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Thread: [SOLVED] vectors, i,j notation- dont understand!

  1. #1
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    [SOLVED] vectors, i,j notation- dont understand!

    I really dont understnd this vector problem. please help!

    In this question, i and j are unit vectors in the direction of the positive x- and y-axes respectively.

    Given that a = i – 2j and b = –3i + j find



    (a) λ, if a + λb is parallel to – i – 3j

    (b) μ, if μa + b is parallel to 2i + j.

    in the solution bank it shows the answer worked out like this but I just dont get the working.

    $\displaystyle xi + yj $ is parallel to $\displaystyle Xi + Yj $ if $\displaystyle \frac{x}{y} = \frac{X}{Y} $



    $\displaystyle a + \lambda b = (i - 2j) + \lambda ( -3i + j ) $



    $\displaystyle = (1 - 3\lambda)i + ( -2 + \lambda)j $ I dont understand how they got this expression ?


    This is parallel to – i – 3j if

    $\displaystyle \frac{ -2 + \lambda}{1-3\lambda} = \frac{-3}{-1} $
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  2. #2
    Member u2_wa's Avatar
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    Quote Originally Posted by Tweety View Post
    I really dont understnd this vector problem. please help!

    In this question, i and j are unit vectors in the direction of the positive x- and y-axes respectively.

    Given that a = i 2j and b = 3i + j find



    (a) λ, if a + λb is parallel to i 3j

    (b) μ, if μa + b is parallel to 2i + j.

    in the solution bank it shows the answer worked out like this but I just dont get the working.

    $\displaystyle xi + yj $ is parallel to $\displaystyle Xi + Yj $ if $\displaystyle \frac{x}{y} = \frac{X}{Y} $



    $\displaystyle a + \lambda b = (i - 2j) + \lambda ( -3i + j ) $



    $\displaystyle = (1 - 3\lambda)i + ( -2 + \lambda)j $ I dont understand how they got this expression ?

    $\displaystyle a + \lambda b = (i - 2j) + \lambda ( -3i + j ) $
    $\displaystyle i - 2j + -3i\lambda + j\lambda
    $
    $\displaystyle i-3i\lambda -2j +j\lambda
    $
    Sum i and j vector separately
    $\displaystyle = (1 - 3\lambda)i + ( -2 + \lambda)j $
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  3. #3
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    Quote Originally Posted by u2_wa View Post
    $\displaystyle a + \lambda b = (i - 2j) + \lambda ( -3i + j ) $
    $\displaystyle i - 2j + -3i\lambda + j\lambda
    $
    $\displaystyle i-3i\lambda -2j +j\lambda
    $
    Sum i and j vector separately
    $\displaystyle = (1 - 3\lambda)i + ( -2 + \lambda)j $
    thanks a lot!
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  4. #4
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    Quote Originally Posted by Tweety View Post
    $\displaystyle a + \lambda b = (i - 2j) + \lambda ( -3i + j ) $
    $\displaystyle = (1 - 3\lambda)i + ( -2 + \lambda)j $ I dont understand how they got this expression ? This is parallel to i 3j if
    $\displaystyle \frac{ -2 + \lambda}{1-3\lambda} = \frac{-3}{-1} $
    $\displaystyle ({\color{red}i} - {\color{blue} 2j}) + \lambda ({\color{red}-3i} + {\color{blue}j} ) $
    Just collect like terms.
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