Results 1 to 5 of 5

Math Help - vector (simplified b1 and b2 )

  1. #1
    Member
    Joined
    Apr 2014
    From
    canada
    Posts
    165

    vector (simplified b1 and b2 )

    we know that r=a +(/lambda b ). /lambda is a constant. the above is the ans given while the below is my working. my b1 and b2 is not the same with the ans. is the ans given acceptable ? (b1 and b2) here's only the partial working. i'm having problem with tht first part. accorfing to the ans given, the b1 and b2 are simplified . in this case, why the b1 and b2 can be simplified?
    https://www.flickr.com/photos/123101...3/14283658984/


    https://www.flickr.com/photos/123101...3/14097487299/


    https://www.flickr.com/photos/123101...3/14097497469/

    sorry i cant upload the photo directly, so i choose to upload it onto flickr
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,881
    Thanks
    743

    Re: vector (simplified b1 and b2 )

    For line one, let's call the variable you use t and the variable given in the answer t'. Then let's see when the two vectors are equal.

    \begin{pmatrix}2 \\ 2 \\ 4\end{pmatrix} + \begin{pmatrix}-\tfrac{1}{2} \\ \tfrac{1}{2} \\ 1\end{pmatrix}t = \begin{pmatrix}2 \\ 2 \\ 4\end{pmatrix} + \begin{pmatrix}-1 \\ 1 \\ 2\end{pmatrix}t' when t=2t'. Since the map t \mapsto 2t is a bijection, you wind up with the same line either way.

    Same thing for line 2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2014
    From
    canada
    Posts
    165

    Re: vector (simplified b1 and b2 )

    Quote Originally Posted by SlipEternal View Post
    For line one, let's call the variable you use t and the variable given in the answer t'. Then let's see when the two vectors are equal.

    \begin{pmatrix}2 \\ 2 \\ 4\end{pmatrix} + \begin{pmatrix}-\tfrac{1}{2} \\ \tfrac{1}{2} \\ 1\end{pmatrix}t = \begin{pmatrix}2 \\ 2 \\ 4\end{pmatrix} + \begin{pmatrix}-1 \\ 1 \\ 2\end{pmatrix}t' when t=2t'. Since the map t \mapsto 2t is a bijection, you wind up with the same line either way.

    Same thing for line 2.
    so if i i do my ans in this way. wouldnt my ans be wrong?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,881
    Thanks
    743

    Re: vector (simplified b1 and b2 )

    Quote Originally Posted by delso View Post
    so if i i do my ans in this way. wouldnt my ans be wrong?
    I just showed that you get the same line either way. Why would you get the wrong answer that way? If two line intersect and you have two different formulas for each line, it is still the same two lines. It doesn't matter that the formulas are not identical. They still represent the same lines, so they will still intersect in the same place.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2014
    From
    canada
    Posts
    165

    Re: vector (simplified b1 and b2 )

    do u mean t=2t" just different in terms of length ? but it's still the same line with same gradient?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Can it be simplified more?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 29th 2010, 06:21 AM
  2. Can this be simplified?
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: July 22nd 2010, 02:29 AM
  3. Can this be simplified?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 20th 2009, 09:17 PM
  4. I need to see simplified
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 5th 2008, 04:04 PM
  5. Can 484/243 be simplified any more?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: November 2nd 2005, 06:16 AM

Search Tags


/mathhelpforum @mathhelpforum