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Thread: "Right-handed" rule with vectors?

  1. #1
    Junior Member
    Mar 2011

    "Right-handed" rule with vectors?

    Could someone please explain to me how the "right-handed" rule works and how you would use it to appropriate vector questions with it? I understand that you have your right thumb pointing upright and move your thumb the way that the vector moves (or something like that). Although, what I particularly do not understand is how you would be able to answer it with questions. Such as, "True or false: a x (b x a) is perpendicular to a but not to b. (The given diagram that the 'end' of both b and a meet at a high point, whereas the beginning of both b and a come outwards ( /\ <-- basically, with b on the left, a on the right.) I have tried to make sense out of the explanation and examples my textbook has provided although, I am still confused.
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  2. #2
    MHF Contributor

    Apr 2005
    To find the direction of $\displaystyle \vec{u}\times\vec{v}$, hold your right hand so that your index finger is pointing in the direction of $\displaystyle \vec{u}$ and your other fingers curl toward $\displaystyle \vec{v}$, the your thumb will be pointing in the direction of $\displaystyle \vec{u}\times\vec{v}$. Basically that just tells you that $\displaystyle \vec{u}\times\vec{v}$ is perpendicular to both $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ and $\displaystyle \vec{u}\times\vec{v}= -\vec{v}\times\vec{u}$.

    $\displaystyle \vec{b}\times\vec{a}$ is perpendicular to both $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ $\displaystyle \vec{a}\times (\vec{a}\times\vec{b})$ is perpendicular to $\displaystyle \vec{a}$ and $\displaystyle \vec{a}\times\vec{b}$ but not $\displaystyle \vec{b}$. To see that more clearly, take a specific example: if $\displaystyle \vec{b}= \vec{i}$ and $\displaystyle \vec{a}= \vec{j}$, then $\displaystyle \vec{b}\times\vec{\a}= \vec{i}\times\vec{j}= \vec{k}$ so that $\displaystyle \vec{a}\times(\vec{b}\times\vec{a})= \vec{j}\times\vec{k}= \vec{i}= \vec{b}$.
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