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Thread: vector help

  1. #1
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    vector help

    AB is a diameter of a circle centred at the origin O, and P is any point on the circumference of the circle.
    Using the position vectors of A, B and P, prove (using a scalar product) that AP is perpendicular to BP (i.e. the angle in the semicircle is a right angle).

    Not sure where to even start. I know that the product of vectors has to equal 0 for them the be perpendicular, but I dont know the magnitude for each vector, so how can you prove they are perpendicular?

    Any help appreciated.

    thank you
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  2. #2
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    Quote Originally Posted by Tweety View Post
    Not sure where to even start. I know that the product of vectors has to equal 0 for them the be perpendicular, but I dont know the magnitude for each vector, so how can you prove they are perpendicular?

    Any help appreciated.

    thank you
    Hi Tweety,

    You may not know the lengths of $\displaystyle \vec{AP}$ and $\displaystyle \vec{BP}$

    however, in using the scalar product we can work with the vectors, $\displaystyle \vec{OA},\ \vec{OP},\ \vec{OB}$

    Since A and B are a radius away from O, then $\displaystyle \vec{A}$ and $\displaystyle \vec{B}$ are going in opposite directions,

    and have the same length, hence $\displaystyle \vec{A}=-\vec{B}$

    Then calculate the scalar product of $\displaystyle \vec{AP}$ and $\displaystyle \vec{BP}$

    using $\displaystyle \vec{AP}=\vec{P}-\vec{A}$ and $\displaystyle \vec{BP}=\vec{P}-\vec{B}$

    $\displaystyle \vec{AP}.\vec{BP}=\left(\vec{P}-\vec{A}\right).\left(\vec{P}-\vec{B}\right)$

    $\displaystyle =|p|^2-|p||b|cos\theta-|a||p|cos({\pi}-\theta)+|a||b|cos({\pi})$

    $\displaystyle =r^2-r^2cos\theta+r^2cos\theta-r^2$
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  3. #3
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    Quote Originally Posted by Tweety View Post
    Not sure where to even start. I know that the product of vectors has to equal 0 for them the be perpendicular, but I dont know the magnitude for each vector, so how can you prove they are perpendicular?

    Any help appreciated.

    thank you
    let AB lie on the x-axis from -1 to 1

    P has position vector $\displaystyle \vec{r_P} = \cos{t} \vec{i} + \sin{t} \vec{j}$

    $\displaystyle AP = (\cos{t}+1) \vec{i} + \sin{t} \vec{j}$

    $\displaystyle PB = (1-\cos{t})\vec{i} - \sin{t} \vec{j}$

    $\displaystyle AP \cdot PB = (1-\cos^2{t}) - \sin^2{t} = 0$
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  4. #4
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    Quote Originally Posted by Tweety View Post
    Not sure where to even start. I know that the product of vectors has to equal 0 for them the be perpendicular, but I dont know the magnitude for each vector, so how can you prove they are perpendicular?

    Any help appreciated.

    thank you
    First, make a sketch.

    So you wan't to show that $\displaystyle \vec{PB}\perp \vec{PA}$, by showing that $\displaystyle \vec{PB}\cdot \vec{PA}=0$.

    Express the two vectors as sums like this:

    $\displaystyle \vec{PB}\cdot \vec{PA} = (\vec{PO}+\vec{OA})\cdot (\vec{PO}+\vec{OB})$
    (where $\displaystyle O$ is the origin / center of the circle)

    now use the fact that $\displaystyle \vec{OB}=-\vec{OA}$ to get

    $\displaystyle =(\vec{PO}+\vec{OA})\cdot (\vec{PO}-\vec{OA})$

    finally expand the dot-product to get

    $\displaystyle =\vec{PO}^2-\vec{OA}^2=|\vec{PO}|^2-|\vec{OA}|^2=r^2-r^2=0$
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