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Math Help - multiplying two vectors - the dot product

  1. #1
    Member Mr Rayon's Avatar
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    multiplying two vectors - the dot product

    Find the dot product of the vectors 3i + 3j and 6i + 2j using the equation u\cdot v = <br />
\left | u\left | \right |v \right |cos \theta <br />



    Please show complete working out of how to do the above. Any attempt to do so will be appreciated.
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  2. #2
    JML
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    To do this question in the specified way you need to find the lengths of both vectors and then the angle between the vectors. If u = x i + y j then \left| u \right| = \sqrt{x^2  + y^2} the same method is used for the length of v.

    If the question did not specify the method you had to use for the dot product it would likely be easier to calculate the dot product as follows:

    say we have u = a_1 i + a_2 j
    v = b_1 i + b_2 j
    then:
    u \cdot v = a_1 b_1 +a_2 b_2
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  3. #3
    Member Mr Rayon's Avatar
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    Quote Originally Posted by JML View Post
    To do this question in the specified way you need to find the lengths of both vectors and then the angle between the vectors.
    But how do you find the angle between the two vectors?
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  4. #4
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    Quote Originally Posted by Mr Rayon View Post
    But how do you find the angle between the two vectors?
    Draw the two vectors and calculate the angle that each makes with the i direction.
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  5. #5
    JML
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    You can find the angle by noting the coordinates that the vector would end at if it started at the origin.

    so let's say \textbf{a} = 3i + 3j and \textbf{b} = 6i + 2j. So for a we get the point A =(3,3) and for b we get the point B=(6,2) . First we find the angle between the line x=0 (the i direction) and the line that passes through the origin and A, this angle = \frac{\pi}{4} . You can do something similar for the other vector. You can then find the angle between the vectors by subtracting the angle to B from the angle to A.
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  6. #6
    Member Mr Rayon's Avatar
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    u = 3i + 3j and v = 6i + 2j
     <br />
\left | u \right | = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}<br />
\left | v \right | = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}<br />

    Okay, I've drawn a diagram...and then I find the difference between the angles:


     cos \theta = \frac{3}{3\sqrt{2}}<br />
, cos \theta = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2}<br />

     cos\theta = \frac{6}{2\sqrt{10}}, cos\theta = \frac{6\sqrt{10}}{20} = \frac{3\sqrt{10}}{10}

    \frac{3\sqrt{10}}{10} - \frac{5sqrt{2}}{10}
    u\cdot v = \left |u  \right | \left | v \right |cos\theta


    I am lost from here, and is the above right? Can somebody provide complete working out?

    EDIT: I don't know what I'm doing here.
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  7. #7
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    Quote Originally Posted by Mr Rayon View Post
    u = 3i + 3j and v = 6i + 2j
     <br />
\left | u \right | = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}<br />
\left | v \right | = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}<br />

    Okay, I've drawn a diagram...and then I find the difference between the angles:


     cos \theta = \frac{3}{3\sqrt{2}}<br />
, cos \theta = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2}<br />

     cos\theta = \frac{6}{2\sqrt{10}}, cos\theta = \frac{6\sqrt{10}}{20} = \frac{3\sqrt{10}}{10}

    \frac{3\sqrt{10}}{10} - \frac{5sqrt{2}}{10}
    u\cdot v = \left |u \right | \left | v \right |cos\theta


    I am lost from here, and is the above right? Can somebody provide complete working out?

    EDIT: I don't know what I'm doing here.
    Let \alpha be the angle between 3i + 3j and the i direction. Let \beta be the angle between 6i + 2j and the i direction.

    The angle between 3i + 3j and 6i + 2j is \theta = \alpha - \beta.

    Calculate \cos \theta using the compound angle formula. This will require that you first get \sin \alpha, \cos \alpha, \sin \beta and \cos \beta.
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  8. #8
    JML
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     cos \theta_A = \frac{3}{3\sqrt{2}}<br />
, cos \theta_A = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2}<br />
    so \theta_A = arccos \frac{\sqrt{2}}{2} = \frac{\pi}{4}


    \tan \theta_b = \frac{2}{6} = \frac{1}{3}

    \theta_b = arctan \frac{1}{3}

    so \theta = \theta_A - \theta_B
    \theta = \frac{\pi}{4} - arctan \frac{1}{3}
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  9. #9
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    Quote Originally Posted by Mr Rayon View Post
    Find the dot product of the vectors 3i + 3j and 6i + 2j using the equation u\cdot v = <br />
\left | u\left | \right |v \right |cos \theta <br />



    Please show complete working out of how to do the above. Any attempt to do so will be appreciated.
    Draw a picture showing \vec{u}= 3\vec{i}+ 3\vec{j}, \vec{v}= 6\vec{i}+2\vec{j} and \vec{u}- \vec{v}= -3\vec{i}+ \vec{j}, the vector from the tip of \vec{v} to the tip of \vec{u}.

    Now you have a triangle with sides of length a= |\vec{u}|= 3\sqrt{2}, b= |\vec{v}|= 2\sqrt{10}, and c= |\vec{u}-\vec{v}|= \sqrt{10}. You can use the cosine law, c^2= a^2+ b^2- 2ab \cos(C) to find C, the angle between the vectors.
    Last edited by mr fantastic; September 6th 2009 at 04:12 AM. Reason: Fixed a latex tag
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