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Thread: Vector proof

  1. #1
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    Vector proof

    Given that three vectors $\displaystyle \vec a, \vec b, \vec c$, are perpendicular to each other (where $\displaystyle \vec a, \vec b, \vec c$ are not $\displaystyle \vec 0$ and given that $\displaystyle p\vec a+q\vec b+r\vec c=\vec 0$, show that p=q=r=0.

    I get the solution, but I don't understand something written in my solution book.

    It says

    Since $\displaystyle p\vec a+q\vec b+r\vec c=\vec 0$

    $\displaystyle \vec a(p\vec a+q\vec b+r\vec c)=0$

    Notice that in the second line it is just 0, not $\displaystyle \vec 0$

    Why?
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  2. #2
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    Quote Originally Posted by chengbin View Post
    Given that three vectors $\displaystyle \vec a, \vec b, \vec c$, are perpendicular to each other (where $\displaystyle \vec a, \vec b, \vec c$ are not $\displaystyle \vec 0$ and given that $\displaystyle p\vec a+q\vec b+r\vec c=\vec 0$, show that p=q=r=0.

    I get the solution, but I don't understand something written in my solution book.

    It says

    Since $\displaystyle p\vec a+q\vec b+r\vec c=\vec 0$ <<<<< the result is the null vector which is perpendicular to any other vector

    $\displaystyle \vec a(p\vec a+q\vec b+r\vec c)=0$ <<<<<< since the null vector is perpendicular to any other vector the scalar (dot) product must be zero. (A dot product yields a real number not a vector!)

    Notice that in the second line it is just 0, not $\displaystyle \vec 0$

    Why?
    ...
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  3. #3
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    Scalar Product

    Hello chengbin
    Quote Originally Posted by chengbin View Post
    Given that three vectors $\displaystyle \vec a, \vec b, \vec c$, are perpendicular to each other (where $\displaystyle \vec a, \vec b, \vec c$ are not $\displaystyle \vec 0$ and given that $\displaystyle p\vec a+q\vec b+r\vec c=\vec 0$, show that p=q=r=0.

    I get the solution, but I don't understand something written in my solution book.

    It says

    Since $\displaystyle p\vec a+q\vec b+r\vec c=\vec 0$

    $\displaystyle \vec a(p\vec a+q\vec b+r\vec c)=0$

    Notice that in the second line it is just 0, not $\displaystyle \vec 0$

    Why?
    $\displaystyle \vec a(p\vec a+q\vec b+r\vec c)$ is the scalar or dot product of the vector $\displaystyle \vec a$ with the vector $\displaystyle p\vec a+q\vec b+r\vec c$, and is more correctly written $\displaystyle \vec a.(p\vec a+q\vec b+r\vec c)$

    Since the vector $\displaystyle (p\vec a+q\vec b+r\vec c) = \vec 0$, this scalar product is the scalar 0.

    The proof then (presumably) continues by noting that $\displaystyle \vec a.\vec b = 0 = \vec a . \vec c$, since the scalar product of perpendicular vectors is zero.

    Hence $\displaystyle p \vec a.\vec a = 0$, and hence (if $\displaystyle \vec a$ is a non-zero vector) $\displaystyle p = 0$.

    Similarly q = r = 0.

    Grandad
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