Results 1 to 4 of 4

Math Help - vectors

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    8

    vectors

    Find the values of t for which the velocity of the car is parallel to the vector (i + j)

    v = (3r^2 - 2t + 8)i + (5t + 6)j ms^-1

    Its doing my head in not being able to work this out
    Last edited by Carl Feltham; September 26th 2008 at 08:10 AM. Reason: wrong thing written
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Sep 2008
    Posts
    12
    For vectors to be parallel they need to have the same slope.

    vector i+j has slope: 1/1= 1

    so slope of v has to be 1 which is:



    (5t+6)/(3t^2-2t+8)=1

    So it is a second degree polynomial equation to solve.

    civodul
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,660
    Thanks
    600
    Hello, Carl!

    Find the values of t for which the velocity of the car
    is parallel to the vector \vec i + \vec j

    . . \vec v \;= \;(3t^2 - 2t + 8)\vec i + (5t + 6)\vec j

    A vector parallel to \vec i + \vec j has the form: . \vec v \:=\:a\vec i + a\vec j
    . . That is, the coefficients are equal.

    So we have: . 3t^2 - 2t + 8 \;=\;5t + 6 \quad\Rightarrow\quad 3t^2 - 7t + 2 \:=\:0

    Factor: . (t - 2)(3t - 1) \:=\:0 \quad\Rightarrow\quad\boxed{ t \;=\;2,\:\frac{1}{3}}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,329
    Thanks
    1243
    Quote Originally Posted by Carl Feltham View Post
    Find the values of t for which the velocity of the car is parallel to the vector (i + j)

    v = (3r^2 - 2t + 8)i + (5t + 6)j ms^-1

    Its doing my head in not being able to work this out
    Another way may be to use dot products.

    Recall that

    \mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta}.

    Since the vectors will be parallel, the angle between them is 0, and so \cos{\theta}=\cos{0} = 1.

    So in other words, in this case...

    \mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|

    So evaluate the dot product, evaluate the product of the moduli, and solve for t.
    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