Results 1 to 11 of 11

Math Help - solve for variable when vector a and b have a 45 degree angle between them

  1. #1
    Member
    Joined
    Jan 2011
    Posts
    75

    solve for variable when vector a and b have a 45 degree angle between them

    vector a = (2,5) and vector b = (-1,t)

    solve for t when the vectors have a 45 degree angle between them

    any help would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    If I remember correctly you can use the dot product

    a.b = |a||b|cos \theta

    -2+5t = \sqrt{29} \cdot \sqrt{1+t^2} \cdot \dfrac{\sqrt{2}}{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2011
    Posts
    75
    hm, yeah thats what I figured. I can't figure out how to solve from there on.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    Where are you having trouble? It just ends up becoming a quadratic after you square both sides.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    -2+5t = \dfrac{\sqrt{58(1+t^2)}}{2}

    10t-4 = \sqrt{58+58t^2}

    100t^2-80t+16 = 58+58t^2

    42t^2-80t-42 = 0

    21t^2-40t-21 = 0

    This one does factor but it might be easier to use the quadratic formula
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jan 2011
    Posts
    75
    I tried that but Ididn't get the correct answer. the answer is t=7/3
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Yes, that is the positive answer you get when you solve the quadratic.

    21t^2 - 40 t - 21 = 0

    t = \displaystyle\frac{40 \pm \sqrt{(-40)^ 2- 4(21)(-21)}}{2\times 21}

    = \displaystyle\frac{40 \pm 58}{42}
    Last edited by Quacky; January 11th 2011 at 09:39 AM. Reason: Clarified formula usage
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jan 2011
    Posts
    75
    ah okay, i'll give it another go
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    I've editted the post. Where was your mistake? I assume it was probably just a sign error or something.
    Last edited by Quacky; January 11th 2011 at 09:38 AM. Reason: Politeness.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by colorado View Post
    I tried that but Ididn't get the correct answer. the answer is t=7/3
    Please show your working rather than just giving an answer. Then it is easier to diagnose what help you actually need.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Jan 2011
    Posts
    75
    noted
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help finding 90 degree angle in sphere
    Posted in the Geometry Forum
    Replies: 3
    Last Post: December 7th 2011, 06:37 PM
  2. 90-135 degree angle on vertice
    Posted in the Geometry Forum
    Replies: 0
    Last Post: August 6th 2011, 03:51 PM
  3. known circle inside of a 60 degree angle?
    Posted in the Geometry Forum
    Replies: 3
    Last Post: April 21st 2010, 07:05 AM
  4. 18 degree angle
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: November 27th 2009, 02:39 PM
  5. Replies: 1
    Last Post: May 28th 2007, 04:25 PM

Search Tags


/mathhelpforum @mathhelpforum