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Thread: solve perpendicular vectors problem

  1. #1
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    Exclamation solve perpendicular vectors problem

    (the vectors in the problem are really in column form, but i don't know how to type that so i put them in component form)

    vectorP = 2i +5j
    vectorQ = -i+ 2j

    ☆find the value of p·q
    (I know the dot product is 8, that was ez) but the next part i don't get...

    ☆ if... s = kp - (p·q)q
    find the value of the constant k such that s is perpendicular to q

    plz help
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  2. #2
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    The dot product $\displaystyle \mathbf{p}\cdot\mathbf{q}$ increases when $\displaystyle \mathbf{p}$ and $\displaystyle \mathbf{q}$ point in the same direction, and drops to $\displaystyle 0$ exactly when $\displaystyle \mathbf{p}$ and $\displaystyle \mathbf{q}$ are perpendicular.

    Therefore, $\displaystyle \mathbf{s}$ will be perpendicular to $\displaystyle \mathbf{q}$ when

    $\displaystyle \begin{aligned}
    \mathbf{s}\cdot\mathbf{q}&=(k\mathbf{p}-(\mathbf{p}\cdot\mathbf{q})\mathbf{q})\cdot\mathbf {q}\\
    &=k(\mathbf{p}\cdot\mathbf{q})-(\mathbf{p}\cdot\mathbf{q})|\mathbf{q}|^2\\
    &=(\mathbf{p}\cdot\mathbf{q})(k-|\mathbf{q}|^2)\\
    &=0.
    \end{aligned}$

    Here, we have used the formula $\displaystyle \mathbf{q}\cdot\mathbf{q}=|\mathbf{q}|^2$.
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  3. #3
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    Quote Originally Posted by Scott H View Post
    The dot product $\displaystyle \mathbf{p}\cdot\mathbf{q}$ increases when $\displaystyle \mathbf{p}$ and $\displaystyle \mathbf{q}$ point in the same direction, and drops to $\displaystyle 0$ exactly when $\displaystyle \mathbf{p}$ and $\displaystyle \mathbf{q}$ are perpendicular.

    Therefore, $\displaystyle \mathbf{s}$ will be perpendicular to $\displaystyle \mathbf{q}$ when

    $\displaystyle \begin{aligned}
    \mathbf{s}\cdot\mathbf{q}&=(k\mathbf{p}-(\mathbf{p}\cdot\mathbf{q})\mathbf{q})\cdot\mathbf {q}\\
    &=k(\mathbf{p}\cdot\mathbf{q})-(\mathbf{p}\cdot\mathbf{q})|\mathbf{q}|^2\\
    &=(\mathbf{p}\cdot\mathbf{q})(k-|\mathbf{q}|^2)\\
    &=0.
    \end{aligned}$

    Here, we have used the formula $\displaystyle \mathbf{q}\cdot\mathbf{q}=|\mathbf{q}|^2$.
    ummm... sorry i feel kinda lame, but im in high school and i kinda got lost in the first step, plz explain, im not so good with the math...i still don't see how to get the value of k
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  4. #4
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    Given that

    $\displaystyle 8(k-|\mathbf{q}|^2)=0,$

    we may divide both sides by $\displaystyle 8$, giving

    $\displaystyle k-|\mathbf{q}|^2=0.$

    Now, we may add $\displaystyle |\mathbf{q}|^2$ to both sides, giving

    $\displaystyle k=|\mathbf{q}|^2.$

    Hope this helps!
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  5. #5
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    i get it now, grazie!
    Last edited by 3k1yp2; Dec 7th 2009 at 06:26 PM. Reason: now i got it
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