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Math Help - Cross product mystery (stewart 12.4 #12)

  1. #1
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    Cross product mystery (stewart 12.4 #12)

    The figure shows a vector, a, in the xy plane and a vector, b, in the direction of k.
    Their lengths are /a\ = 3 and /b\ = 2.

    find /a x b\ (cross product)

    My explanation of the figure:
    The figure is such that b is <0,0,2> and the angle of a to the positive y axis is about 30degrees

    I am totally stuck because i need the coordinates of a, to cross it with b? b is right on the pos x axis so obviously it must be <0,0,2> but all i know about a is its magnitude and I can only approximates its angle to the y axis.
    Any hints or help is really appreciated!
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by adam21 View Post
    The figure shows a vector, a, in the xy plane and a vector, b, in the direction of k.
    Their lengths are /a\ = 3 and /b\ = 2.

    find /a x b\ (cross product)

    My explanation of the figure:
    The figure is such that b is <0,0,2> and the angle of a to the positive y axis is about 30degrees

    I am totally stuck because i need the coordinates of a, to cross it with b? b is right on the pos x axis so obviously it must be <0,0,2> but all i know about a is its magnitude and I can only approximates its angle to the y axis.
    Any hints or help is really appreciated!
    |a x b| = |a|*|b|*sin(theta) ... where theta is the angle between the two vectors.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by adam21 View Post
    The figure shows a vector, a, in the xy plane and a vector, b, in the direction of k.
    Their lengths are /a\ = 3 and /b\ = 2.

    find /a x b\ (cross product)

    My explanation of the figure:
    The figure is such that b is <0,0,2> and the angle of a to the positive y axis is about 30degrees

    I am totally stuck because i need the coordinates of a, to cross it with b? b is right on the pos x axis so obviously it must be <0,0,2> but all i know about a is its magnitude and I can only approximates its angle to the y axis.
    Any hints or help is really appreciated!
    remember |a x b| = |a||b|sin(theta), where theta is the angle between the two vectors.

    obviously the angle between them will be pi/2 radians, so:

    |a x b| = |3||2|sin(pi/2) = 6

    EDIT: again! that's two times in a row ecMathGeek! It's all good though, do your thing
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    Why is it obviously pi/2 radians though? I dont know what angle between the 2 vectors is, --when i said 30degrees between pos y-axis and vector a, thats just my eyeball approx. The problem doesnt tell me what theta is?? --thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by adam21 View Post
    Why is it obviously pi/2 radians though? I dont know what angle between the 2 vectors is, --when i said 30degrees between pos y-axis and vector a, thats just my eyeball approx. The problem doesnt tell me what theta is?? --thanks
    first of all, math is precise, we don't do eyeball approx around here, especially since mathematicians are often too lazy to draw diagrams to scale, we just use them to think about the solution.

    anyway, the xy, xz, and yz planes are all at 90 degrees to each other. one vector is on the z-axis, so it is parallel to the xz and yz-planes and therefore perpendicular to the xy-plane. the other vector is in the xy-plane, so the vectors have an angle of 90 degrees between them
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    i'll make this comment in a new post to make sure you don't miss it.

    i cannot emphasize enough how much of a no-no is "eye-balling" in math. save yourself a lot of grief and never do it again. there are two main reasons why you should not eye-ball a diagram and guess dimensions:
    1) you cannot measure angles and distances with your naked eye accurately enough to satisfy the standards of math
    2) mathematicians almost never (or probably never) draw diagrams to scale. we are not architects, we don't go for a t-square, and protractor when we are sketching diagrams, we just draw one object in relation to another, not necessarilly in their correct scale.

    so please, no more eye-ball approximations, it's for your own good. if it is not stated in the problem, or you have not calculated it, then you don't know, end of story

    hehe, here i was saying "we" as if i'm a mathematician
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