Results 1 to 7 of 7

Math Help - Vectors

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    56

    Vectors

    Hi everyone.
    Does anybody kno how to create seven vectors in R^2 with integer components with norm 5 and from this list of vectors must be given two which are parallel and two which are orthogonal.?
    Thx for any help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1
    Is this what you mean?
    \begin{array}{cccc}   {\left\langle {3,4} \right\rangle } & {\left\langle { - 3, - 4} \right\rangle } & {\left\langle { - 3,4} \right\rangle } & {\left\langle {3, - 4} \right\rangle }  \\   {\left\langle { - 4, - 3} \right\rangle } & {\left\langle {4,3} \right\rangle } & {\left\langle { - 4,3} \right\rangle } & {}  \\\end{array}<br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2008
    Posts
    56
    i think so. So which one are parallel and which one are orthogonal?

    thank you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1
    Quote Originally Posted by Snowboarder View Post
    So which one are parallel and which one are orthogonal?
    OH! Come on, you tell us that much at least.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    732
    To work out which ones are orthogonal (that's at 90 degrees to each other) o or parallel.

    Take a pair of vectors, e.g. (1, 2) and (3, 4).

    Multiply the 1st numbers or each pair together (that gives you 1 x 3 = 3) in this case.

    Multiply the 2nd numbers together (that gives you 2 x 4 = 8).

    Add them together (that's 3 + 8 = 11).

    If that comes to zero, they're orthogonal. (Here they're not.)

    e.g. (1, 2) and (-4, 2) are orthogonal because (1x(-4)) + (2 x 2) = -4 + 4 = 0.

    Parallel's more complicated except when the vectors are the same length (i.e. have the same norm).

    If, when you add the 2 bits you get after you've done the multiplication, you get the same as the square of the norm of one of them (or minus the square of one of them), they're parallel.

    E.g. (3, 4) and (4, 3) both have norm 5 (square root of 3^2 + 4^2). So the square of their norm is 25.

    But (3x4) + (4x3) is 24, so they're not parallel.

    So (3, 4), (-3, -4) are parallel because you get (3 x -3) + (4 x -4) = -9 + -16 = -25 which is minus the square of the norm.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1
    Quote Originally Posted by Matt Westwood View Post
    Parallel's more complicated except when the vectors are the same length (i.e. have the same norm).
    That is not correct.
    Two vectors are parallel if each is a nonzero multiple of each other.
    There is no requirement for the vectors to have the same length.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    732
    Yes I *know* they don't have to be the same length to be parallel! I was just keeping it *really really simple* for the context of the question.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 15th 2011, 05:10 PM
  2. Replies: 3
    Last Post: June 30th 2011, 08:05 PM
  3. Replies: 2
    Last Post: June 18th 2011, 10:31 AM
  4. [SOLVED] Vectors: Finding coefficients to scalars with given vectors.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 23rd 2011, 12:47 AM
  5. Replies: 4
    Last Post: May 10th 2009, 06:03 PM

Search Tags


/mathhelpforum @mathhelpforum