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Thread: vector dot product

  1. #1
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    vector dot product

    Hello
    I have a question which I am not happy with

    Find the angle between a and b given

    |a|= 4
    |b|= 4
    and
    |a-b|= 7

    I have drawn it out and just got myself confused.
    Thank you
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  2. #2
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    Magnitude of (a-b) is given as

    $\displaystyle R^2 = a^2 + b^2 - 2abcos(\theta)$

    Substitute the values and find θ.
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  3. #3
    MHF Contributor red_dog's Avatar
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    Method I.
    Let $\displaystyle \overrightarrow{a}=\overrightarrow{AB}, \ |\overrightarrow{a}|=|\overrightarrow{AB}|=AB=4$

    $\displaystyle \overrightarrow{b}=\overrightarrow{AC}, \ |\overrightarrow{b}|=|\overrightarrow{AC}|=AC=4$

    Then $\displaystyle \overrightarrow{a}-\overrightarrow{b}=\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{CB}$

    and $\displaystyle |\overrightarrow{a}-\overrightarrow{b}|=|\overrightarrow{CB}|=CB=7$

    So, $\displaystyle \overrightarrow{a}, \ \overrightarrow{b}, \ \overrightarrow{a}-\overrightarrow{b}$ are the sides of the triangle $\displaystyle ABC$.

    From the cosine law we have

    $\displaystyle \cos A=\displaystyle\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=-\frac{17}{32}$

    Method II
    $\displaystyle |\overrightarrow{a}-\overrightarrow{b}|^2=(\overrightarrow{a}-\overrightarrow{b})^2=|\overrightarrow{a}|^2+|\ove rrightarrow{b}|^2-2\cdot\overrightarrow{a}\cdot\overrightarrow{b}$

    Then $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=-\displaystyle\frac{17}{2}$

    Now, $\displaystyle \cos\left(\widehat{\overrightarrow{a},\overrightar row{b}}\right)=\displaystyle\frac{\overrightarrow{ a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\cd ot|\overrightarrow{b}|}=-\frac{17}{32}$
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  4. #4
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    Thanks to both
    I shall have a go at this and come back if I am struggling
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