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Math Help - Simple vectors

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Simple vectors

    My friend asked me to help him with his homework, and I think I might have forgotten how to do it

    They were all like

    "Find a vector with given magnitude and in the same direction as the given vector"

    So I think what you would do is this

    Given the position vector

    <a,b,c>

    and the new vector in the same position must be of magnitude d

    Would I just do this

    the new vector must be of the form

    \varpi<a,b,c>

    where \varpi is an abritrary constant, so now I just have to find that constant such that the new magnitude is

    Would I do this by solving

    d=\sqrt{(a\varpi)^2+(b\varpi)^2+(c\varpi)^2}

    For \varpi?

    And then the answer will be of the form |\varpi=|\text{whatever}<br />
\Rightarrow\varpi=\pm\text{whatever}

    So should I then take the \varpi that when plotted goes in the same direction, because won't one of them be parallel but in the opposite direction?
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by Mathstud28 View Post
    My friend asked me to help him with his homework, and I think I might have forgotten how to do it

    They were all like

    "Find a vector with given magnitude and in the same direction as the given vector"

    So I think what you would do is this

    Given the position vector

    <a,b,c>

    and the new vector in the same position must be of magnitude d

    Would I just do this

    the new vector must be of the form

    \varpi<a,b,c>

    where \varpi is an abritrary constant, so now I just have to find that constant such that the new magnitude is

    Would I do this by solving

    d=\sqrt{(a\varpi)^2+(b\varpi)^2+(c\varpi)^2}

    For \varpi?
    Same direction + same sense would yield \varpi>0

    So d=\varpi \sqrt{a^2+b^2+c^2}

    Your method looks ok to me..
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    My friend asked me to help him with his homework, and I think I might have forgotten how to do it

    They were all like

    "Find a vector with given magnitude and in the same direction as the given vector"

    So I think what you would do is this

    Given the position vector

    <a,b,c>

    and the new vector in the same position must be of magnitude d

    Would I just do this

    the new vector must be of the form

    \varpi<a,b,c>

    where \varpi is an abritrary constant, so now I just have to find that constant such that the new magnitude is

    Would I do this by solving

    d=\sqrt{(a\varpi)^2+(b\varpi)^2+(c\varpi)^2}

    For \varpi?
    Just make your vector a unit vector then multiply the unit by the magnitude that you want

     \vec v=<a,b,c> and you want a vector in the same direction with magnitude M then


    M\frac{\vec v}{|\vec v|}=\frac{M}{\sqrt{a^2+b^2+c^2}}\vec v
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,


    Same direction + same sense would yield \varpi>0

    So d=\varpi \sqrt{a^2+b^2+c^2}

    Your method looks ok to me..
    So how would I discern between the |\varpi| giving me a plus or minus answer, can I tell which I want without graphing them?

    Thanks Moo
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  5. #5
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    So how would I discern between the |\varpi| giving me a plus or minus answer, can I tell which I want without graphing them?

    Thanks Moo
    If it's stated that the direction and the sense (for example, to the top) are the same as the original vector (which would be directed to the top too), you can say that \varpi>0. Otherwise, I'm afraid we would need more information.

    But I'm not sure I understood correctly the question that your friend asked, what exactly is given, etc
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Just make your vector a unit vector then multiply the unit by the magnitude that you want

     \vec v=<a,b,c> and you want a vector in the same direction with magnitude M then


    M\frac{\vec v}{|\vec v|}=\frac{M}{\sqrt{a^2+b^2+c^2}}\vec v
    ...good point....I guess this is payback for the odd functions thing
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by Mathstud28 View Post
    ...good point....I guess this is payback for the odd functions thing
    No thanks for that!
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    No thanks for that!
    Haha, I was just kidding, thanks for giving me an easier way to do this thing above

    And thank you Moo for validating my semi-correct method.
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