Results 1 to 6 of 6

Math Help - Questions about Dot Product (Vectors)

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    3

    Questions about Dot Product (Vectors)

    This is my first post here so I'm sorry if this isn't the right section (didn't see a vectors subforum). I am also not sure how to put arrows on top of letters, so assume that every letter in the following question is a vector. I will also use | | for magnitude.

    1. |a| = 3 , |b| = 2 , the angle in between these vectors is 60
    Determine the numerical value of (3a + 2b) dot (4a - 3b). - In order to do this I tried distributing.

    I know that (3a dot 4a) = 12|a| and (2b dot -3b) = -6|b|
    I however don't know what to do in the case of (3a dot -3b) and (2b dot 4a)

    The formula given is ( |a||b|cosθ = a dot b ) and the final answer is 81 as listed in the back of the book.


    2. A regular hexagon has sides of 3cm, as shown below. Determine a dot b.



    This is my drawing - I slid b over resulting in an angle of 120 between a and b.
    I used the above formula to yield: |3||3|cos120 = -4.5

    The answer in the back of the book is however 4.5 (positive, not negative). This has also occurred in other questions where I am getting negative answers where they should be positive - is there something I'm not doing right?

    All help would be greatly appreciated.
    Last edited by StuckOnVectors; March 9th 2011 at 12:23 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,957
    Thanks
    1780
    Awards
    1
    Quote Originally Posted by StuckOnVectors View Post
    1. |a| = 3 , |b| = 2 , the angle in between these vectors is 60
    Determine the numerical value of (3a + 2b) dot (4a - 3b). - In order to do this I tried distributing.

    I know that (3a dot 4a) = 12|a| and (2b dot -3b) = -6|b|
    I however don't know what to do in the case of (3a dot -3b) and (2b dot 4a)

    The formula given is ( |a||b|cosθ = a dot b ) and the final answer is 81 as listed in the back of the book.
    \frac{a\cdot b}{\|a\|\|b\|}=\cos(60^o)=\frac{1}{2}

    \alpha a\cdot \beta b=(\alpha\beta)a\cdot b

    a\cdot a=\|a\|^2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    So, roughly speaking (!), we have 12a^2 - ab - 6b^2
    after distributing.
    "a" = 3, "b" = 2, and "ab" = 3 by following Plato's computation.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2011
    Posts
    3
    I'm still just not understanding.. How did you come up with -ab = -3? I know that you have to use the formula, I just don't get how.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by StuckOnVectors View Post

    2. A regular hexagon has sides of 3cm, as shown below. Determine a dot b.

    This is my drawing - I slid b over resulting in an angle of 120 between a and b.
    I used the above formula to yield: |3||3|cos120 = -4.5

    The answer in the back of the book is however 4.5 (positive, not negative). This has also occurred in other questions where I am getting negative answers where they should be positive - is there something I'm not doing right?

    All help would be greatly appreciated.
    The vectors including an angle had to be placed tail to tail and not - as you did - head to tail. Or in other words: The arrows representing vectors had to start at the vertex of the angle.
    Attached Thumbnails Attached Thumbnails Questions about Dot Product (Vectors)-eingeschlwinklvector.png  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2011
    Posts
    3
    earboth thank you for the help, I see where I went wrong with the hexagon

    thechaz the problem was that I didn't know how to get to  12a^2 - a \cdot b - 6b^2, but i figured it out

    anyone reading this: 3a \cdot -3b is the same thing as -9a \cdot b - not knowing this was my problem
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 15th 2011, 06:10 PM
  2. vectors and dot product in 3-d
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 25th 2009, 04:52 PM
  3. Replies: 1
    Last Post: May 14th 2008, 12:31 PM
  4. dot product and vectors.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 11th 2008, 03:04 AM
  5. Vectors (Dot product?)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 6th 2008, 09:23 PM

Search Tags


/mathhelpforum @mathhelpforum