead of |ad-bc|/2. Any help is greatly appreciated. Thank you.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Paperwings View Post
    Hello, I am having trouble. I would appreciate if someone could help me with part b of this problem, I have solved part a.

    Problem:
    Let u = (a,b) and v = (c, d) be nonzero points in the plane R^n and let \theta be the radian measure of the angle with vertex 0 formed by 0, u, and v.
    **notations: ||u|| is the norm and <u,v> is the scalar/dot product

    a. Prove that \left\| u \right\|^2 \left\| v \right\|^2 - \left( <u,v> \right)^2 = \left( ||u|| ||v|| sin \theta \right) ^2

    (1) LHS = \left\| u \right\|^2 \left\| v \right\|^2 - \left( ||u|| ||v|| cos \theta \right)^2 by definition of dot product
    (2) LHS = <u,u><v,v>-<u,u><v,v>cos^2 by distribution and the property that <u,u> = ||u||^2
    (3) LHS = [<u,u><v,v>](1-cos^2 \theta) by factoring
    (4) LHS = RHS by sine cosine rule that sin^2 \theta + cos^2 \theta = 1

    b. I am given that |ad-bc|= \left\| u \right\| \left\| v \right\| |sin \theta |
    I have to use this to verify that |ad-bc|/2 is the area of the triangle with vertices 0, u, and v and that that as a consequence |ad-bc| is the area of the parallelogram with vertices 0, u, u+v, and v.

    I know that the area of a triangle is \frac {1}{2}bh, so if the area of the triangle with vertices 0, u, v. Then shouldn't the area be |ac-bd|/2 instead of |ad-bc|/2. Any help is greatly appreciated. Thank you.

    See the attachment (then think about the sign of sin(\theta) when its not less than a right angle)

    RonL
    Attached Thumbnails Attached Thumbnails Scalar Product/Norm proof-gash.png  
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  3. #3
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    If  \theta is a right angle, then the area would be  \frac{1}{2}bh = ||u||||v||sin(90) = ||u||||v|| = |ad-bc|/2

    If the angle is less than a right angle, then area would be ||u|| ||v||sin \theta = |ad-bc|/2

    Else if the angle is greater than a right angle, then the area would be  \frac {1}{2} ||u|| ||v|| sin(180- \theta) = ||u|| ||v|| sin( \theta) = |ad-bc|/2 since  sin( \theta) = sin(180 - \theta)

    Thank you for the help.
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