Results 1 to 3 of 3

Thread: Question on vectors

  1. #1
    Member
    Joined
    May 2015
    From
    United Kingdom
    Posts
    81

    Question on vectors

    I have attached my question

    here is my attempt:

    I have attached my question


    Here's my attempt :
     \overrightarrow{OA}= 2\mathbf{i}+3\mathbf{j}


     \overrightarrow{OB}= 3\mathbf{i} - \mathbf{j} + k(\mathbf{i}+2\mathbf{j})


     \overrightarrow{OC}= 4\mathbf{i} + 13\mathbf{j} + k(\mathbf{4i}-\mathbf{j})


     \overrightarrow{OB}= (3+k)\mathbf{i} + (2k-1)\mathbf{j}
     \overrightarrow{OC}= (4+4k)\mathbf{i}+(13-k)\mathbf{j}
     (4+4k)\mathbf{i}+(13-k)\mathbf{j} = (3+k)\mathbf{i} + (2k-1)\mathbf{j} since OB is parallel to OC
    thus
     (4+4k)\mathbf{i} = (3+k)\mathbf{i}


     (13-k)\mathbf{j} =(2k-1)\mathbf{j}




     (4+4k) = (3+k)


     4k-k = 3-4


     k = -\frac{1}{3}




     (13-k) =(2k-1)


     13 +1 = 2k+k


     k = \frac{14}{3}


    but in the book the answer is 3 or -2/3


    please tell me where i did wrong?


    thank you for helping
    Question on vectors-1.jpg
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,155
    Thanks
    3672

    Re: Question on vectors

    positions after motion ...

    $A(2,3)$

    $B(k+3,2k-1)$

    $C(4k+4,13-k)$

    for A, B, and C to be collinear, vectors AB and BC need to have the same direction.

    slope for AB ... $\dfrac{2k-4}{1+k}$

    slope for BC ... $\dfrac{14-3k}{3k+1}$

    $\dfrac{2k-4}{1+k} = \dfrac{14-3k}{3k+1}$

    solve the resulting quadratic for $k$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,421
    Thanks
    2889

    Re: Question on vectors

    Quote Originally Posted by bigmansouf View Post
    I have attached my question

    here is my attempt:

    I have attached my question


    Here's my attempt :
     \overrightarrow{OA}= 2\mathbf{i}+3\mathbf{j}


     \overrightarrow{OB}= 3\mathbf{i} - \mathbf{j} + k(\mathbf{i}+2\mathbf{j})


     \overrightarrow{OC}= 4\mathbf{i} + 13\mathbf{j} + k(\mathbf{4i}-\mathbf{j})


     \overrightarrow{OB}= (3+k)\mathbf{i} + (2k-1)\mathbf{j}
     \overrightarrow{OC}= (4+4k)\mathbf{i}+(13-k)\mathbf{j}
     (4+4k)\mathbf{i}+(13-k)\mathbf{j} = (3+k)\mathbf{i} + (2k-1)\mathbf{j} since OB is parallel to OC
    No! Saying to vectors are "parallel" does not say they are equal. It says one is a multiple of the other.
    4+ 4k= c(3+ k) and 13- k= c(2k- 1), two equations to solve for c and k

    thus
     (4+4k)\mathbf{i} = (3+k)\mathbf{i}


     (13-k)\mathbf{j} =(2k-1)\mathbf{j}




     (4+4k) = (3+k)


     4k-k = 3-4


     k = -\frac{1}{3}




     (13-k) =(2k-1)


     13 +1 = 2k+k


     k = \frac{14}{3}


    but in the book the answer is 3 or -2/3


    please tell me where i did wrong?


    thank you for helping
    Click image for larger version. 

Name:	1.jpg 
Views:	4 
Size:	152.0 KB 
ID:	38000[/QUOTE]
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Sep 21st 2011, 08:21 PM
  2. Vectors question
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Apr 25th 2010, 08:27 AM
  3. Vectors question!
    Posted in the Geometry Forum
    Replies: 12
    Last Post: Mar 14th 2009, 09:19 AM
  4. Vectors Question -- Please help
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Jan 19th 2009, 11:38 PM
  5. Vectors question
    Posted in the Calculus Forum
    Replies: 12
    Last Post: May 31st 2008, 06:52 AM

/mathhelpforum @mathhelpforum